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Elementary co-ordinate geometry for collegiate use and private study / By William Benjamin Smith.

Elementary co-ordinate geometry for collegiate use and private study / By William Benjamin Smith. Smith, William Benjamin, 1850-1934. 400dpi TIFF G4 page images University of Kentucky, Electronic Information Access & Management Center Lexington, Kentucky 2002 b92-276-32008276 Electronic reproduction. 2002. (Beyond the shelf, serving historic Kentuckiana through virtual access (IMLS LG-03-02-0012-02) ; These pages may be freely searched and displayed. Permission must be received for subsequent distribution in print or electronically. Elementary co-ordinate geometry for collegiate use and private study / By William Benjamin Smith. Smith, William Benjamin, 1850-1934. Ginn, Boston : 1886. xxx, 281, [1] p. : diagram ; 22 cm. Coleman Microfilm. Atlanta, Ga. : SOLINET, 1995. 1 microfilm reel ; 35 mm. (SOLINET/ASERL Cooperative Microfilming Project (NEH PS-20317) ; SOL MN05121.06 KUK) Printing Master B92-276. IMLS This electronic text file was created by Optical Character Recognition (OCR). No corrections have been made to the OCR-ed text and no editing has been done to the content of the original document. Encoding has been done through an automated process using the recommendations for Level 1 of the TEI in Libraries Guidelines. Digital page images are linked to the text file. Geometry, Analytic. ELEMENTARY CO-ORDINATE GEOMETRY, FOR COLLEGIATE USE AND PRIVATE STUDY. BY WILLIAM BENJAMIN SM1IITH, PAi.D. (GOrrINGEN), PROFESSOR OF PHYSICS, MISSOURI STATE UNIVERSITY. MISSOURI. MAXIMUM REASONING, MINIMUM RECKONING. BOSTON: GINN & COMPANY. 1886. Entered according to Act of Congress, in the year 1885, by WILLIAM BENJAMIN SMITH, in the Office of the Librarian of Congress, at Washington J. S. CUSHING & CO., PAtTwmts, BoSTON. PREFACE. IN the study of Analytic Geometry, as of almost anything else, either or both of two ends may be had in view: gain of knowledge, culture of mind. While the first is in itself worthy enough, and for mathematical devotees all suffi- cient, it is certainly of only secondary importance to the mass of college students. For these the subject can be wisely prescribed in a curriculum only in case the mental drill it affords be very high in order of excellence. The worth of mere calculation as an exercise of reason can hardly be considerable, for reason is exercised only in a tread-mill fashion. Even the solution of problems by al- gebraic processes is a very inferior discipline of reason, for onlv in forming the analytic statement does the reasoning rise clearly into consciousness; the operations that follow conduct one to the conclusion, but -with his eyes shut. In this respect Geometry is certainly a better discipline than Algebra, and the Euclidean than the Cartesian Geometry. But not in any kind of reasoning is the very best discipline found. No argument presents difficulty or calls for much mental effort to follow it, when once its terms are clearly understood; for no such argument can be harder to under- stand than the general syllogism of whice it is a special case, and that is of well-known simplicity. The real diffi- cultv lies in forming clear notions of things; in doing this all the higher faculties are brought into play. It is this formation of concepts, too, that is the really important part of mental training. He who forms them clearly and accu- rately may be safely trusted to put them together correctly. PREFACE. Logical blunders are comparatively. rare. Nearly every seeming mistake in reasoning is really a mistake in concep- tion. If this be false, that will be invalid. It is considerations like the above that have guided the composition of this book. Concepts have been introduced in abundance, and the proofs made to hinge directly upon them. Treated in this way, the subject seems adapted as hardly any other to develop the power of thought. The correlation of algebraic and geometric facts has been kept clearly and steadily in view. While each may be taken as pictures of the other, the former have generally been treated as originals, lending themselves much more readily to classification. Only natural logical order has been aimed at in the devel- opment of the subject; no attempt has been made to keep up the distinctions of ancient and modern, analytic and synthetic. With every step forward in Geometry the difficulty and tedium of graphical representation increases, while more and more the reasoning turns upon the form of the algebraic expressions. Accordingly, pains have been taken to make the notation throughout consistent and suggestive, and Deter- minants have been used freely. By all this effort to make the book an instrument of culture, its worth as a repertory of mathematical facts has scarcely suffered; in this regard, as in others, comparison with other texts is invited. AUTHOR. CENTRAL COLLEGE, MO. March 25, 1885. 1Y TABLE OF CONTENTS. INTRODUCTION. DETERMII4ANTS. Article. Page. 1. Permutations, Straight and Circular, of n Things ............ xiii 2. Inversions in Permutations ......... ...................... xiv 3. Exchange of Elements in a Permutation ...... ............. xiv 4. Definition of Determinant ......... ....................... xv 5. Ways of writing Determinants ....... ..................... xvi 6. Simplest Properties of Determinants ...... ................ xvii 7. Co-factor of an Element of a Determinant .................. xviii 8. Products of Elements by their Co-factors ................... x ix 9. Decomposition of a Determinant ....... ................... xx 10. Evaluation of a Determinant. Examples ...... ............ xxi 11. Multiplication of Determinants ............................ xxiii 12. Determinant of the Co-factors ............................. xxiv 13. Solution of a System of Linear Equations ...... ............ xxv 14. Conditions of Consistence of Equations ...... .............. xxvi 15. Elimination between Equations of Higher Degree ..... ...... xxvii PART I. THE PLANE. CHAPTER I. FIRST NOTIONS. 1. Definition of Functions ...................................... 1 2. Algebraic Expression of a Functional Relation. Argument ..... 2 3. Classification of Functions ............. ...................... 3 4. Definition of Continuity. Illustrations ..... .................. 4 5. Numbers pictured by Tracts; Addition and Subtraction ........ 6 6. Pairs of Numbers pictured by Points ....... .................. 7 7. Equations pictured by Curves ..........................:.. 8 Vi TABLE OF CONTENTS. Article. Page. 8. Determination of Position on a Surface ...... ................ 9 9. Points pictured by Pairs of Numbers. Co-ordinates ........... 10 10. Polar Co-ordinates ................... ................. 11 11. Polar Equations pictured by Curves ....... .................. 12 12. Degrees of Freedom. The Picture changes with the System .. . 12 13. Summary................................................. 14 14. Definition of Co-ordinate Geometry ....... ................... 14 N.B. As to Real and Imaginary Numbers .................... 15 CHAPTER Il. THE RIGHT LINE. 15. Distance between Two Points ........ ................. ..... 16 16. Intersection of Curves .......... ........................... 17 17. Division of a Tract in any Ratio ............................. 18 18. Parallel Projection of a Tract ........ ....................... 20 19. Projection of a Polygon ................... ................ 21 20. Co-ordinates as Projections ........ ......................... 22 21. Formulke of Transformation ................................. 22 22. Note on the Formula .......... ............................. 25 23. Linear Substitution.................. 25 24. Equation of 1st Degree pictures a Right Line ............. ... 28 25. Special Forms of the Equation ....... ....................... 29 26. Angle between Two Right Lines ............................. 32 27. Distance from a Point to a Right Line ...... ................. 34 28. Families of Right Lines ......... .......................... 35 29. Pencils of Right Lines ............................ ......... 38 30. The Equation M1L1+tL2+:,3L3=0. TriangularCds .......... 39 31. Constant Relation between Triangular Co-ordinates ............ 40 32. Geometric CMeaning of the Parameter A ...... ....... ........ 42 33. Halvers of the Angles between Two Right Lines ..... ......... 43 34. Illustrations of Abridged Notation..................... ..... 43 35. Polar Equation of the Right Line ....... .................... 45 36. Area of a Triangle given by its Vertices ................ ... 46 37. Area of a Polygon given by its Vertices ...................... 47 38. Area of a Triangle given by its Sides ...... .................. 48 39. Ratio in which a Tract is cut by a Right Line ..... ........... 48 40. Theorems on Transversals .................................. 49 41. Cross Ratios of Ranges and Pencils ................... ...... 50 42. Cross Ratios of Rays given by their Equations ........ ........ 54 43. Equation of Condition that Four Rays be Harmonic ..... ...... 55 TABLE OF CONTENTS. Vii Article. Page. 44. Relative Position of Four Harmonics ....... ................. 55 45. Homographic Pencils ....................................... 56 46. Common Harmonics ........... ............................ 57 47. Involution of Rays (Points) ........................... ..... 57 48. Centre and Foci of Involution ............................... 58 49. Cross Ratio in Involutions ............s..................... 59 50. Homogeneous Equation of Nth Degree ...... ................ 60 51. Resolution of Equations of Higher Degree ...... ............. 60 52. Angles between the Pair kx2 + 2hxy +jy2 = 0 .................. 62 53. Halvers of these Angles .......... .......................... 63 54. Intersection of Right Line and Curve of 2d Degree ..... ...... 63 55. The Right Line as a Locus. Examples ...... ................ 64 66. Families of Right Lines through a Point ...... ............... 69 CHAPTER III. THE CIRCLE. 57. Equation of the Circle ...................................... 71 58. Determination of a Circle ................................... 71 59. Normal Form of the Equation of a Circle ..................... 72 60. The Circle conditioned ................... ................. 73 N.B. On the Quadratic Equation ....... .................... 74 61. Axial Intercepts of the Circle . .............................. 75 62. Polar Equation of the Circle .. ............................. 76 63. Intersection of Circle and Right Line ....... ................. 76 64. Coincident Points. . . ............ 77 65. Tangent to the Circle ........... ........................... 79 66. Tangent to Curve of 2d Degree ........ ..................... 80 67. Tangents from a Point to Curve of 2d Degree ..... ........... 81 68. Outside and Inside of a Curve ............................... 82 69. Polar and Pole. Conjugates ................................ 84 70. Construction of Polar ........... ........................... 85 71. Polar as Locus of a 4th Harmonic ....... .................... 85 72. Equation of a Pair of Tangents ........ ..................... 87 73. Diameter and Normal ........... ........................... 88 74. Power of a Point as to a Circle .............................. 89 .5. Relative Position of Pole and Polar .......................... 89 76. Power-Line............................... 92 77. Power-Centre ........ ....................... 93 78. Power-Line of a System ................................ 94 79. Equation of a System . ..................................... 94 Viii TABLE OF CONTENTS. Article. Page. 80. Limiting Points ........................................... 94 81. Orthogonal Circles ............ ............................ 95 82. Similar Figures .................. 99 83. Circles are Similar ........... ............................. 100 84. Chords of Contact ........... ............................ 101 85. Axes of Similitude ........... ............................. 101 86. Centrode of a Circle cutting 3 Circles under W a ............. 102 87. The Taction-Problem proposed ............................. 103 88. The Taction-Problem solved ....... ........................ 104 89. The Circle as Locus. Examples .......... ................ 106 CHAPTER IV. GENERAL, PROPERTIES OF CONICS. 90. Results recalled ........................................... 113 91. Distance from a Point to a Conic ..... ..................... 114 92. Diameter and Centre ........... ........................... 114 93. Centre and Diameter.................. .. 115 94. Direction of Diameters ....... ............................ 116 95. Central Equation of the Conic ....... ...................... 117 96. Axes of the Conic ............ ............................ 117 97. Criterion of the Conic .......... ........................... 118 98. Asymptotes of the Conic ......... .......................... 119 99. Central Distances of Pole and Polar ...... .................. 119 100. Equation of P reduced............. ........ ........ 120 101. Ratio of Distance Products ........ ....................... 120 102 Constant Functions of k, hj, ............................. 123 CHAPTER V. SPECIA.L PROPERTIES OF CONICS. 103. Equations of Centrics in Intercept Form ..... ............... 125 104. Relation of a Centric to its Axes ...... ..................... 126 105. Central Polar Equations of Centrics ...... ................. 127 106. The Centrics traced ........... ............................ 127 107. The Equations solved as to y ....... ....................... 129 108. Direction of Conjugate Diameters ...... .................... 1,0 109. Co-ordinates of Ends of Conjugate Diameters ........ ....... 131 110. Squared Half-Diameters . .................................. 132 111. Central Distance of Tangent ....... ........................ 132 TABLE OF CONTENTS. ix Article. Page. 112. Tangents, Subtangents, Normals, Subnormals .134 113. Perpoles and Perpolars .136 114. Foci and Directrices .137 115. Polar and Perpolar as Bisectors .138 116. Focal Radii .138 117. Asymptotic Properties .141 118. Asymptotic Equation of H .142 119. Polar Equations of E and H .143 120. P as the Limit of E and H .145 121. Properties of P as such Limit .146 122. Tangent and Normal to P .148 123. Interpretation of 4q' ..................................... 149 CHAPTER VI. SPECIAL METHODS AND PROBLEMS. 124. Magic Equation of Tangent .151 125. Magic Equation of Normal .152 126. Illustration of Use of Magic Equations .153 127. Eccentric Equation of E .153 128. Eccentric Equation of Chord, Tangent, Normal .154 129. Quasi-Eccentric Equation of H .155 130. Hyperbolic Functions .157 131. Supplemental Chords .158 132. Auxiliary Circles of E and H. 158 133. Their Correspondents in P .159 134. Vertical Equation of the Conic .160 135. Tangent Lengths from a Point to a P .161 136. Areas in the E .162 137. Areas in the P ....................................... .... 162 138. Areas in the H .163 139. Varieties of Conics .166 CHAPTER VII. SPECIAL METHODS AND PROBLEMS.-(Continued.) 140. Conditions fixing a Conic .168 141. Conic through 5 Points .169 142. Conics through 4 Points. 170 143. Pascal's Theorem .171 X TABLE OF CONTENTS. Article. Page. 144. Construction of the Conic by Pascal's Theorem ..... ......... 173 145. Elements of the Centric ........ ........................... 173 146. Elements of the Non-Centric ....... ........................ 176 147. Construction of Conics ............... .................... 177 148. Confocal Conics ........... ............................... 179 149. Confocals as Co-ordinate Lines ....... ...................... 180 1.50. Similar Conics ............................................ 181 151. Central Projection .......... .............................. 183 152. Projection of 4 s into 4 s given in Size ..... ............... 184 153. The Conic a Central Projection of a Circle ..... ............. 185 CHAPTER VIII. THE CONIC AS ENVELOPE. 154. Homogeneous Co-ordinates ................................. 187 155. Line-Co-ordinates ......................................... 188 156. Equations between Line-Co-ordinates ...... ................. 189 157. Interchange of Equations ...... ........................... 190 158. Tangential Equation of 2d Degree .......................... 191 159. Double Interpretation ......... ............................ 193 160. Brianchon's Theorem ...................................... 193 161. Loci of Poles and Envelopes of Polars ..... ................. 195 Note on Points and Lines at X ...... ....................... 196 Examples ................................................ 197 PART II. OF SPACE. CHAPTER I. 1. Triplanar Co-ordinates ......... ........................... 223 2. Cylindric Co-ordinates ......... ........................... 224 3. Spheric Co-ordinates ............... ...................... 225 4. Direction-Cosines ......................................... 226 4 Projections with Oblique Axes ...... ....................... 228 5. Division of a Tract ........................................ 231 6. Transformation of Co-ordinates ...... ...................... 231 7. General Theorems .......... ............................. 233 8. Equation of the Right Line. ................................ 234 TABLE OF CONTENTS. xi Article. Page. 9. Intersecting Right Lines .................................... 235 10. Common Perpendicular to Two Right Lines ................... 237 11. Co-ordinates of a Right Line ............... ................ 238 12. Equation of the Plane ...................................... 2 38 13. Normal Form of the Equation ............................... 239 13 The Normal Form with Oblique Co-ordinates .... r ............ 239 14. The Triangle in Space ...................................... 240 15. Position-Cosines .. ...................................... 241 15 Position-Cosines with Oblique Co-ordinates ....... ............ 241 16. Intersecting Planes ................. ............... 241 17. Lines on a Surface................. 243 18. Three Planes .................. ............................ 243 19. Four Planes . ............................................... 243 20. Meaning of X......................2 ....................... 244 21. Right Lines halving 4 between Two Right Lines .............. 244 22. Pencils and Clusters of Planes ....... ...................... 245 23. Distance between Two Right Lines ........................... 246 24. Tetraeder fixed by Planes .............. .................... 247 25. Meaning of Co-ordinates of a Right Line .247 CHAPTER II. 26. Generation of Surfaces .249 27. Cylindric Surfaces .250 28. Conic Surfaces ........................................ . 250 29. Surfaces of Revolution. 251 30. Discriminant of the Quadric. 251 31. Centre of the Quadric. 252 32. Tangents to the Quadric. 253 33. Tangent Cones . 254 34. Poles and Polars. 254 35. Diameters ............................................ . 255 36. The Right Line and the Quadric. 255 37. Perpendicular Conjugates. 256 38. Rectangular Chief Planes. 258 39. Special Case. 259 40. Classification of Quadris .......... ............. -260 41. Ratios of Distances to the Quadric ....... ................... 261 42. Invariants of the Quadric ......... ........................ 261 43. Geometric Interpretation ......... .......................... 262 44. The Ellipsoid .......................---.-..... 263 45. Cyclic Planes .............. ....-.-.-.. 264 Xii TABLE OF CONTENTS. Article. Page. 46. The Ellipsoid a Strained Sphere ....... ..................... 265 47. Eccentric Equation of the Tangent Plane ..... ............... 266 48. Normal Equation of the Tangent Plane ...................... 266 49. The Simple Hyperboloid ................... ................ 267 50. The Double Hyperboloid ........ ........................... 267 51. The Hyperboloids as Strained Equiaxials ..................... 268 52. The Elliptic Paraboloid ..................................... 268 53. The Hyperbolic Paraboloid ........................ ........ 269 54. Imaginary Cyclic Planes .................................... 269 55. The Quadric as ruled ........... ............................ 270 56. Right Lines on the Simple Hyperboloid ...... ................ 271 57. Imaginary Right Lines on the Ellipsoid ...... ................ 272 58. Right Lines on the Hyperbolic Paraboloid ..... ............... 272 59. Foci and Confocals ........ ...................--. 273 60. Confocal Co-ordinates .................................... 274 61. Cubature of the Hyperboloids ............................... 274 62. Cubature of the Ellipsoid .................................. 276 63. Cubature of the Elliptic Paraboloid ..-.-..-.-.-.. 276 64. Cubature of the Hyperbolic Paraboloid .......... ..... 277 65. Determination of the Quadric .......................... 278 INTRODUCTION. DETERMINANTS. Permutations. 1. Two things, as a and b, may be arranged straight, in the order of before and after, in but twco ways: a b, b a. A third thing, as c, may be introduced into each of these arrangements in three ways: just before each or after all. Like may be said of any arrangement of n things: an (71 + 1)th thing may be introduced in n + 1 ways, namely, before each or after all. Hence the number of arrangements of n + 1 things is n + 1 times the number of arrangements of it things. Or Pn+l = P.- it + 1. Writing n ! for the product of the natural numbers up to n, we have n + 1 ! = i n + 1 Hence, if P,,, = n!, P.-,= it + 1!, and so on. Now P2=2=1-2=2!; hence, P3 = 1 .2 - 3 = 3 !,and P, = it! The various arrangements of things in the order of before and after are called straight Permutations or simply Permuta- tions of the things. The number of permutations of n things is n ! (read factorial it or n factorial). If the things be arranged not straight but around, in a ring, we may suppose them strung on a string; if there are n of them, there are also ut spaces between them. We may suppose the string cut at any one of the 7i spaces and then stretched straight; this will turn the circular permutation into a strigltt one; and since we may make n different cuts, each yielding a CO-ORDINATE GEOMETRY. distinct straight permutation, the number of straight permuta- tions of it things is n times the number of circular ones. Or P.,- C", n ; .-. C", =n n-1 ! 2. The things, whatever they be, are most conveniently marked or named by letters or numbers. Of letters the alpha- betic order is the natural order; of numbers the order of size is the natural order; as: a, b, c, .. z; 1, 2, 3, 4, ... n. If any change be made in either of these orders, say in the last, then some less number must appear after some greater, some greater before some less. Every such change from the natural order is called an Inversion. The number of inversions in any permutation is found by counting the number of numbers less than a number and placed after it, and taking the sum of the numbers so counted. A permutation is named even or odd, according as the number of inversions in it is even or odd. Thus 2 5 3 1 6 4 is an even permutation containing 6 inversions; 3 1 2 .5 4 6 is an odd permutation containing 3 inversions. The natural order, 1, 2, 3, -- n, contains 0 inversions and is even; the counter order, n ... 3, 2. lcontains 1 + 2 + 3 + -- + n- or " 1.2 inversions and is even when the remainder on division of n by 4 is 0 or 1, odd when the remainder is 2 or 3. It is plain that in any permutation any thing, symbol, or ele- ment may be brought to any place or next to any other one by exchanging it in turn with each of the ones between it and that other one. Thus, in 3 7 4 5 1 6 2, 7 may be brought next to 6 by exchanging it in turn with 4, 5, 1. Hence any permutation may be produced from any other by exchanges of odJacents. 3. By an exchange of any two adjacents, as p), q, the rela- tions of each to all the others, and the relations of all the others among themselves, are not changed; only the relation of those two is changed. Now if pq be an inversion, qIP is not; and if pq be nlot an inversion, qp is one; hence in either case, by this Xiv INTRODUCTION. exchange of two adjacents, the number of inversions is changed by 1; hence the permutation chcnges from even to odd or from odd to even. If p and q be non-adjacent, and there be k elements between them, then p is brought next to q by k exchanges in turn with adjacents, and then q is brought to p's former place by c + 1 exchanges with adjacents: p is carried over k elements and q over k + 1; thus p and q are made to exchange places by 2 k + 1 exchanges of adjacents. The permutation meanwhile changes, from even to odd or from odd to even, 2 k+ 1 times; and an odd number of changes back and forth leaves it changed. Hence, an exchange of any tzco elements in a permutation changes the permutation from even to odd or from odd to even. Plainly all the permutations may be parted into pairs, the mem- bers of each pair being alike except as to p and q, which are exchanged in each pair; hence. one permutation of each pair will be even, and one odd; hence, of all the permutations, half are even, half odd. Determinants. 4. It is plain that n2 things may be parted into n classes of n each. We may mark these classes by letters: a, b, c, ... n, where it is understood that n is the nth letter; the rank of each in its class may be denoted by a subscript; thus p, will be the kth member of class p. Clearly all members of rank k will also form a class of n members. The whole number may be thought arranged in a square of n rows and n columns, as in the special case n = 5, thus: a, b, Cl dl el a2 b. C9 c12 e2 a. b3 C3 d., e3 a4 b4 (4 d4 e4 a5 b5 c. d5 e5 This arrangement is not at all necessary to our reasoning, but is quite convenient. TV Xvi CO-ORDINATE GEOMETRY. Suppose we pick out of these n- things n of them, taking one of each class and one of each -anlk (clearly, then, we take only one of each). This we may do in ni! ways; for we mav write off the n letters in natural order, a, b, c, .. n, and then suffix the subscripts in as many ways as we can permute them, i.e., in n! ways. Now suppose these n2 things symbolized by letters to be mag- nitudes or numbers, and form the continued product of each set of n picked out as above: write off the sum of these products, giving each the sign + or - according as the permutation of the subscripts be even or odd: the result is called the Deter- mint of the magnitudes so classified. Accordingly a Deter- minant may be defined as A sum of products of n2 symbols assorted into n classes of n ranks each, formed of factors taken one from each class and each rankt each product marked + or - according as the order of ranks (or classes) is an even or an odd permutation, the order of classes (or ranks) being natural. The symbols are called elements of the Determinant; each product, a term; the number of the degree of the Determinant is the number of factors in each product. The classes may be denoted bv letters anld the ranks by subscripts, or vice rersa. The definition shows that classes and ranks stand on exactly like footing; in any reasoning they may be exchanged. 5. There are several ways of writing Determinants. In the square way, exemplified in Art. 4, the classes are written in columns and the ranks in rows, or vice versa. Hence rows and columns are always interchangeable. This is a very vivid way of writing them, but is tedious. It is shorter to write simply the diagonal term, thus: Y. : alb2c, n.. The sign of summation E refers to the different terms got by permuting the subscripts, -there are a! of them; the double INTRODUCTION. sign means that each term is to be taken + or - according as the permutation of the subscripts is even or odd. Still another way is to write the diagonal between bars: a al2 c3 ... n, This is very convenient when there can be no doubt as to what are the elements not written: otherwise, the square form is best. 6. To exchange two rows in the square form would clearly be the same as to exchance in every term the indices or sub- scripts that mark those rows; but by Art. 3 this would change each permutation of the subscripts from even to odd or from odd to even; and this, by the definition, would change the sign of each term, and hence of the whole Determinant. Moreover, since rows and columns stand on like footing, the same holds of exchanging two columns: hence, To exchange two columns (or rows) changes the sign of the Determinant. If the two rows (or columns) exchanged be identical or con- gruent, i.e., if the elements corresponding in position in the two be equal each to each, clearly exchanging them can have no effect on the value of the Determinant, although it changes the sign; now the only number whose value is not changed by changing its sign is 0: hence, T7he value of a Determinant with two congruent rows (or col- umns) is 0. Every term of a Determinant contains one and only one element out of each row and column; hence a common factor in every element of a row (or column) must appear as a factor of every term of the Determinant and hence of the Determinant itself; hence, we may divide each element of the row (or col- umn) by it, if at the same time we multiply the whole Determi- nant by it; i.e., any factor of every element of a row (or col- Xvii CO-ORDINATE GEOTMETRY. umn) of a Determinant may be set out aside as a factor of the whole Determinant. It is equally plain that any factor may be introduced into each element of a row (or column), if at the same time the whole Determinant be clivided by that factor. 7. If we will find all the terms that contain any one element of the Determinant, say a,, we may suppose all the other elements in its column and row to be 0; this will make vanish no term containing a, and all terms 7tot containing a,. The Determi- nant, say of 5th degree, will then be a10 0 0 0 o b2 C2 (11 e2 o b3 c3 d3 e3 o b4 c4 (14 e4 o b; c, c. e3 Setting aside ac as the first factor in each product, we find all the part-products by holding the order bede fast and permuting the subscripts 2 3 4 ; but this is the way we form the Deter- minant jb2 c3c14e,5; also the sign of each whole product, after multiplying by a,, will be the same as the sign of the corre- sponding part-product, since 1 being in its natural l)lace, the only possible inversion will be in the subscripts 2 3 4 ' -ence, the sum of the part-products or multipliers of a, is the Deter- minant bvecld4e.' It is called the co-factor (or sub-(determinant, or minor) of al and is the Determinant left after destroying the row and column of a,. If now we will find the co-factor of Pk, i.e., of the element in the pth column and kth row, we may bring its row to the first place by exchanging its row in turn with each row before it, and its column in turn with each column before it. By these p + k - 2 exchanges the positions of the other rows and col- umns as to each other are not changed at all; they stand exactly as if the pth column and kth row had been destroyed. The new Determinant got by these exchanges will be the old one term for term with like or unlike sign according as 2r + 7 - 2 (the :2ziii INTRODUCTION. number of changes of sign) is even or odd, or, what amounts to the same, according as p + k is even or odd; hence the co- factor of Pk in the new Determinant will be the co-factor of p, in the old with like or unlike sign according as p + k is even or odd. But pk is in the first column and first row of the new Determinant, hence, by the foregoing, its co-factor is got by destroying its column and row, i.e., by destroying the pth col- umn and kth row of the old Determinant; hence the co-factor of pk in the old Determinant is the Determinant left on destroying the row and column of pv but taken + or - according as p + k is even or odd. The co-factor of any element may be denoted by the same symbol written large; thus, the co-factor of pk is Pa. 8. By definition, all the terms containing any element are got by multiplying that element by its co-factor; if, then, we mul- tiply each element of a row (or column) by its co-factor and form the sum, we shall get all the terms of the Determinant that contain any element of that row (or column); but every term contains one element of that row (or column); hence we get all terms of the Determinant; and since no term contains two elements of that row (or column), we get each term but once. Hence, the sum of products of each element of a rote (or column) by its own co-factor is the Determinant itself; a, b2c3... n I = aAl + blBl + cl Cl + - 4- +t N,,l = etc. It is to note that the co-factors subscribed contain everv other subscript in their values but 1, and every other letter but their own. Now change the subscript 1 to 2 on both sides of the equation ; there results I ma b2c3 -- -n, j = a22Al + b2B1 + C2 CO + -- + n2-1. The subscripts of the co-factors are not changed, because the subscript 1 does not appear in their values. Now the left side of this equation is a Determinant with two rows congruent, namely, the 1st and 2d, since the subscripts of 1xi CO-ORDINATE GEO-METRY-. the 1st, which were all w were changed to 2, the subscripts of the 2d(; hence its value is 0 by Art. 6; i.e., a 2A,b+ b2B + 1c2 C1 + - + n2NP= 0- The small letters are the elements of the 2d row, the large letters are the co-factors of the corresponding elements of the 1st row; plainly the reasoning about columns or about any other pair of subscripts would be the same; hence the sum (f produacts of each element of a row (or column) by the co-factor of the corresponding element of any other row (or column) is 0. 9. If the elements of any row (or column), as the 1st, be regarded each as the sam of two part-elements, so that cU1 = c1u+ at,', b1=bj'+ bj, ..., 7n1 = nj== n, then we shall have albo-1 23Ao it, = a ,a'A + bltBI + - + god Nj1 +bac'Ajl.+b..B, =+ 1b 1'It. The first bracket is clearly the Determinant ,alybc3- 1 j, the second is j a,"b2c3 . n,, 1; hence, it is plain that (a,' + a1") a c3 nb , j = a'bc3.. n + 1"b+C;,... In this way one Determinant may always be expressed as the samn of two part-Determinants which have all their rows (or columns) the same as in the whole Determinant, but one pair of corresponding ones, while the sum of ally two corresponding elements in this pair equals the corresponding element in the corresponding row (or column) in the whole Determinant. It is now clear that any Determinant may be broken up into 3, or, indeed, into any number of part-Determinants, each ele- ment of a row being supposed made up of 3, or any number of parts, thus: a, = al'+ -la,"+ -4l-'. bi-+ bill+ blr', etc. If every row (or colimnn) l)e broken up into parts this way, a p)art-Deterlninant may be formed by taking for a 1st column any part-column of the 1st column, for a 2d column any part- 2Xx INTRODUCTION. column of the 2d column, and so on throughout. Hence th total namber of part-Determinants will be the product of the cmlflbers of part-columns for all the columns. Evaluation of Determinants. 10. In the Determinant I albc3 ... nI add the 2d column to the 1st, each element to its correspondent; we get i(al + bj)b.c3---nn I =1 ajbc,---nj+jbib2C... nnl The 2d Determinant on the right has two identical columns of b's, hence its value is 0; the 1st on the right is the original one; hence the new Determinant on the left equals the old one. If instead of adding the 2cd column we had added its rn-fold, we should have got m times the 2d Determinant on the right, which would still be 0; plainly, too, the reasoning about any other pair of columns or about rows would be the same. Hence, the value of a Determinant is not chanqed b)!/ addinq to each of its elements in one row(or column) any fixed multiple of the cor- respondent elements in any other row (or column). This theorem furnishes a ready method of reducing the degree of a Determinant. For by proper additions all the elements of a row (or column), say the 1st, may be made 0 but one; then the whole Determinant will be equal to this one multiplied by its co-factor, for we shall have aj17J2c,---nn! = aAj + 0 B, + O-c+ -- +O -Vj = a,1 I2j 3 ... 'n. The degree of this co-factor is clearly one less than the degree of the original Determinant; by repeating this process the degree of the Determinant may be brought down to 2 or even to 1. Thus far the reasoning has been so closely eonnecte(l and withal so simple that it has been deemed best not to interrupt it in any way. The following examples will amply illustrate all the foregoing. Xxi Xxii CO-ORDINATE GEOMETRY. 1. 4 6 3 1 5 2 7. In this permutation the inversions are: 43, 41, 42, 63, 61, 65, 62, 31, 32, 52, ten in number, the permu- tation is even. b c a efg d. Here the inversions are: ba, ca, ed, fd, gd, five in number, the permutation is odd. On exchang- ing 6 and 2, the number of inversions falls to five, the permu- tation becomes odd; on exchanging a and f, the number of inversions rises to eight, the permutation becomes even. 2. The permutations of 1 2 3 are, in pairs: 123, 132; 213, 312; 231, 321; one of each pair is even, one odd. Which 3. The Determinant of 2d degree a b = a, b2 - a2b,. a2 bl 4. al b, cl = a b2c3 - a,b3c2 + a2bC, - a2 bc3 + a3bC2 a2 b6 c2 - a3b2 cl. Ct b3 C3 The number of terms is 3 ! = 6, so we write off the combina- tion a b c six times and suffix the subscripts permuted. The numbers of inversions in the permutations are resp. : 0, 1, 2, 1, 2, 3, and the signs are prefixed accordingly. A simple mechanical rule for calculating the Determinant of 3d degree as shown in this diagram: The arrows turning up are drawn through the + combina- tions; those turning down, through the - ones. But this does not hold for higher degrees. 5. Ia, 1 I, cl , s ., d., -b, a2 c2 d.2+ --C, a. b2 d2 a, b-2 C-) d2 b. c; 4 a I c, ,a363 d3 a,3 bc3 dC b4C444 a4C4d4 c464d4 a4 b4 C4 d4 -d, a2 6 C2 a: b3 C3 a4 b4 C4 INTIIODUCTION. Xxiii 6. a,+ a, bi c = a, b1 c1l +a b1 cl a2+ a2 b2 C2 a2 b2 C2 a2 b2 C9 a3+ a3 b3 c3 a3 b3 c3 a3 b3 C3 7. 0a,+a, b +,81 cl+7yl a2 + a2 b2 +/32 C2 + 72 C3+ a3 b3 +133 c3 + y3 =Iab2c31+Iab273I+al82c31+Iai/2y3l +Iab2c31 + I ab2Y31 + I al/32C31 + I al /323 1. 8. a,+mb1 b1 c1 ='al b& cl + nmb b1 c1 ct2 + vib9 b2 C= a2 b2 C2 rnb2 b2 C2 a3 + mnb3 b3 C4 a3 b3 c3 1fmb3 b3 C3 = 01 b2c3j + ml b62c3 = a6b2C3l- 9. b6 b, cl =blb2c3-blb3c2+b2b3cl-b2bicb3+b3blC2-b3b2CI b2 b2 c2 =0. b3 b3 c3 10. a, c1 bi = aic2b3 - 0ac3b2 + a2c3b, - a2clb3 + a3c1b. a2 C2 b.2 - a.,c26b a3 c3 b3 =-/, b1 c1. a02 b2 c2 a3 b3 C3 11. 4 5 =32-30 = 2 1 2 5 = 2(16-15) =2j 4 5 6 8 38 3 4 - 2(16 - 15) = 2. 12. 4 5 2=96+12+100-32-40-90=46. 6 8 10 2 1 3 Also 4 5 2 =2 2 5 2 =2 0 3 -4 =213-41 6 8 10 3 8 10 0 5 1 15 1 2 1 3 1 1 3 1 1 3 = 2(3 + 20) =46. CO-ORDINATE GEOMETRY. 13. 2 3 5 7 '= -1 5 4 2 6;1 , 7 2 3 8! 1 7 3 4 2 51 3 =11 17 1 1 a 24 2 j-1 11 1 -1 4 2 4 6 C=! 1 .1 I 3 2 --1 2 6l 0 3 81 0 2 5 0 1 17 16 !=1 0 79 7l ()28 27 -1 3 2 ' -1 17 16 -_ 24 22 1 11 11 2 771=-23. 1 271 into the form k h g !, and find the co-factors If, H, G, F, C, J. h jfl g f c 5 3 21,1 3 !4 7 81 6 9 2 61 2 -2 4 -7 5 ,18 3 -2 , a b c . -3 5 -7 61 b c a 6 4 9 -5f lc a b r y 1 =0 - r2x, rjy2 - 22YI 1 r2- ri - r9_ -r3X2 r2 Y,-r y2 1 - r3 r2-1 - rs 1 x- r, r2 r3l' XI X. X3 1 11 1, 1 y i+1 7'12 r, j x1 x2 x3 I= 0. Multiplication of Determinants. 11. An interesting case of Art. 9, as illustrated in Example 7, is this: a a,+bf B1+c, y1+,+n v1, a a2+b1O2+... +nlv2 . ... a. ..ablja L+..f--+nljVn a2aj+b201+C2'Y+1+n2v1, a2c a2+b22+'+n2v2,. ... --, a2c&+b,)9i+7&+n2.V.n I afal+bfl+c.,yj+---+n.vj, aCC2+b-J32+--+nev2 . ... ...+ We notice here that the Greek and the Roman letters enter this Determinant in the same way: the Greek appear in the columns as the Roman do in the rows, and vice versa. Again, this whole Determinant, which we may write AGR, clearlv breaks up into the sum of 7t, part-Determinants; for any of 14. Reckon 15. Reckon 16. Bring the equation ri x2, - ri - 9'2 x - r, r2 r3 Y1 Y2 Y3 1 1 lji xxiv INT RO)DUCTIIN. O the I st n part-columns may be coml)ined with any one of the 2(1 n part-columns, and so on. Blt all btt i ! of these l)art-Determi- nants vudWsh; for, if the Roman letter of one part-columin be the same as the Roman letter of any other, then on setting out the Greek factor the part-determinant will have two columns identical, and vanishes by Art. 6. Accordingly, to get part- l)eterrninants not = 0, we must pick out each time all the Roman letters for our part-Determinant, one for each part-column hence, on setting out the Greek factors, the part-columns become the columns of the Determinant of Roman letters, C Ib c3 ... n, 1, which we may write AR ; hence, our part-Determi- nant contains as a factor a determinant that can differ from AR only in the order of its columns, i.e., only in sign; hence. being a factor of each part-Determinant, AR is a factor of the whole Determinant. Hence, since the Greek and the Roman letters enter AGR alike, AG is also a factor of AGR. Now the terms of the product AG- AR are of 2Sth degree, and so are the terms of AGGR; also the number of terms both in AGR and in AG AR is the same, n! n! Hence, AGR can differ from AG -AR, if at all, only in sign; i.e., AGR= AG- AR. To decide as to the sign, consider the product of the diagonal terms; it is + in AG AR; the like term is also + in AG]R. For it is got by taking the 1st part-column of the 1st column, the 2d of the 2&, etc. the factors set out are aC1 3293 ... ,, and the order of the columns of Roman letters is natural, as in AR; and in AR the diagonal alb2c3 ... no is +, as is the diagonal term in every Determinant. Now one pair of corresponding terms being like-signed in AG AR and AG], all are like-signed; i.e., AG AR= AGR. ILLUSTRATION: la1 jj+ bj/3 + cjly ala2 + b, 32 + C1Y21 aa3 + bi133 C7y3 1 (a2u + b.,1+ c2y7, a2a2 b2 22+ C.,7, 2(2a3+ b2/3+c.,73 ata, + b3j31 + c3 y7 1((3a2 + b,342 + c3702, a3a3 + b3/33 + C3 t3 = a,2/2y3 I aib2c3 I +a1233 I a, c2b3 + 112a3 bc2Co a:, + 13jlaY3ba2c3 +yILa293 ICca2b3 !+yi/2a3 1cb2aa31 xxv CO-ORDINATE GEOMETRY. Here the inversions in the order of the columns of Roman letters are those got by permuting those letters; if we restore them to natural order, the indices, which are now in natural order, will be permuted as were the letters; also, restoring them will change or not change the sign of the term according as the number of inversions is even or odd; but the inversions in the order of Roman letters are the same as in the order of the Greek; hence, the Greek products are + or - according as the permutations of the letters are even or odd; hence, the sum of the Greek products is the Determinant of the Greek letters. As no reference has been made to the number of letters, 3, this proof is quite as general as the one already given. Actually bringing the Roman letters into natural order, we get AGR = al,82y3AR -alf3y2AR + a3f, y2AR - a2/f313AR + a2/31Y AR - a3f32yAR = AG. AR. 12. If a, = Al, a2 =A2, ..., I i= B1, ..., the Greek letters being the co-factors of the corresponding Roman letters, then, in AGR all the elements vanish but the diagonal ones, bv Art. 8, and these are each AR; hence AGR has but one term, the diagonal term, and that is AR'. Writing A for AR and A for AG, and remembering AGR = AG AR, we have A.A=An or A=W-1; the Determinant of the co-factors of the elements of a Determi- nant of nth degree is the (n -- 1) th power of this Determi-ant. In particular, if n = 3, A =A2. If the co-factor of A, be a1t, and so on, and if the Determinant of the co-factors a,', etc., of the co-factors A1, etc., be written A', then A = A 1 = (A-) -I = 2-2n+l Now if we multiply a1, bi, ..., a2, .... each element of A by A-2 or A: A, it will be the same as to multiply A by A(,-,2) since it multiplies each row (or column) by An-2; the resulting Determinant is then An'2_+1, or is A'. Hence, we may con- xxvi INTRODUCTION. dude that a'- a-A 2 a proposition we have not space to prove more rigorously. In particular, if n = 3, a' = a A. Applications. 13. If there be given a system of n Equations containing n Unknowns in first degree only, it is possible by successive elimination to get one Equation with one Unknown, whence this Unknown may be found; then, by substitution, all the others may be found, one by one. But this process is verv tedious. The theorems of Art. 8 furnish a direct solution of the problem. Call the Unknowns u1, U2, "', U.; then a2ul + blu2 .+ 2-- -2 a.U1 + bn12 . + ... ... ... + nn = kn. If the Equations be multiplied by Al, A2, ... A,, in turn and summed, where Al, A2, ..., An are the co-factors of a,, a9, ... an in the determinant of the coefficients I alb2c3 ... n1 , the coeffl- cient of ul in the sum will be aAl + a2A2 + -- + a,, = A ab2c3 -... n,, 1, by Art. 8, while the coefficient of any other u, as lt2, will be b6Aj + b2A2 + --+ bA, =0, by Art. 8, and the absolute term in the sum will be kAl + 12A2 + + k.A. = I ki b2C3... n.; or, [alb2c...nflnul =kb2C3e... n., whence vl Ik bc3 a - n al b.,c, n I Note that the denominator is the Determinant of the coeffi- cients in order; the numerator is got from the denominator by xxvii CO-ORDINATE GEOMETRY. replacing the column of coefficients of the Unknown in question by the column of absolutes. Plainly the reasoning holds alike for all the Unknowns. 14. If the absolutes, the k's, be all 0, then the numerator of the value of each Unknown, of each u. has a column of 0's, hence itself is 0; hence the value of each u must he 0 uinless the (denominator also be 0; when the k's are 0, the Equations are homogeneous in the it's; hence The condition that i hoinoyeneous equations of 1st degree in n Unknowins may consist, is that the Determinant of the coefficients of the Unknowns vanish. Where the Equations are homogeneous in the it's, each may be divided by one of the u's, say it.; the quotients of the it's being considered as newt Unknowns, new i's, there are n Equa- tions and only (n - 1) Unknowns, while the coefficients of ue are absolutes; the condition of consistence of the Equations is unchanged; hence Th7e condition that n Equations among (v - 1) Unknowns in 1st degree may consist, is that the Determinant of the coefficients and the absolutes vanish. 15. Often it is required to eliminate one Unknown between two Equations of higher degree in that Unknown; or, what is the same, to find what relation must hold among the coefficients in the two Equations, if the two are to consist, i.e., hold for the same values of the Unknown. An example will make this clear. Given ax3+ ba2+cx+d=0 and ex2+fx +g=0 find the condition that these Equations consist, i.e., that the roots of the 2d be also roots of the 1st. When the first Equation is satisfied, so is this: ax' + bx +cx2+dx = 0; when the second is, so are these: ex3 +fx2 + gx = 0 and ex4 +fx3 + gx = 0; .x .i. INTRODUCTION. hence, when both are satisfied, so are these five: 0 . a + ur" + bx2'+ cx + d = 0, a.x4+W3 +cx2+dx +0=0, 0.x4+0- +ex2 +fx +g=0, 0 X4+e x2+fx2 +gX +0=0, e x4+fxr + yx2 +0-x+0=0. Here are 5 Equations containing 4 Unknowns: x4, x3, x2, x; by Art. 14 they consist when and only when '0 a b c d =0. a b c d1 0 0 0 e f g 0 e f g 0 e f g () 0 Clearly this method is always applicable. Be one Equation of nth degree, the other of (n + d)th degree; bv multiplying the 1st by x (I times successively we raise it to the (n + d)th degree, and get in all d + 2 Equations and n + d Unknowns; then by multiplying each of the two Equations of (n + d)th decree by x we get two more Equations and one more Unknown, the next higher power of x, xnTd-l; by n - 1 such successive multiplica- tions we get in all 2 a + cl Equations and 2 n + cl -1 Unknowns; then Art. 14 is to be applied. Under the hands of British and Continental masters the Theory of Determinants has been of late years built up to colos- sal size and applied to almost every branch of mathematics; in fact, it has become well-nigh indispensable to higher research. An excellent English work is Muir's Theory of Determinants. EXERCISES. 1. Solve the systems of equations: 3x+4y-5z=7, 2x-3y-4z=9, 4x-5y+z=8; x-y+2z+5v=10, 2x+3y-z+v=7, 44y-3x+8z-2v= 5, 3z-2y-5x+ 7v=3. xxix xxx CO-ORDINATE GEOMIETRPY. 2. Do these systems of equations consist 5x-3y=7, 8x+5y=9, 3x-2y=-4; 2x+3y-4z=5, 5x-2y+3z=7, x+ by-5z=-8, 4x+3y-3z=2. CO-ORDINATE GEOMETRY. PART I. THE PLANE. CHAPTER I. INTRODUCTORY: FIRST NOTIONS. Function and Argument. 1. In a table of logarithms are found two series of corre- sponding values: one of natural numbers and one of logarithms. Given any number, we may find from such table a corresponding logarithm; given ally logarithm, we may find the corresponding number. Like may be said of a table of natural sines: given any number (expressed commonly in degrees), we can find the corresponding sine; given any sine, we can find a correspond- ing number. Such tables are calculated to a greater or less degree of exactness by rules or formula; other like tables are found in works on Physics, calculated, however, not by rule, but by experiment. From such a table we may find (within certain limits) e.g. the tension of saturated vapor of water for every degree of temperature, and conversely; but we lknow no rule to calculate one from the other. Two magnitudes, such that to any value of one corresponds a value of the other, are called functions of each other. Such are a number and its logarithm; a number and its sine; the surface or volume of a sphere and its radius; the velocity of a wave-motion, as of sound, and the elasticity of the medium the density of pure water and its temperature; etc. CO-ORDINATE GEOMETRY. 2. As is seen, in Physics the interdependent magnitudes called functions in general define physical states, and the nature of their interdependence cannot generally be stated in a formula or rule; in Mathematics the interdependents are empty forms, symbols: x, y, z, a, b, c, and the nature of their interdepend- ence is expressed by a formula or equation. Take, for example, the Eq., 2x + 3y = 12. Assigning arbitrary values to x resspectively y, we reckon the corresponding values of y resp. x by the formulce (rules) y=12-2x: 3 resp. x=12-3y: 2. A series of pairs of corresponding values is (xy)= l(3,6); (2,; (- 1, 3); (0, 4) ; (1, 1-A); (2,); (3, 2); -. In the unsolved Eq., 2x+3y= 12, x and y stand on precisely like footing: each is an implicit function of the other, but in the Eq. solved as to one of them, say y, thus y= 12-2x: 3, they no longer stand on like footing; contrariwise, this Eq. gives a rule for reckoning the value of y for any assignied value of x, but not conversely. 77tat symbol to which wie assign arbi- trary values is called the Argument; the symbol whose values are reckoned is called specifically the function; if the Eq. be solved as to any symbol, that symbol is called an explicit function. ILLUSTRATION. x' + y2 = 25: here x nid y are each an implicit function of the other; solved, V25 - x2: here y is an explicit function of the argument x. x= + a/25 - y2: here x is an ecplicit fitnction of the argument y. Short for Equation. 2 KINDS OF FUNCTIONS. N. B. Thoughll the relation between two symbols l)e -ccurately expressed by -an Eq., yet it is not in general possible to state a rule for reckoning one throtu(gh the other, since the general Eq. of d(egree higher tlian the fifth has not yet l)een solved. ,None the less, the symbol to which we suppose arbitrary values .issig,-ed is still called the argument; that, as to which we sul)- pose the Eq. solved, the function. 3. If in the Eq. connecting two symbols each be operated upon lwv a finite number of algebraic ol)erations, additions, mul- til)lications, involutions, an(d their inverses, theen is each called ,in flqJbric fanotion of the other; but if the number of such oI)erations upon either be ihnfiite, then are thfey called transcen- dlental functions of each other; as y = sinx=x_ Z--+ - - + - ,( _ )1 - 3 ! 5 ! 7 ! -=O 2n+1! X2 X3 x4 x. 2 ! 3 ! 4 ! -0 n. Alore imlrortant for us is this (listinction: when of the argument correspond one, two, three, resp. of the function, the latter is called a one-, twvo-, miany-valued function of the argument; as to onie value ma(o/ values three-, resp. 2x+3y= 12: X 2+ ,2 =-r2. J = 4px: x and y each one-valued functions of the other; x and y each tco-valued functions of the other; x a one-valued function of y, y a two- valued function of x; 2n. y= cosx=(-1), .-: y a one-valued function of x, x an in- 2n ! finitely many-valued function of y (since, if x = r be any root of the Eq. for any assigned value of y, then is also x = 2n7r r a root, n being any natural number). 3 CO-ORDINATE GEOMETRY. EXERCISES. What functions of each other are x and y in as + '=1; y3-3axy + x3 = 0; ( + y2)3 = 4a2Xq2; '2 6 Y=2 (ev +e a) From the last Eq. express x as an explicit function of y. 4. To different values of the argument correspond in general different values of the function; if the difference between two argument-values be very small, in general the difference be- tween the corresponding function-values will be very small: as the argument changes gradually, so does the function. Be now x2 - x-x the difference between two argument-values, y2 - =Ay the difference between the corresponding function- values; if then, by taking Axo-, we can make Ay& for all values of Ax a-, where the oi's mean magnitudes small at will, then is y called a continuousfunction of x. ILLUSTRATIONS. 2x +3y = 12. If (X11y,), (X2,y2) be two pairs of corresponding values, then 2x + 3y1= 12, and 2x- + 32 = 12; whence x,2-x, =-2 (yI,-yjj), or Ay= - '2Ax. The sign - shows that, as either x or y increases, the other decreases; but we have here to dleal only with the absolute (or signless) values of the differences Ax and Ay, and since their ratio 2: 3 is finite, clearly we can make and keep either small ait will by making and keeping the other small at will. This is so for every finite value of x or y; hence, each is a continuous function of the other for -X x oc, -so y x . Like mav be shown of the functions y = sin x, y = cos x, but not of their quotient sin x y-tanxZ _ cos x 4 KINDS OF FUNCTIONS. 5 For x=-2 +h, h very small, sinx=-1 nearly, cosx 2 is + and = 0 nearly; hence, y _ tan x is very great and neg- ative. As h increases toward , sinx increases toward -0 2 (i.e., nears 0 from the - side), cosx increases toward +1, tan x increases toward - 0 gradually; as h nears 7r, x nears - sin x nears + 1, cos x nears + 0, tan x becomes + and very great, but changes throughout gradually with x. For x = - -, 2 a- small at will and +, tan x is very great and +; for 2 + a,, cosx changes from the + to the- side of 0, sinx changes not, tan x changes from being very great and + to being very great and -. Hence, for - x 7, tan x is 2 2 a continuous function of x; but for x = -, tan x is a discon- tinuous function of x: if x1 a, X2 2- we can not make Y2- y-tan x2- tan x1 _ Ay small at will by making x2- x1 -Ax small at will; as x passes through the value x= 2, tanx springs from + o to -o . The value x = 2 is called a point of cdiscontinuity for the function y tan x. Since wr is the period of the tangent, i.e., tanx=tan(x 7r), x=(2n+l)72r is 2 also a point of discontinuity. EXERCISE. Show that y = c. is discontinuous for x=a. HINT. As x nears a, increasing, y nears -c; as x nears a, decreasing, y nears +c; as x passes through a, increasing resp. decreasing, y springs from-c to +c resp. from +c to -c. 6 CO-ORDINATE GEOMETRY. Geometric Representation of Magnitudes. 5. Any measurable magnitude may be represented by a number, called its metric number: the ratio of the magnitude to an assumed unit-magnitude; a number may be represented, or pictured, by some definite part of a line, some arbitrary tract (i.e., definite part of a right line) being taken to picture 1. Thus, if AB picture 1, BD will l)ictulre 2, BC will picture V2, a b C"1 C' C 0 A S BCD will picture -r. Instead of the tract whose metric number is a may be said briefly the tract a. On the ray OD lay off from 0 any number of tracts picturing as many (positive) numbers. To picture the sum of any two numbers, a and b, lay off a tract s equal to the sum of the tracts a and b. To picture the difference of two numbers, a and b, lay off a tract d such that the sum of the tracts b and d shall equal the tract a. To do this, lay off from the end A of tract a, toward 0, i.e., counter to the direction a was laid off in, a tract b to C; then is OC the sought tract d. Three cases may arise: (1) a b, then the point C fails between 0 and A; (2) a = b', then the point CI falls on 0, the difference d is 0; (3) ab", then the point C" falls beyond 0 leftward; but then the difference a -b" or d is negative; hence negative numbers are pictured by tracts laid off counter to the direction in which are laid off tracts picturing positive numbers. But in this way was laid off a positive tract, to subtract it; hence, to PICTURES OF NUMBERS. add a negative tract, subtract an equal positive one; hence, too, to subtract a negative tract, add an equal positive one. N.B. To add to a tract, or to subtract from it, we lay off from its end: forwards, from it, to add a positive, or subtract a negative, tract; backwards on it, to subtract a positive, or add a negative, tract. Like reasoning and conclusions hold for angles and arcs. 6. Assumning, then, any RL. (short for right line), we may l)icture all real numibers by tracts laid off on it rightward and leftward from any assumed l)oint 0. As all such tracts have 0 a A the same beginning, 0, each is fully defined by its end. Ac- cordingly, not only the tract OA by its length and direction, but its end, the point A1, by its position as to 0, pictures the num- ber a; so, every point on the RL. OD pictures some real number, the RL. itself pictures the whole of real numbers. If we take two RLs., as OX, 0 Y, intersecting under any 4 w, each will picture by its Y points the whole of real numbers, the section 0 naturally taken to pict- tire zero in each picture. Any point P in the plane of the RLs. will picture A then not simply one number but a pair of numbers, as (a, b) ; for its distances from the zero-point 0, measured along these RLs., are a and b. N.B. The choice of RLs. and of positive and negative direc- tions is arbitrary; it is common to treat rightward and upward as +, leftward and downward as 7 8CO-ORDINATE GEOMETRY. In the figure, by them. 1. Find the (-2, 2) EXERCISES. the corner-points picture the pairs of numbers bracketed points that picture these pairs: (1, 3), (-3,-1), (-2, 1), (4,_-5), (0, a,(0,65), (5, 0), (6,0), (0,0). 2. Where are all points picturing pairs whose first 3,1) /'term is 0; i.e., of the form X (0,1b) 3. Of the form (a,0)3 4. Where are all pict- tures of pairs whose terms (4, -2) are equal; i.e., of the form (aa) S. Of the form (a,-a) 6. Where are all pictures of pairs whose first terms are all alike, second terms unlike 7. Whose second terms are all alike, the first unlike 7. Be now f(x, y) 0 (read: the f-function of x and y equals 0) any Eq. determining x and y as functions of each other, and suppose the functions continuous and one-valued. Be (xl, y,), (x, 2y2), . . (xa, y.,) pairs of corresponding values; i.e., bef(xl,y,)==, f(X9Y2.) =0,.. .f(xL,,y,) = 0 Each pair is pictured by a point in the plane of OX and 0 Y Suppose the x's subscribed in the order of size, thus: x,. x2 X3 x:. By taking consecutive values of x very close to each other, we make the consecutive values of y very close to each other; this series of pairs of values of x and y will then be pictured by a series of points consecutively very close to each other. Now, it is true, however close together we may heap these points, we can never make out of them a line. But if the function be continuous, by teking xk ,1 -xs = a we can make Ik1+ - = At, and for every value of x between x,,, tand x,. the corresponding value of g/ will differ from y, by a', positively or negatively; i.e., all points 8 DETERMINATION OF POSITION. picturing pairs of values for x between x,+, and x. will lie in the double parallelogram whose sides are cr and 2 d'. Hence all points picturing pairs of values that satisfy the Eq. f(Ax, y) = 0 lie in a series of con- tiguous double palrallelo- / grains, whose sides are 7 small at will; so, too, are the diagonals of the single parallelograms. The train of diagonals (that one of each couple X lbeing taken that joins two picturing points) will form a polygon (opeu or closed) whose vertices are picturing points, andl by taking the values of x, and therefore of y, ever closer and closer together, we make this polygon near as its limit a definite curve, of which each diagonal is a chord. Every pair of values satisfying the Eq. f(x, y) = 0 is pictured by a point on this curve ; and conversely, every point of this curve pictures a pair of values satisfying the Eq. (x, y) = 0. Hence, A yeometric )ieteare qf the Eq. f(x, y) = 0 is a plane curve. We say indifferently the Eq. f(x, y) 0 and the curve f(., y) = 0. N. B. The curve breaks up in ease: of a many-valued function, into mnany branches ; of a discontinuous function, into distinct parts. But the reasoning needs but slight ehange. If some other than a RL. be used to picture a series of numbers, like reasoning and conclusions hold. Determination of Position on a Surface. 8. We say of a surface; it is doubly extended, or has two dilnensions, meaning that two independent measurements are necessary and sufficient to fix any point on the surface. Thus we lnow ay lplace on the earth's surface, kuowingY its latitude and longitude. In this familiar example we suppose the sur- 9 CO-O1DINATE GEOMETRY. face covered with a double system of lines: half-meridians and perpendicular parallel small circles. Through any point of the surface passes one and only one half-meridian, one and only one parallel; hence, knowing any point, we know its meridian and parallel. Also, any half-meridian cuts any parallel in one and only one point; hence, knowing any meridian and any par- allel, we know their junction-point. The parallels are named from their angular distance from the mid-parallel (equator) the half-meridians from their angular distance from an assumed fixed half-meridian. Likewise we may think a plane covered with a double system of (say parallel right) lines, making any angle with each other. Through any point in the plane passes one and only one line of each system; hence, knowing any point, we know what pair of lines meet in it. Conversely, any pair of lines meet in one and only one point of the plane; ases / r hence, knowing an+ pair of --lines, we know their june- tion-point. We name and know each line of a system by its distance from an assumed fixed line of the system (measured on any line of the other system). If this measurement be rightward or up, the metric number of the distance is marked +; if leftward or down, -. The assumed fixed RLs., as OX, o Y, are called Co-ordinate Axes, or axes of X and Y, or X- and Y-axes. The angle to, reck- oned from the + X- to the + Y-axis, is called the co-ordinate angle; for = 900, the axes are rectangular; otherwise, oblique. The junction-point, 0, of the axes is called the Origin. 9. A point is fixed as the junction of a pair of parallels, or co-ordinate lines; conversely, a pair of co-ordinate lines are 10 POLAR CO-ORDINATES. fixed as meeting in a point. The distance from the origin 0 at which a co-ordinate line cuts the X- resp. Y-axis is called its intercept on that axis. Such intercepts are denoted by the symbols x resp. y. If a be the metric number of the x - intercept of any parallel to the Y-axis, then is this par- allel known. completely from the Eq. x = a, which is there- fore called its Eq. So y = b is the Eq. of a parallel to the X-axis making an intercept b on the I-axis. The junction-point of this pair is known completely from the two Eqs. x = a, y = b, which are therefore called the Eqs. of the point. The point itself is spoken of as the point (a, b) or as 1P (a, b) ; a is called the abscissa or x of the point, b, its ordi- nate or y; a and b, its co-ordinates or its x and y. We may now convert the proposition of Art. t;, thus: Any point in a plawe may be represented by a pair of numbers: the rectilinear co-ordinates of the point. 10. We may think the plane covered with some other double system of lines: as a system of rays from the centre of a sys- tem of concentric cir- cles. Each point is pI fixed as the junction of a, pair of co-ordi- nate lines: ray and circle; each pair of such co-ordinate lines is fixed as having o such a junction-point. Each ray is known and named from its angle with some assumed fixed rat, as OD, called base-line or polar-axis; angles reckoned clockwise, as 0', are marked those reckoned counter-clockwise, as 0, are marked + ; the direction of a ray which bounds the ray's angle, as OP, is taken + ; the counter-direction, as OP", -. Each circle is known and named from the radius p. The angle 0 of a ray and the 11 CO-ORDINATE GEOMETRY. radius p of a circle are called the polar co-ordinates of the junction-point of ray an(1 circle; i.e., of the point (p, 0). p = C is the Eq. of a circle: it declares that every point of the circle is distant c from 0; 0 =d is the Eq. of a ray : it declares that the radius vector from 0 to any point of thie ray is sloped d to the polar axis. Ray and circle meet in the poinlt whose Eqs. are p= c, O = d, i.e., in the point (p, 0). N.B. To fix all points in the plane by rectilinear co-ordinates, x andl y, it is necessary to let each range in value ftom - x to + x ; bWt it is sufficient to let p range from 0 to + x . 0 from o to 27r. So confining p and 0 we have but one pair of values to fix any one point; but if we let each range from - x to + , then any point having co-ordinates (p, 0) will also have co- ordinates (-pI 0 + Tr), (-pI-.. + 0), (pi-2-. + 0), and in each of these four pairs we may suppose 0 increased or de- creased by 2 n-,, n being any natural number. In polar co- ordinates with this range, to any pair of values corresponds but one point, but to any point correspond foutr ih/inities of pairs of values. 11. By reasoning quite like that of Art. 7 it may now be shown that the geometric picture of any Eq. between polar co-ordinates, as (PI, 0) = 0, is a plane curve. Every point of the curve pictures a pair of values of p and 0 satis- fying the Eq. (p, 0)=0; conversely, every pair of values of p and 0 satisfying the Eq. O(p, 0) = 0 is pictured by a point of the curve. We speak indifferently of the Eq. (p, ) = O and of the curve (p, 0) =0. 12. There are various other kinds of co-ordinates, as bi- polar, trilinear, homogeneous, elliptic; but rectilinear (called also Cartesian, from Descartes, the inventor) and polar are the most common and important. A point that may be anywhere in a plane and a pair of co- ordinates that may have any values may be said to have two degrees of freedom; a point that may be anywhere on sonic 12 CO-ORDINATES. curve in a plane and a pair of co-ordinates whose values must satisfy some Eq. may be said to have one degree of freedom; a point that must have one of several definite positions and a pair of co-ordinates that must have one of several definite sets of values may be said to have no degree of freedom. Thus it is seen that mobility in the point corresponds to variability in the pair of values. Y .1" It is to be noted that the same pair of values will in general be pictured by different points, not only in systems of different co-ordinates, as rectilinear and polar, but also in different systems of the same co-ordinates. Thus the pair (2, 1) is pictured by the point P in the system OX, 0 F. but by the point P in the system O'X', O'V Y. Conversely, in different co-ordi- miate systems the same point will picture different pairs of values. See Art. 21. EXERCISES. Assume a system of rectangular axes, also take the + X-axis as a polar axis; then, 1. Find the point (p = 2, 0 =!'r, and show that its rectangular co-ords. are (x/3,1). 2. Show that the rectang. co-ords. of (p, 0) are x = p cos 0, y psin 0. 3. Hence, express p and 0 through x and y. 4. Find the rectang. and polar Eqs. of the axes and of a circle about 0, radius 5. 13 CO-ORDINATE GEOMETRY. SUM-mary. 13. The results reached so far may thus be summed: A. A pa;r of numbers (as corresponding values of argument and function) may be pictutred geometrically by a point ia a plane. A'. A point in a plane may be pictured algebraically by a pair of numbers (the co-ordinates of the point). The point picturing a number-pair and the number-pair pic- turing a point will vary with the co-ordinate system chosen. B. An equation (or functional relation) between two symbols (as f(x, y) = 0, -(p, 6), may be pictured geometrically by a plane curve- B'. A plane curve may be pictured algebraically by an equation between two symbols (called current co-ordinates of a point of the curve). The curve picturing an equation and the equation picturing a curve will varv with the co-ordinate system chosen. Strict proof of B' is neither in place nor needed here; as occasion may offer it will be verified as we go on. 14 The doctrine based on these facts is named Co-ordinate (or Analytic, or Algebraic) Geometry of the Plane. Its problem is twofold: I. Given any algebraic form (as f(x, y) = 0), to picture it by a geometric form in a plane (a plane curve) and to inter- pret its properties as geometric properties (of the curve). II. Given any geometric form in a plane (a plane curve), to picture it by an algebraic form (as f(x, y) = 0) and thence deduce its properties algebraically. N.B. Not to repel by over-subtlety, by a number has thus far been meant a (so-called) real number. But, as the student 14 SUBJECT DEFINED. 15 mav know, there are equations whose roots (some or all) are (so-called) imaqiiary numbers, i.e., numbers involving the symbol i or V-i; as x2+y2= - 1. This Eq. is satisfied by no pair of real values of x and y; hence it cannot be pictured geometrically in the plane of the axes OX, OY, in which, e.g.. x-+y2=1 is pictured by a circle about 0 with radius 1. since every point of this plane pictures and pictures only a lair of real values of x and y. If, then, X2+y2= _1 can be pictured geometrically at all, it must be by some geometric form not in the plane of OX, o Y. The whole question of the depiction of pairs of imaginary numbers must be reserved. A real number may be defined as one whose second power is positive ; an imaginary, as one whose second power is negative ; a complex number, as made up of a real and an imaginary part. CO-ORDINATE GEOMETRY. CHAPTER II. THE RIGHT LINE. Before attacking the problem proper of Co-ordinate Geome- try it may be well to establish certain useful elementary relations. I X Y=AZ 1 1 15. Distance between two points in terms of their rectilinear co-ordinates. Let the points P,. P2 picture the pairs written beside them. Then, at once, 12 = x-1+ 2-Y - 2x2-xI, y-Y1.cos (r-o), or, = =x2-x,+y2-y,2+2x2- l Y2-Y1 -cos W- N.B. We may as well write X1 - x2 and y, - Y2 COROLLARY 1. For o = 900, cos o = 0; . rl2 =X-1-2 - XI+ Y2 - YI ;i.e., The squared distance between two points equals the sum of the squared rectauidqlar co-ordinate differences of the points. CoR. 2. If one of the points, as PI, be the origin, then x1= 0, y=0; d1= x22+y292+2x2y2cosw0; i.e., ll) INTERSECTION OF CURVES. The squared distance of a point from the origin equals the sum of its squared rectilinear co-ordinate differences plus twice their product by the cosine of the co-ordinate angle.' N.B. It will be convenient to use certain self-explaining symbols and abbreviations: as, I for perpendicular; A for triangle; 4 for angle; L for right angle; 11 for parallel; Eq. for equation ; Cd. for co-ordinate; RL. for right line; P (x, y) or simply P, or simply (x, y) for the point whose co-ordinates are x and y; P1 or (xj, ,2J) for the point whose co-ordinates are x, and 1i; and so for other subscripts, the unform use of the subscript being to limit a general symbol; also P1P2 or (x1, Y2) (X2, Yl2) for the tract from P, (x,, yj) to P2 (x2, Y2). EXERCISES. 1. If (PiOi,, (P2, 02) be two points, 8 their distance apart, show that 51 = pL2 + p., - 2P p2 COS 01 -2. 2. The vertices of a A are (2,4), (-2, 7), (-6,-8); draw it, and find lengths of its sides, for w=900, and for cw=600. 3. Draw the 4-side (quadrilateral) whose vertices are (7, 2), (0, 9), (-3, -1), (-G,4), and find lengths of sides and diagonals, for = 900, and co= 45'0 4. Find length of tract between (17, 30 11') and (19, 480 26'). 5. Find points on Y-axis distant d from (xj, .y), for co = 00. 6. Say (by an Eq.) that (x,, y,) is distant 11 from (7, -2); c =600. 7. Say that (x, y) is equidistant from (2, 5) and (-11,1), =456. 8. Find (x, y) equidistant from (2, -13), (-9, 5), (17, 23), cv900. 9. The tract (.5,-3)(22, y) is v'31l long ; find y, for w= 90. 10. When is (4, 5) equidistant from (-3, 1) and (9, - 2) 16. If the Cds. (x', y') satisfy an Eq. fi (x, ) = 0, then is thle point (Xv' y') on the curve .f; (x, y) = 0 ; if the same pair satisfv a second Eq. f. (x, y) 0, then the same point (x', y') is on a second curve f, (x, y) = 0; conversely, if (x', y') be a common, or junction, point of two curves: fA (x, y) = 0, f2(x,y) = 0, then the pair (x',y') satisfies both Eqs., i.e., 17 CO-ORDINATE GEOMETRY. f(x',y') = 0 and f2 (x', y') = 0. Hence, to find the Cds. of thejunction-points of two curres, solve their Eqs. as simultaneous. From Algebra we know that the solution of two simultaneous Eqs., one of pth and one of qth degree, involves in general the solution of one Eq. of pvqth degree; such an Eq. has pq roots; therefore, there will be in general pq pairs of values of x and y satisfying both Eqs. ; i.e., two curves, one of pth and one of qth degree, meet in pq points. Two, three, or many pairs of values may be equal, in which case there is a double, triple, or multiple common point; two or 2 n pairs may be imaginary, where two or 2n common points are imaginary, not in our plane of OX, 0 Y. ILLUSTRATIONS. 1. 3 X-7 ya 55 and 5 x + 2 y =-4 meet in (2, 7). 2. x+(Q-y)(5--x-y)=0 and x+y=7meetin (4.3) and (-2,9). 4 3. 9 x2 + 10xy + y2 = 273 meets 9 x2 - l0xy + y2 = 33 in (1, 12), (-1, -12), (4,3), (- 4, -3). 4. y2=4x, x2=4y; hence, y4-64y=O; or y(y-4)(y2+4y+16)=O; common are (0.0), (4,4), (-2-i20, -2+ i2V/3), (-2+i2-3,/-37--i2 /-3). The last two pairs cannot be pictured by points in our plane; in what sense they are section-points of the two curves in our plane cannot now be made clear. On review let the student construct the above curves. 17. Mds. of the point that divides a tract P1P2 in a given ratio l: P-2- If P(x,y) be the point, so that PIP: PP2 = 11:JU2 the n, by similar A, x-xi: X2-, = 1X ; P1 X2 + /2 XI or, = 1 + 18 TRACT-DIVISION. likvewrise, y = fIY2 + .2YI Notice the order of the subscripts. /-Lj + /'2 If either term of the ratio, as /L2, be negative, then is P'RP2 to reckon counter to PIP1; i.e., P' falls without the tract, next to 1P2. The division is then called outer. Conversely, if the division be outer, one term of the ratio, and hence the ratio itself, is negative. The formulu y-ield only one pair of values of x and y; hence, only one point divides a tract in a given ratio. Y -v. P1 P and PP2 (or P, P' and PI P2) are called segments of the tract PI P2, and P, P: P P2 (or P1 P': P'P0) is called the distance-ratio of P (or P') to P1 and P_. This distance-ratio is + or - according as the division is i .nner or onter. If P be the (inner) mid-point of the tract, then /,u = P21 .X.=x1--.v9:2, y=/1+!f2:2; i.e., the Cds. of the (inner) mid-point of a tract are the half-sums of the like Ods. of its ends. If " = - P21 PI is the outer mid-point, and x and y are infinite; the outer mid-point of a tract is at Oo on the RL. the tract is part of. EXERCISES. 1. Find the Cds. of the points which divide the tract (7, 11)(3, 5) in the ratio 2: 3, and the tract (3, 13)(- 7,i- 1) in the ratio 3: -4. See note, page 196. 19 20 CO-ORDINATE GEOMETRY. 2. The vertices of a A are (x1, y1), (x2, y2), (x3, y,); find the points that cut its medials in the ratio 2 :1, reckoning from the vertices. HINT. Take care to think as much, and reckon as little, as possible. Here, taking any vertex, we find the x of the division-point on the medial is x1 + x2 + X3: 3; . . the y is Yl + Y2 +.Y3: 3. These expressions, being sym- metric, like-formed, as to the subscripts, hold for all the medials; .-. the 3 points fall together, are one. This point is named mass-centre of the A. 3. A point P starts from (xl,y,) and moves half-way toward (x,,Y2), then turns and moves one-third of the way toward (x3, y,), then one-fourth of the way toward (x, ,y,), and so on, till at last it moves one nth of the way toward (xToy.); where does it stop The final position of P is called mid-centre or mean point of the n points. 4. The point P starts from P1 and moves over /2 of the way toward P2 then over a3 of the way toward P3, then over 4 of the PIF +e A241 + '2 + P3 way toward P4, and so on; where does it stop The final position of P is called centre of proportional distances. 5. Three vertices of a parallelogram are (x1, y,), (X21 Y2), (X3Y3) ; find the Cds. of the 4th and of mid-points of the diagonals. What do the results mean geometrically Parallel Projections. 18. The intercept, on any RL., made by two 11 planes through the ends of a tract is named parallel projection of the tract on the RL. If the planes be I to the RL., the projection is orthogonal; otherwise, oblique. Thus, OR and OD are projec- tions of OB: OR, orthogonal; OD, oblique. PARALLEL PROJECTIONS. Clearly, projections of the same tract by the same planes on i1 RLs. are equal. Accordingly, in comparing the lengths of a tract and its projections, we may suppose all the lines of pro- jection to pass through one end of the tract. Calling the tract t, its projection p, and denoting by dd' the angle from any direction d reckoned around to anv other direction d', we have by the Law of Sines, sin tl p: t=sintl: sinpl, whence p = t sin pl i.e., the projection of a tract equals the product of the tract and the quotient of the sine of the slope of the tract to the direction of pro- jection by the sine of the slope of the projection to the same direction. By odds the most important 1I projection is the orthogonal. The 4 of a tract with a line of orth. proj. is named direction- angle, its cosine is the direction-cosine of the tract. Since the 4 of a RL. with a I)lane is the complement of its 4 with a I to the plane, we have p = t cos pt; i.e., orth. proj. of a tract = product of the tract by its direction-cosine. 19. If we project on any RL. the sides of any closed poly- gon, taken in order, the end of the projection of the lIast side will fall on the beginning of the projection of thefirst; i.e., The sum of the projections of the sides of a closed polygon is 0. Hence, The projection of any side equals the negative sum of the projections of the other sides. Or, The sum of the projections of a train of tracts between two points equals the projection of the one tract between them. If the tracts projected and the RLs. they are projected on lie all in one plane, we may put projecting RLs. for projecting planes. For this, the figure illustrates the above proposi- tions. 21 CO-ORDINATE GEOMIETRY. We see A'B': AB = sin A'BB': sin AB'B. If AB'B = 90, A'B' = AB- cos B.A'B'. A'B' is the projection both of AB and of AFEDCB; the projection of BCDEFA is B'A' or -A'B'. 20. The tract from. the origin to any point is called the radius vector of that point. In the light of the above we may now define: The rectilinear cds. of a point are the projections of its radius vector on each of two axes in its plane, II to the other axis. The polar Cds. of a point are its radius vector and the direc- tion-angle of its radius vector reckoned from the polar axis. The doctrine of projections is of prime importance in mathe- matics. It is here used to treat the Transformation of Co-ordinates. 21. The Cds. of a point vary with the system of Cds. (Art. 12). To express the Cds. of a point, in one system, through the Cds. of the same point, in another system, is to transform the Cds. Several cases arise. I. To pass from rectilinear to polar Cds., the origin being the same for both. From the above definitions, or from the figure, we have at once: z: p = sin py: sin xy = sin (- 0) : sin w y: p = sin .: sill xy = sin 0: sin W; 22 TRANSFORMATION OF CO-ORDINATES. = p sin(o-) sin ( sin9 sln (a) 1, X a These Eqs. presume that the X-axis is the polar axis; if the X-axis be sloped a to the polar axis, put 9- a for 0. For co = 900, x=p cos0, y=p sin0. II. To pass from one rectilinear system to another with same origin. By Art. 19 the proj. of p, on OX 11 to OY, equals the sum of the projs. of x' and y'; .-.X=X .ln sin S and -y + .nY -7/; or, x.sin y=x'sinX y+y siny'y, xy sin xy y Sill y = XI Sill XX' + yF sin xy'. 23 CO-ORDINATE GEOMETRY. This last Eq. is got by first exchanging x and y in the Eq. above it, which amounts to projecting on 0 Y II to OX, and then exchanging the letters in the angles, it being remembered that ab -ba and sin ab =-sin ba. Angles are best reckoned from the + X-axis or toward the + Y-axis. If the X'- resp. Y'-axis be sloped a resp. fP to the X-axis, we may write x sin X = xI sin (W -a) + ye sin (W -,8), y sin f = x' sin a + y'sin fl. From these general formulae the student may find special ones: (a) For passing from rectangular to oblique axes. (b) For passing from oblique to rectangular axes. (c) For passing from rectangular to rectangular axes. The results are not so symmetric and easy to recall as the general formulae. Let the student draw the figures and inter- pret geometrically each term in each Eq. III. To pass to parallel axes through a new origin. Be OX, 0 Ythe old axes, O'Xr 0 Y' the new ones; x', y', the old Cds. of the new origin 0'. If x, y resp. x', y' be the old resp. /f / new Cds. of P, we have I /x X=X'+Xl, Y=Y'+Ym, i.e., for the old Cds. put the new Ods. plus the old Cds. of the new origin. X If we will change both origin and axial directions, we can change first either, then the other, or both at once, by adding to the ex- pressions for the old Cds. the old Cds. of the new origin. Calling these latter, as above, X1, Yi, and putting qj, q2, q,1, qua 24 'LINEAR SUBSTITUTION. for the sine-quotients in II., we get as the most general relations between the Cds. of a point referred to two systems: x =XI+ qtq2yt, Y=Y1+ql'X'+q2Y- Conversely, such a pair of Eqs. may always be interpreted as a transformation of Cds. For xi, yj may be taken as old Cds. of a new origin (or negative new Cds. of an old origin), and we can find w, a, and P such that sin(w -a) : sinl qw, sin (o)-,3): sin W = q21 etc. 22. Note that the general Eqs. of transformation are homo- geneous of 1st degree in Cds. So much might have been assumed, it being clear that a length, as x or y, can be ex- pressed only as made up of lengths. Under this assumption, by determining the values of xi, Y', and the q's, the student may now get the Eqs. already found; this is recommended as a use- ful exercise. The magnitudes x1, Yl, and the q's, are not of the same class; the latter are pure numbers, trigonometric ratios, while the former are Cds., metric numbers of tracts. A number is said to have as many dimensions as the geometric magnitude it stands for: the metric number of a length, area, resp. volume has one, two rest). three dimensions. A pure number, the ratio of two like metric numbers, is of 0th degree or dimension. Thns, 641 =_ 82_ 41 has one, two, three dimensions, according as it is the metric number of a length. an area, a volume. It is plain that any Eq. may be thought as homogeneous by thinking the numeral coefficients of proper dimensions. 23. If in any Eq. f(x, y) = 0, we put for x and y any linear functions (i.e., functions of 1st degree) of x' and y', we are said to make a linear substitution or transformation. Such a substitution may, of course, change the form of the Eq., but it will not change its degree in x and y. For it cannot raise the degree, since any term or factor of a term, as xT, will be replaced by a series of terms, none of degree higher than the 26 CO-ORDINATE GEOMETRY. rth, in x' and y'; neither can it lower it, since then by express- ing x' and y' as clearly as we can, linearly through x and y, and re-substituting, we should get the original Eq., an(d so raise the degree by a linear substitution, which is imnp)ossible. This again we might have foreseen. For the picture of f (x, y) 0 is a curve whose degree tells the number of points in which it may be cut by a RL. (see Arts. 24, 16); a linear substitution is interpreted geometrically as a change of axes; by such a change we in no wise affect the curve, hence (lo not chance the number of points in which a RL. cuts it; hence we do not change the degree of the Eq. The doctrine of Transformation of Cds. is of special impor- tance to TMechanics. Any, motion of a plane system of points may be resolved into a push and a turn. A push corresponds to a change of origin simply, a turn to a change of axial directions; a twist corresponds to a change of both. ILLESTRATION-S. 1. Transform x2'+ 14x+ y2 _-10 + 49 = 0 to 11 axes through (-7,5). We have x = x'- y=y7+5; (x'- )2+14(x'- 7) + (y+5)2_10(y'+,) +49 =0; or, ,.2 +jF2 = 25. 2. x2 - y2 = a2. Pass to axes halving the As between the old ones. We have 90 a = 450 / =135 w(i-a=45', w-,3=-45' x = x' sin 450-y' sin 450 = Y V2 y= x' sin 45+y' sin 1350 = x + ; -1/2 whence, substituting and dropping accents, 2 x =-a2. 26 LINEART SUBSTITUTION. 3. 3x2+4xy+5y2-5x-7y-21=0= .f(xy). Change the origin and turn the axes, keeping them rectangular, so as to make the terms containing the first powers and the product of x and y vanish. Putting x'+ x1 for x, y'+ y, for y, and collecting, we get 3,X2+42y'+ +y'2+ 2(3 x+ 2 y-; )x + 2(2x1-+ y, - 7,)y'+-f(X, y) = 0. If the terms in x' and y' vanish, then 3x1+2y-4 =0, 2 x,+5y1-7 =0; whence, x, 1l: 2, yi= l: 2. Hence, f(31,Y1) =f( -4)-3 +4.+54---7 -5-- -21=-24. Accordingly, on passing to new 11 axes, through (1, ), the Eq. becomes 3x'2 +4x'y'+5yl2_ 24=0. Now turn the axes through an W a; then, X and Y being new axes, XI= X COS a-y Sill a, y'-= sin a + y cos a; (3 cos a +5 sinl a +4 sinl a cos a)x2 + (3 sin a + O e -4 sin a COS a)y2 ., - .2 + (4 sin a- cos a +4 cos a - 4 sin a ) xy= 24. If the term in xy vanishes, then Sina.COSa+COSct -sinl =0l , or, - sin 2a =-cos 2 a, or, tan 2a =-2. Hence, on reduction, (4 +V5) X+(4 -V)y2 24, i 1S 58 16' 57" or - 131' 43 3"; but it is needless to find a from the tables ; it is much better to construct it geometrically. P- CO-ORDINATE GEOMETRY. We pass now to the geometric interpretation of Eqs., and naturally begin with the general Equation of First Degree in x and y. 24. This has the form lx + my + n = 0. To interpret it, let us pass to a new system of Cds. x' and y', such that X'= lx + my + n. Every point whose old Cds., x and y, satisfy Ix + my + = 0 has its new Cd. x' = 0 and clearly all such points are on the Y'-axis; again, every point on the Y'-axis has its x'= 0, and hence has such old Cds., x and y, as satisfy Ix+my + n = 0; this Y'-axis is a RL.; hence every point whose rectilinear Cds., x and y, satisfy an Eq. of 1st degree in x and y lies on a certain RL., and the Cds., x and y, of every point on that RL. satisfy that Eq. Conversely, suppose given any RL., and seek the form of its Eq. Assume it as a new Y'-axis; for all points on it and for no others x' = 0; but any Cd., x', is a linear function of the old Cds.x x and y; i.e., x' = lx + my + na; hence for all points on this RL. and for no others lx + my + n = 0. Therefore, I. The geometric picture of any Eq. of 1st degree in rectilinear Cds., x and y, is a RL. II. The algebraic picture of any RL. is an Eq. of 1st degree in rectilinear Ods., x and y. We speak indifferently of the Eq. and of the RL.: lx + my + n = 0. A convenient abbreviation for lix + ny + n is L; if 1, m, i, have any subscript, L has the same subscript, and conversely. so that Lk. x + iny +uk. n We shall also speak of the right line L, meaning the RL. whose Eq. is L=0. The above proof is the most natural, and presents no diffi- culty; but owing to the great importance of the proposition, it will be well for the student to' frame a proof fiom the figure: showing that the Eqs. of the 4 RLs. are really such as are 28 EQUATION OF THE RIGHT LINE. written by them; thus showing that by changing a, b, s one of these Eqs. may be made to fit any RL. that may be drawn in the plane. 1/ y=b / 7 X 25. The values of x and y range in pairs picturing the points of the RL. each from - X to + c ; hence x and y are called running or current Cds. -For any one RL. 1, m, n are not definite, since we may multiply the Eq. by any expres- sion we will, without changing the relation between x and y. But the ratios 1: n, n: n, n: 1 are fixed: for any one RL., different for different RLs. They are called arbitraries or parameters. Their number is apparently three, really two, for the third is but the quotient of the other two. Clearly they are not changed by multiplying the Eq. at will. To interpret them, assume any axes, and construct the RL. To do this it suffices to know two points of the RL. or one point and the direction. To find a point, we must find a pair of values of x and y satis- fying the Eq. To do this, we may assign any value, sav, to x, and reckon the corresponding value of y. The simplest value we can assign to x is 0; the corresponding point will be on the Y-axis, since only on the Y-axis are points whose x is 0. The corresponding value of y is -i: in. This, then, is the distance OIy from the origin at which the RL. cuts the Y-axis. Likewise, putting y = 0, we find the distance from the origin at which the RL. cuts the X-axis to be - n :1. The RL. making these inter- 29 CO-ORDINATE GEO-METRY. cepts, - n: m resp. - n :1 on the Y- resp. X-axis, is the RL. lx+miny+n=0. Two of the ratios are thus seen to represent negative inter- cepts made on the axes hy the RL. Denote these intercepts on the X- resp. Ylaxis by a resp. b, so that a_-n:l, b n: m. Then, on transposing n and dividing by it, the x Y= Eq. becomes - + - = 1, which is the Intercept Form (I.F.). a b g IL b The 3d ratio -l: m -- let us denote by the symbol s and name the Direction-Coefficient of the RL. Note that s is the negative ratio of the coefficients of x and y. If 0 be the slope of the RL. to the X-axis, then clearly s=sin 0:sin(W-0). Hence for w = 90, i.e., for rectang. axes, s= tan 0. Solving the Intercept Form as to y, we get y = sx+ b, which is the Directional Form (D.F.). Drop from 0 on lx + my + n = O, a I p sloped a resp. / to the X- resp. Y-axis; then p = a cos a = b cos /8. Substituting in the I.F. for a and b, we get x cos a + y cos /3p-p = 0, which is the Nornial Form (N.F..). 30 THE -NORMTAL FOR.M. 31 For w = 90, cos/3= sin a; hence, we get x cos a + y sin a -p = 0, an important special form. Hence we have the following Rules: To bring the Eq. of a RL. to the I.F., divide hy the absolute term talen to the right member. This is l)ossible unless the absolute 1)e 0; then the pair (0, 0) satisfies the Eq., and the RL. goes through the origir. - In general, if the absolute in any Eq., of any degree, betiveent Cds. be 0, the curve goes throvgh the origin. -To construct a RL. through the origin, assign either Cd. any convenient value, then reckon the other Cd. ; this pair pictulres a point, which with the origin fixes the RL. To being to the I).F., solve as to y. Trris is possible unless the coefficient of y be 0, i.e., unless y does not appear at all; then the Eq. reduces to x = a constant, - the RL. is 11 to the Y-axis. Likewise, if xr does not appear, y = a constant,- the RL. is 1I to the X-axis. To bring to the N.F., multiply by the normalizing factor F. To find F, we note that, since by hypothesis Fix + rny + Ft x cos a +y cosf --p, Ft = cos a, Fm = cos 3, Fn =-i) whence, F- -p: n. I) + n _ 2n- n cos Z = 2 sin ), + l'it Rzz since each is the double area of the A OII Hence F= sin wt): V12 2 mn 2 biV Ccos co. Before the V' we have choice of signs; we agree to take always that sign which will make the absolute negative. For v = 90, the most important case, F= I1: 2+/C +m2 CO-ORDINATE GEOMETRY. EXERCISES. Construct the following RLs., reduce the Eqs. to the various forms, for 90', unless otherwise stated: 1. 5x-3y+30=0. The I.F. is 2G + --' = 1, got by dividing by -30; - 6 10 the D.F. is y = 5x + 10, got by solving as to y; the N.F. is-- 5 x + 3 = 0, got by multiplying by F= -1: A. a=-6, b=10, s-=5:3=tanB, p3=30: 034. For w = 60, a, b, s are the same, but F= -V'3: 14, p = 15X/.3: 7. 2. 3x+7=0. 4. 3x-2y=0. 6. 2x+4y=9(w=60'). 3. 7y-9=0. 5. 3x-4y= 12. 7. Howarex +y=0 and x-y=0, x-y+a=0 and x+ y+b=0 related (i at will) Y/ A'A 26. Angle between two RBs.: 11x + my4- nl 0 'and l.,.x + my + n2 = 0. The 4 4 between the RLs. equals the a - a2 between the Is PI, P2 let fall on them from the origin. By Art. 25, Cos a, = 11 sin w: V/I12 + mr2 - 2 11 ml cos as coB a2 12 siln w( V/12 + 'M22 - 2 12m2 in s c . 32 ANGLE BETWEEN TWO RIGHT LINES. 33 Square, take from 1, extract square roots; there results, sin a, = Ml- 11 cos w: -lIm -2 11rml cos o, sin a2 = M 12 cos 1 : C (2). From applying the Addition-Theorems of Sine and Cosine there results: sin 0 = sin (a,- a2) = (11m2 - 12m1) sin w (0: COS 0 = COS (a,- a2) -= I1112 + m1 -2 - 11n2 -+ 12mI cos -W - - tan 0 = tan (a, - a2) = (11m2-12mI)sinoj :l1l2+mIm2-11 In2+ 12 II COswt SPECIATL CASES. 1. If the RLs. be , =0, .-. tan= 0, .11n2-l2m1t= 0; or, 11: In = 12:m2; or, 81-82; i.e., RLs. are I when, and only when, their Direction-Coefficients are equal. 2. If the RLs. are I, 4=90, .-. tan 4=oo, 1112 + Mlm2 - (11m2 + 12mI)cos = 0. For w = 900, 111l2 + MIMm2 = 0, or = or sl 1112+mm2= - -S ; 2 . In rectangular Cds. two RLs. are I when, and only when, their Direction--Coefficients are negative reciprocals of each other. This is the case when the coefficients of x and y in the one Eq. are the exchanged or inverted coefficients of x and y in the other, with the sign of one of them changed. The absolute term affects neither l)erpendicularity nor paral- lelism. The rectang. Eq. of the RL. through (xj, y,) I to lx +ny +n = 0 is I (Y-Y1) = III (x- x'). Why ILLUSTRATIONS. 1. The sides of a A are: 3x - 4y + 12 = 0, 5x + 2y + 10 = 0, x + 5y + 5 = 0; find its angles ( = 60). CO-ORDINATE GEOMETRY. tan A (25 -2) A,/3 23 5+ 10 - (25 + 2).- 7-3 whence = 850 47' 28". Find 02 and (A, check by 01 + 02 + ,3 180, and construct the A. 2. Find Eq. of RL. through (1,2) 1 to 3x+5y=7, and draw it, for w - 900 and 60. 27. Distance from a point (x', y') to the RL. xcosa+ycosf3-p= 0. The Eq. of a RL. 11 to the given RL. is xcosa+ycosB -p' = 0. If this RL. goes through (x',y'), then x cos a + y' cos 8 = pp. Subtracting p =p, we get x' cos ( + y' Cos -p =p'-p, which is clearly the distance sought. y x -x This result is + or -according as (x', y) lies on the outer or inner side of the RL. (the inner side being next to the origin). Hence the Rule: Reduce the Eq. to the -X.F., put for the current Cds. the Cds. of the point; the result is the metric :34 THE RIGHT LINE UNDER CONDITIONS. number of the distance from the point to the RL., and is + or -accordling as the point lies on the outer or inner side of the RL. If we put N_ x cos a + y cos -p, then N is the distance of (x, y) from N = 0. N changes sign, passing through 0, as (r, y) changes sides, passing throurh Y-= (0. Carefully distinguish between the edyression X and the Eq. IV= 0. In NY, . and(l y are Cds. of any point in the plane; in Y ) 0, ; and y are CUs. of any point on the RL. N= 0. It is this double use, 1)ewilder though it may at first, that gives the N.F. its importance. ILIJ-7STRATIONS. 1. Find the distance from (3, -4) to 4 x + 29 7. For 90', N.F. is 2 2X ___ =0; hence, the dis- tance is -3: 2V'5. X V 2-/ For = 60, the distance is- .l-. The point is on the inner side of the RL. 2. Fin(l the distance from (2, 3) to 2x +y = 4. 3. Fin(d tlc distanice from the origin to a(x-a) +b(y-b)=0. The Right Line under Conditions. 28. By Art. 25 the Eq. of the PRL. contains two arbitraries or parameters. If we hold one of these fast, for every value of the other we get a RL., and for the totality of values from -C to + o we get a family or system of RLs. Thus, in y = sx + b, holding b fast and assigning s the whole series of real values from - c to + x, we get a family of RLs.. all cutting the Y-axis b from the origin. Loosing b, assigning it the same series of values, we get affamily of families, one through every l)oint of' the Y-axis. Except this Y axis, which is common to all the families, no RL. of one family is a RL. of another: all are (lifferent IlLs. ; i.e., all possible different RLs. in a l)lane, enough to.fill the lplane, form a fanly of fairilies, an infinity of ik/inities of RLs. ; i.e., the plane viewed as full of ilLs. is doubly 35 CO-ORDINATE GEOMETRY. extended. Hence, to know a RL., we must know two things about it: what member of what family it is; and to fix a PRL. we must put it under two conditions: we must determine the two parameters in its Eq. We consider here some simplest cases. e17 I. To find the Eq. of a family of RLs. through a point (x17y l) The Eq. of any RL. is y=sx + b, and the Eq. Y1 = sXl+ b says that the RL. passes through (xl,yi). This last is not an Eq. of a line, since it contains no current Cds.. but an Eq. of condition. By its help we can eliminate s or b, better b, and get y - y= s(x - x). For any one value of s this is the Eq. of a RL. (since it con- tains current Cds. in 1st degree only) through (xl,Y,), since that pair of values satisfies it. To any slope 0 of such a RE. there corresponds a value of s; viz., s = sin 9: sin X-9, and conversely; hence, letting s range from -x to + , we get all RLs., the family of RLs., through (x1,,y). To determine a member of this family, we may impose various conditions; as, that it go through (x2y2). Hence, 36 THE RIGHT LINE UNDER CONDITIONS. II. To find the Eq. of a RL. through two points. Since it goes through both (x,yI) and (x2,Y2), it is com- mon to the families y-y,=s(x-x,) and y-y2=s(x-x2), and for it s has the same value in the two Eqs.; hence, elimi- nIating s, we get the Eq. soug'ht: J- Y: Y -Y2 = X - X1: X - X2- Let the student interpret this proportion geometrically. We may reason otherwise; thus: Be lx + my + n = 0 the Eq. of the RL.; since it goes through (x1, Yi), 1x, + my, + n =0; and, for like reason, lx. + my, + n = 0. By Lntro(dtction. Art. 14, these Eqs. consist when, and only when, x y 1 =0. X1 y' 1 Ix2 Y2 This Eq. of the RL. is equivalent to the other, and very con- venient. COROLLARY Y. _ yJ = - or I/' 1 2 Y3-Y2 X-X2 X1 Y3 is the Eq. of condition that (xI.YI), (x2, Y2), (x2,y3) lie on a RL. III. To find the Eq. of a RL. through a given point (x,,Iy) and, sloped 0 to a RL. whose Direction-coefficient is sI (W = 900). The RL. is of the family y - y, = s(x - x1) ; also, by Art. 26, tan - mlS-l=s or s= -tan 111 + mm1- 1 + SS1' 1 + SI tan + '. - = s +- tan ; (x- XI) - I 4+ SI tan 0 xx) 87 38 CO-ORDINATE GEOMETRY. CORO.LARY. If 4a = 90, i.e., if the RL. is to be I to the given RL., _!h=- (x-h) or Q1(y-y,)=m,(x-x1). Let the student solve this problem for X not = 90. EXERCISES. 1. The vertices of a A are (5,- 7), (1,11), (-4, 13); find the Eqs. of: its sides, its medials, Is through the mid-points of its sides, Is from its vertices on the counter sides; find the lengths of the last Is. 2. Find the Eq. of the RL. through (x3,y3) cutting PIP., in the ratio Al: 2. 3. Find the RL. through (5,4) forming with + axes a A of area 80. (W = 90O). 4. The sides of a 4-side, taken in order, are y- 5x, 6y + 5x = 35, 3x-y =21, 4y1+ 9x =0; find the Cds. of its vertices, Eqs. of its diag- onals, and of the junction-lines of their mid-points. 5. Find Eqs. of RLs. through the junction-point of 3x -4 y = 7 and 2 x+ 5y+ 8=0, and sloped 60 to y=4x+3. 6. Find Eqs. of RLs. sloped 300 to X-axis and cutting Y-axis 7.5 from origin. 7. Are (6,2), (7,-3), (-a,-5), on a RL. Are (3,-1), (1,2), (7, -7) X 8. Two counter sides of a 4-side being axes, the other sides are _ + Y = 1, X + b = 1; find the mid-points of the diagonals. 2 a' 2 b 2a' 2b 29. The parameter of a family of RLs. is the arbitrary in its Eq. THEOREM. When the parameter appears in 1st degree only, all the RLs. of the family go through a point. Solve the Eq. as to the parameter A, we get X=(lx+ 7ny+ n,): (1,2x+ m72y+n2), or 71x + mly +'nI-A(2x + my +2) =0, or L1-XL.=0. PENCILS OF RIGHT LINES. Whatever X be, this RL. goes through the junction-point of the RLs. L,= 0 and L2= 0, since the pair (x,y) which satis- fies both these Eqs. also satisfies L1 - XL2 = 0. Conversely, all RlLs. throuyh the junction-point of L, = 0 and L2 = 0 are of the family L1-X L2 =0. For the Direction-coeffi- cient is s = (X12 - 1l) : (mi - XAm2), and this ranges with A throturh all real values. hence, to find the Eq. of any special RL. through the junc- tion of L1 = 0 and Lo = 0, which let us call the point (L,. L2), it suffices to find the corresponding special value of A, which we may call the parameter of that RL. Thus, if the RL. is to pass through the origin, then n, - ,2 = 0, A = n1: 'a2; if it is to be 11 to the X- resp. P axis, then 11 - Al2 = 0 resp. ml-Xm2=0, X=11:12 resp. X=m1:m2. The student may substitute these values of A, and find the Eqs. of the RlLs. 30. The above is a special case of the important theorem: the curve C0 - A2 = 0 goes through all junction-points of the curves C = 0, 02= 0 (A being any constant). For any pair (x, y) that satisfies both C, = 0 and C2= 0, sat- isfies also C1-A Q = 0. This again is a special case of the still higher theorem: If. when two Eqs. are satisfied, a third is also satisfied, the third curve passes through all junction-points of the other two. The proposition is evident as soon as its terms are under- stood. Hence, it is plain that the RL. IL1 + ,M2L2 = 0 goes through the junction of L1= 0 and L = 0, since its Eq. is satisfied when the others ale. If the expression p4ILI + pA2L2 + P3L3 0, i.e., vanish identically, then the three RLs. L, = 0. L2 = 0, L3= 0, pass through a point, since the pair that satisfies two of the Eqs. must satisfy the third. Hence, if the sum of some multiples of the Eqs. of three RLs., vanish identically, the three RLs. go through a point. 39 CO-ORDINATE GEOMETRY. If Mil + 12 2L2 + pX3LI = 0, but not identically, then this Eq. imposes some condition on the symbols x and y; hence it is the Eq. of some line, and in fact of some RL., since x and y enter the Eq. in first degree only. We may go further, and say that this Eq. may be made the Eq. of any RL. by choosing the fL's properly. For lx + my + it = 0 is any RL., and we may choose the three fL's so as to satisfy the three Eqs.: '111 +PLAl2 + /313 = 1, F1m1 + jP2 in + ,x3m3 = M, IL'nl + P-212 + /L313 = n. The RLs. L1 = 0, L2= 0, L3 =0, determine a A, which may be called the A of reference, or referee A; and the expres- sions L, L2, L,, may be called the trilinear, or triangular Cds. of points on the RL. [ L, fL2L2 + W3L3 = 0. If F,, Fe, F3, be the normalizing factors of L1, L.,, L3, then FL,, F2L2., F3L3. are the distances of any point (xy) from the RLs. L, = O, L2 = 0, L3 = 0. Hence, it is seen that the triangular Cds. of a point (xiy) are certain fixed multiples of its distancesfrom the sides of the referee A. Triangular Cds. will be simplest, then, when they are the simplest multiples of the distances from the sides of the A ; i.e., when they are the distances themselves; and these they are when, and only when, the Eqs. of the sides of the A are in the N.F. ; in the N.F. we write them, 1V1 = 0, 2_ = X, = 0, where X, k_ X Cos ak + Y COS 13k -)' Hence, calling now the multipliers of the NT's v's, we have I NIA + l 2Nil + V3 i3 = 0 as the normal Eq. of a RL. in triangular Cds.: Ala 2, N3. 31. Here the point has three Cds., the N's, while we know that two are enough to fix it. But, since the Eq. is homogeneous in N's. we can at once reduce the number of Cds. to two by dividing through by one, as V, and treating the quotients IVl : 2, 40 TRIANGULAR CO-ORDINATES. -Y;7: N7 as the independent Cds. In general, the triangular Cds. of a point are bound together by a certain constant relation. For if we suppose, as we may, that the origin 0 is within the referee A, and put r1, r2, -3 for the tracts between the vertices, A for its area, then for every point (x, y) or (Ab, eN,, NX3) we shall have r+2P- A have al --VI + -fi2 a'+ 73=2 A. This is clear at once when P is within the A, and is equally clear, on proper regard of signs, when P is without. By multi- plying this Eq. appropriately, we can express any constant homogeneously through the triangular Cds. of any point. If the origin ) be without the A, it suffices to change the sign of one of the Y's. A general test of whether three RLs. L, = 0, L2 = 0, Ls = 0, go through a point, is found in the Determinant 11 ml Il i, 1., In2 1, 13 im13 S It which, by Introduction, Art. 14, must vanish when the three Eqs. 11x+mly+nl=0, 12X + rn2Y + n2 = 0, 13x +-4- 3Y + U13 = 0, consist, are all satisfied by the same pair (x, y). andl 41 42CO-ORDINATE GEOMETRY. So too must V 1' 1i't vst, =0, "1 II 2 F 1 3 when through a point go the RLs. V1el + lV2'lX2 + V3N3 = 0,3 VI"XI + 1'2"x2 + 1'3 1Y3 = O. VI IN + 121 II2 + V31 W3 = 0. Geometric Interpretation of A. 32. A family of RLs. through a point we may call a pencil, the point itself, the centre of the l)encil; the two RLs. through which the others are expressed, the base-lines of the pencil. If the Eqs. of the base-lines be in the N.F., then IV - XN2 = 0, for any special value of A, is the Eq. of a RL. of the pencil. ".4 Io j '3 X \ a- X,+2i= o Here A = 1: Nj = ratio of the distances of any point of the RL. N, -X X = 0 fromn the base-lines N, = 0 and x2 = 0; or, A = ratio of the sines of thle slopes of the RL. N, - AX V = 0 to the base-lines IXN = 0 and x2V = 0. If we call the angle containing the origin, and its vertical angle, and the lines in them, all inner, the others all outer, then we see that for inner lines the distances are like-.sign ed, as from PI, P3, and .'. A is + ; for outer lines the distances are uldike-signed, as from PF, P4, and . . A is - 42 ABRIDGED NOTATION. If we call the inner side of a RL. also the - side, the outer also the + side, then angles and sines of angles reckoned from the + resp. - side of a RL. are themselves + resp. -; we agree to reckon angles from the fixed to the variable RL. 33. For the inner resp. outer halvers of the angles at (N1, N2) the distances are equal and like- resp. unlike-signed; hence, A = + 1 resp. -1; hence, the halvers are N1 - IV, = 0 resp. N1 + N2 = - Hence, to find the Eq. of the inner resp. outer halver of the Ws between two RLs., form the difference resp. sum of their Eqs. in the N.F. N.B. Of the two equivalent Eqs., N= 0, we have taken that as the N.F. in which the absolute term is negative. But this test fails, and with it the test of which is the + and which the - side of the RL., when the absolute term is 0, i.e., when the RL. goes through the origin. In this case, we may agree to take always the term in y, or always that in x, as positive. If we make the first agreement, as is common, the + side will lie next to the + Y- axis; for, holding x and letting y increase, we must gret a + result in the left member of the Eq. 34. This Abridged Notation (a single letter N stauding for the left member of the Eq. of a RL. in the N.F.) with its imme- diate outgrowth, tile system of triangular Cds., yields a method of great strength and beauty; we have space for but a few simple ILLUSTRATIONS. 1. The inner halvers of the 4s of a A meet in a point. For, be the origin within the A, and N1 = 0, N = 0, I3 = 0. its sides, then the sum of the Eqs. of the inner halvers, N1 - -ZV = 0, N2 -N3 = 0, O. - N1 = 0, van- ishes identically. 2. Two Quter halvers and the third inner halver meet in a point. They are iV, + 0t 1, + N1=, N1-Y-Y = 0; multiply the second by - 1, and sum. 43 CO-ORDINATE GEOMETRY. 3. The Is from the vertices of a A on the counter sides meet in a point. The 4s being A,, -2, A3, the I from Al on N, = 0 is 1X cos A2 - N3 cos A3 = 0; 1)ermlute the inidices, and sum. 4. If through each vertex of a A be drawn a pair of rays like-sloped to the sides meeting there, the junction-lines of the junction-point of each couple next to a side and the counter vertex ineet iii a point. A2A1 A Be N1 = O. N, = 0, N3 = O the sides, A,, A_. A3 the counter vertices, '/I, 72, y3 the slopes of the pairs at the ver- tices. The Eqs. of 124A1' resp. A3A,' are X1 sin(Y2+A.)+ Xsiny2= 0O resp. i'Vl sin (y3 + A:,) + ZN sil 73 = O. The Eq. of -4,A,' is, therefore, 1N1sin (y2 +A2)i+ s3 ill 2-A[N 1 inly3 +A3 +iNX2sin y3 t=0.- At A1 both XN and 2N1 are 0; hence, X=sin 2+A2: sin y+ A3. . sin y2 sin y3+ A3- XN- sin y3siny+A2= is the Eq. of A1,1A'; Y I N sin y:-siny, + 1- - ill-siny sin y3+A3= 0 is the Eq. of A2A12'; 44 POLAR EQUATION OF THE RIGHT LINE. N2 sin 7 sin y2 + A2- N1. sin y2 sin y, + Al = 0 is the Eq. of A3A37. Multiply by sin -,, sin y2 resp. sin y3, and sum. If A3 A1' meets A1A3' in A2", show that A2A2P, A3A3", AIAI'' meet in a point. 5. If through each vertex of a A be drawn a pair of rays like-sloped to the halver of the 4 at that vertex, and if any three of these rays meet in a point, so will the other three. If N1 sinl A2 +y2-N3siny2=0, N2sinA3+y3-N1siny3 =O, N3 sin A1 + y7 - 12 sin 7y = 0, be any three, the others are got by simply exchanging the N's in each Eq. The value of any N in one of these Eqs. is not in general the same as its value in another; but if the first three RLs. meet in a point, the first three Eqs. must consist for one set of values of the N's; hence we may transpose, multiply, cancel the product N-N2N3, and get as a condition that the three RLs. meet in a point, the Eq., sin A2 + 2 -sin A3 + -y.- sin Al + -y, = sin y2 -sin 73- sin yl. Now, clearly, exchanging the N's in each Eq. affects not this result. 6. If the three Is from the three vertices of one A on the three sides of a second A meet in a point, so do the three Is from the three vertices of the second on the three sides of the first. Be the sides of the one A N1 = 0O X, = 0, 3 = 0, the sides of the other NV1= 0, _N7t=0, N3O =0; then N2 cos N17 N3 = - co os N I', IVy' cos N1 N3'= N37. co s N12VN are a pair of corresponding Is. Let the student complete the proof. Polar Equation of the Right Line. 35. If the 1 from the origin on the RL. be sloped a to the polar axis OD, we have at once, 45 CO-ORDINATE GEOMETRY. p cos(O - a) =p Let the student get this Eq. by transforming to polar Cds. from the N.F. a- x EXERCISES. 1. Reduce p = 2a see + ) to rectang. Cds. 2. Find where p = a see (a-X ) and p cos 0(-) = 2 a meet, and under what angle. 6 2 3. Find where the 1 from the pole 0 meets the RL. through (Pi, 6,), (P2, 02) Miscellany. 36. The Eq. of the RL. through (x2, y2), (x3,y3), in the N.F. is x y 1 sino ./y2-y3-+x2-x32+2y11 - x3 2-x3. COsw(= 0. x2 Y2 1 X3 Y3 1 The left side is the distance from (x, y) to the RL., the V is the length of the tract (x2. 1) (x3,13); hence, their product, (x,,yl), being written for (x, y), or i Y,, 1 n W, X3 Y3 1 is the double area of the A whose vertices are (x1,yl), (x2,y2), (X3, y3) . 46 AREAS. 37. To find the area of a polygon whose vertices (taken in order) are Note that when one vertex of a A is the origin (0, 0), its (loulble area is I I, (X1, Y-), (X2,Y2) being the other vertices, the axes assumed rectangular. Now from the origin, assumed within the polygon. draw rays to the vertices cutting it up into 7 A ; the double area of the polygon then is ,I Xo21 X2 X3, +- + iX. X. n + X1 IYI Y2 2 3 J 1 Y I Y.Yli Now s5p1)pose the origin moved to (x0,y0); the double area of the first A changyes to ' i+ X0, X, + Xo XI' X22 -+I X1 X0 X( X., + x (x) /I'+Y0o Y2+YOI /IY' /2' 1 1 Yo I I Y SY2 y o yo The first of these Determinants is the original primed, the last vanishes, the sum of second and third is XO(Y2l'- Y') + YO(X1'-2') ) If we operate likewise upon the other Determinants, we shall clearly get the original Determinants primed, while the sum of the multipliers of x0 and yo will each vanish identically. Hence wherever the origin be, the double polygonal area is Xl X, + 1 X,, Y + +1 X: 1I Y1 Y2 1 3 Y3 Y i Y the primes being dropped. We thus learn that the sum of the areas of A with one corn- maon vertex, the other vertices being vertices of a( 2 polygon, is thc area of that polygon: a theorem important in mechanics. The practical rule is to write the X's and y's iii order, each in a hori- zontal row, repeating the first of each as the last; thus, X1 X2 , X - n X1. Yl Y; Ya ...3n Y,' 47 CO-ORDINATE GEOMETRY. and take the cross-products of the consecutive pairs with oppo- site signs. The algebraic fact that the sign of the area is + or - according as we take one or the other set of products +, corresponds to the geometric fact that we may compass the polygon clockwoise or counter-clock-wise. For axes not rectangular, multi ply by sin w. 38. To find the area, of the A whose sides are L= 0, L, = 0, L3 = 0, form the Determinant of the coefficients: 11 ml nl 12 m2 n9 1, 13 m3 n. and denote the co-factor of anv element, as 11, by that symbol primed, I,'; be (x,,y,) the junction-point of L2 = 0, L3 = o. Then we have x1 = I,': nl', y, = rnz': n,', and so for (x2, Y2), (xe y3). Form the Determinant of Art. 36; replace the column of 1's by nl': nt, 2t : '/2, n.'f : nf3; set out the divisors of the rows, nl', n2', n.s'; there results 2A i 1,'m2n3 sin w: n1" n2' nf3 =, M1n, 2 3sinX lo 1,n2l' 12n)'31 '13MI I If two of the RLs. be 11, a factor of the denominator van- ishes, the area is so; if the three RLs. meet in a point, the numerator vanishes, the area is 0. 39. To find the ratio P1: P2 in which the tract (XI'y,) (X2,Y2) is cutt by the RL. Ix + my + n = 0. We have at once, on can- celling the normalizing factor F, PI PI 1xa + my1 + n P2 P2 - 1X2 + Iny2 + n When the section is inner, /Al and iU2 are like-signed, pi and P2 unlike-signed ; when the section is outer, the P's are unlike- signed, the p's like-signed; hence the - signs in the above Eq. 48 TRANSVERSALS. If the RL. go through (x, y3) and (x4, y4), its Eq. is x y 1 1=0; hence, /Cl: J2=- X1 Y1 X2 Y2 1 X3y31 Y'Yl x 3y31 X4 Y4 1X4 4 Xy4 14 P1 P.P4 P2 P3 14 40. By the help of these expressions for ILI: 2 we can now prove two fundamental Theorems on Transversals: I. Three points on a RL. and the sides of a A cut its sides into sepnments whose compound ratio is -1. II. Three RLs. through a point and the vertices of a A cut its sides into segments whose compound ratio is +1. Be ABC the A, L _ lx + my + n = 0 the transversal, cutting the sides at A', B', C'; be kL= Ixk + my + n, the result of putting xk, yk in L for x, y. Then, by Art. 39, 49 CO-ORDINATE GEOMETRY. BA': A'C= -2L: ,L. CB': B'A1= -3L:, L. AC': CB=--,L:2L; the product of these ratios is - 1, which proves I. Be (x, y) the Cds. of P, through which are drawn IAA", BB", CC". Then B1" l 2,X2J 1 j2lX I: A"G c . 1 :x I CB" X3 Y` 1 X1 YIi 1 =-.,,XI2 1 1: X2 Y2 1, B.- j fl y J C"B X/., 1 x3 / 1 In the product of these ratios the determinants of the denom- inator are those of the numerator with two rows exchanged; i.e., with sians changed; hence, the product is + 1, which proves II. Note the order of letters and indices. EXERCISES. 1. Find the areas of the A whose vertices are: (2,1), (3, -2), (4,- 1); 12,3), (4,-5), (-3,- ;). Ans. 10; 29. 2. Find the area of the tetragon: (1, 1), (2, 3), (3, 3), (4, 1). Solution. 1 34 1 3+6+3+4-2-9-12-1--8; .. Ans. 4. 3. Find the area of the A: 2x-3yy=5, 4x+a5y--7, y-2 x9. Cross Ratios. 41. A set of points on a RL. is called a range or r-oz.w'; a set of RLs. through a point, a pencil; each (half-) RL., a ray. The RL. is the carrier of the row or range; the point, the centre or carrier of the pencil. 50 CROSS RATIOS. The distance-ratio of Ig2 to P1 and P. is P1 P2: P2g (Art. 1 7); and so the distance-rrtio of 14 to P, and P.) is P1 P1 P4 P . The ratio of the distaince-ratios of twco points of a rmnqe to two other )oints of that range is called the cross ratio of the range, and is written P.P23 P4 =P1 P2. PI P4. P2 PI P4 P3 Observing signs, we see at once, I" I 'P, P4 P P p ., P P P2- P4 P1 or. simply, 1 2 3 4= 1 2 4-, 2 3 -4 1 which is the neatest way to write it. Clearly, the order of the points is essential. The alternates in position, as 1st and 3d, 2d and 4th, are called conjugates; the consecltites are non-conjugates; as 1st and 2d, 2d and 3d, 3d and 4th, 4th and 1st. Any one has one conjugate, two non-conjugates. The four symbols may he permuted in 4!, = 24, ways. Any permutation or order may be got from any other, as 1 2 3 4, by one or more exchanges, which will be either of a pair of conju- gates or of a pair of non-conjugates. To exchange a pair of conjugates inverts the ratio. Be 11234r= 23.41 14.32 23.1 then 11 4 3 2t= 1 4 3 2 2 3 4 1= - 4i3 -2 1 12 :- 34 r To exchange a pair of non-coija gates takes the complement of the ratio to 1. For, exchange (say) 2 and 3 ; then 51 52 CO-ORDINATE GEOMETRY. 11 3 241=-1 3 -2 4 (1 2+2 3).(2 3+34) 3 2-4 1 32 -4 1 12 -3 4 23 (1 2+2 3+3 4) 3 2.- 41 3 2-4 1 2 12 34 23 14 =+ = 1-r. i23.4 1 32 4 1 Hence, to exchange two pairs of conjugates, or two pairs of non-conjugates, keeps the cross ratio unchanged. Hence there are four permutations for which the cross ratio (or CR.) is the same; hence, too, there are six permutations, or six sets of four permutations, for which the CRs. are different. BY invert- ing and complementing to 1, we get these values: 1 1 1 1 r 1-r r 1-r The circle of these six values is complete; any amount of inverting and complementing will reproduce one of them. The last two man I)e written v-i r and r r - The ratio of the sines of the :s into which a third ray cuts the 2 between two other rays is called the sine ratio of the third ray to the other two; the ratio of the sine ratios of two rays to two others is called the cross ratio of the rays, and is written sini 2 sini 4 sin 1 2-sin3 4 S1 234t= - sin2 3 sin4 3 sin2 3-sin4 1 when S is the centre of the pencil, and the rays are 1, 2, 3, 4. When the rays of a pencil go through the points of a range, the two are conjoined. CROSS RATIOS. 53 From the figure, we see at once p-1 2=S1-S2-sinl 2, p-3 4=S3-S4-sin3 4, p-1 4= S1.S4-sin1 4, p.2 3=S2-S3-sin2 3, whence, 1 2 3 4t=S81 2 3 4 ; or The CIRs. of a conjoined range and pencil are equal. Hence, holding the carrier of either range or pencil fast, we may move the other at will. If 1 2 3 4=-1=S81 2 3 4t, the tract between two conjugate points, resp. W between two conjugate rays, is cut innerly and outerly in the same ratio by the other conjugates; the range resp. pencil is then called harmonic, and either pair of conjugates (points or rays) is called harmonic to the other pair, while the foumrth element (point or ray) is called a fourth harmonic to the other three in order. If in an harmonic range one point halve innerly the tract between two conjugates, the other (its conjugate) must halve the some tract outerly, i.e., must be at o. If in an harmonic pencil one ray halve innerly the X between two conjugates, the other (its conjugate) must halve the same outerly; i.e., must be I to thefirst. 54 CO-ORDINATE GEOMETRY. Clearly, both these propositions may be converted by simply exchanging the words innerly and outerly. Also, if of two conjugates in an harmonic range one be at OC, the other halves innerly the tract between the other pair; and, if one pair of conjugate rays in an harmonic be at R. A, they halve innerly and outerly the W of the other pair. EXERCISE. Prove by similar A that 1 2 3 4 = 1' 2' 3' 4'. 42. Cross Ratio of four rays given by their equations. Be the base-rays, XI = 0, N, = 0, a pair of conjugates, and be Y2 = LNI-XA213O=, ON4==1-X4NY3=O, the other pair. Denoting the rays by their proper indices, we see at once, from Art. 32, A-2 -sinl2 A sinl 4; hence I1 2 3 4,=)2:24. sin 2 3 sin 4 3 If L, =0, L3 = 0 be the base-rays, fi, f3 their normalizing factors; then L1 = O, L1-A2L3=O, L3=O, L1-X4L3=O may be written in normal form; thus, ALI = f. A LL-JA L3 = -0 f3 f3L3 = 0, fLI-A4 f L3 =0; whence, by the above, we have again, I1 2 3 4 =X2: X4. If L'= 0, L"= 0 be base-rays; then to find the CR. of any rays, L'-X L"-0 L'-X2L"= 0, L'-A3L"=O, L'-AXL"=O, take either pair of conjvgates as base-rays, say first and third, and express the other pair through them; thus, L'-AL"=Lj, L'-A3L"=L3; whence, finding L' and L", and substituting, we find the other pair are CROSS RATIOS. L1 A-A2 -O L. IWA 4 L. = 0; hence, 11 2 3 4__AA3_-_ A2 -A3 A4 -A1 43. If X2: A4 =-1, the pencil is harmonic; that is, spe- cially, L'= O, L"= O, and L'-AL"=O, L'+AL"1=O are two harmonic pairs. The four rays (1, 2, 3, 4) of the pencil (LI, L") are har- monic when, and only when, A1-A2 - A-A4. 1. A2 - A3. A4 - Al i.e., when A1A3-_. A1 +A3 . A2+A4+A2X4=O- 44. If A, B and P, Q be two pairs of harmonic points, then for 1' midway between A and B, Q is at x ; as P moves out from its mid-position toward B, Q moves out from its mid- position at x into finity toward B. As P falls on B, so does Q. The same remarks hold when A is put for B; also, when rays SA, etc., are plut for points, it is necessary only to note that for Q in x SQ is 11 to the carrier of the range. If the CR. = + 1, clearly a pair of conjugates (rays or l)oints) must fall together. 55 CO-ORDINATE GEOMETRY. 45. If L, = O, L2= 0 meet at S, and Li'= O, L2'= 0 at S", any ray, L, -AL2 = 0, of the onie pencil is said to correspond to the ray, L1'-XL21= 0, of the other; the penl- cils are called homographic. Since the CR. of two pairs of rays depends only on their parameters, the A's, it follows that: Th7e CR. of twvo pairs of rays equals the CR. of the correspond- ing pairs in any other homographic pencil. By eliminating A we get L, L'- L.,L1'= 0; this, then, is a relation between the Uds. (x,y) of any junction-p)oint of two corresponding rays; since the L's are of first dle-)ree in x and y, this Eq. is of second degree in x and y; hence, The junction-points of pairs of corresponding rays in two homographic systems (pencils) lie on a curve of second degree. EXERCISES. 1. N1 =0 and N, =0 enclose an 4 a; find the Eqs. of the Is to them through their intersection (VN, N2). 2. I-ow do 1x-A X2=0 and AN, -. -0 lie in the pencil (N., IV,) 3. Show that the transversals from the vertices of a A to the contact- points of the inscribed resp. escribed circles meet in a point. 4. Nl+ xNV=0 and N1 + A, N, 0 being taken for base-rays in N1 + Ax N = 0, what is the value of h when N1A'-0 is the same as N1\ + A,1N2 + I + (1A2xN2) =0 5. Is 3x-l0y+4=0 of the pencil 5x-7y+ 3 +A(2x+3y-1)=0 If so, express its Eq. through 7x- 4y+2=0 and 19x+14y-4=0 as base-rays. 6. The CR. of a pencil is r; three rays are L, = 0, L2 = 0, L,+xL2 = 0; whatis thefourth E.g., 11x-2y+7=0, 3x+5!1=6, 17y=2x+25, r =9:8. 56 INVOLUTION. 7. Four rays of the pencil 5x-7y+3+A(2x+3g-1)=0 are llx+2ty=O, x+5=13y, 7x+2=4y, 15x+8y=2; find the CR. First find A1, A2, A2, A4i 8. The inner halvers of the Xs of a A are cut harmonically, each by the centres of an escribed an(l the inIscribed circle. 9. Are these two pairs of rays harmonic: 3y+4x=25, 2.y-3x=11; 7y-2x=47, 10x-y=3 10. Find the fourth harmonic to rl1-AL,2=(, IL-A7x2=0, L1-A'L2=0. 11. Findl the harmonic conjugates to each of the three rays in the pencil I1-+Aj-JO" I,'+A2 L"=O, L'+AL"=0. E.g., L'=2x+311-5, L"=7x-211+1; takeasrays, l3y-33= 10, 23x-3y=2, 9x+y=4. Involution. 46. To any one pair of conjugates there is an infinity of har- monic pairs of conjugates ; for the Eq. that says the pair (A,, K,) is harmonic to the plair (A. K) 2 AK -A + K - A1 + cl + 2AIhK= (1) is clearly fulfilled for an oo of values of A and K. If a second pair (A2, K2) be also harmonic to the same pair (A, K), then must hold the second Eq., 21kK - A + K A2+ K2+ 2A2 K2O=. (2) 'These two Eqs. are linear in AK and AX+ K; hence they are both satisfied by one, and only one, pair of real values of AtK and A + K ; this piair will yield one, and only one. l)air of values of A and K, which may le real or imaginary; hence, It ciy pencil there is always one, and only one, p)air of rays, reaul or ibnaginary, harmonic to each of two given, pairs. 47. If, now, there be a third pair (A3, K3) harmonic to the same lpail (A, K), then must hold the third Eq., 2XK-A +K A3+ K3,+2A3K3= O. 57 (3) CO-ORDINATE GEOMETRY. These three Eqs., (1), (2), (3), will be fulfilled by the same pair of values (A, K) when, and onlv when, 1 A + KI A1 K1 0, 1 9+ K2 A2 K2 1 A3 + K3 A3K3 which is therefore the Eq. of condition declarin(g that the three pairs of rays. L'- Al""= 0. L'- K1L"= 0; L'-A2L"= 0, V- K.L"= 0; L'-A3Lf= 0, L'--KL"= 0, have a com- most harmonic pair. Three or more pairs of rays hactrmonic erach to the samne pair form an Involution. The commoii harmonic we may call focal rays. Any transversal is cut by an itvolbtdion of r(qJs in an involution of points, whose foci are the section-points of the focal rcay,t. Pairs of conjugate points correspond to pairs of conjugate rays. The foci and a pair of conjugate points form an har- monic range. The mid-point between the foci is called Centre of the Involution. 48. The product of the central distances of a pair of conjv gate points is a constant: the squared half-distance between the foci. F C F' P! P For FPF'P' = FP F'P': PF' P FF =-1; if FF'= 2 c, CP= d, CP'= d'. then (c+d) (d'-c) =-(c+d') (d-c), or dd'= C2. If P fall on F or F', so must P'; i.e., foci are double points, and focal rays are double rays. If P and P' fall on the same side of C, dI and cl' are like-signed, c2 is +, c is real, the foci and focal rays are real; but if P and P' fall not on the same side of C, d and d' are unlike-signed, c2 is -, c is imaginary, the foci and focal rays are imaginary. The foci fix the centre, an(d so the Involution; but, bv Art. 46, two pairs of conjugates fix the common harmnonic pair; i.e., the focal rays resp. points; hence, tivo jpairs of conjuyates (rays resp. points) fix an Inrolution. 58 DIAGONALS OF A FOUR-SIDE. 49. If '= 0 L"= 0 be focal rays, L'-k XL" = 0, where s = 1l, 2, 3, 4 , any four rays, then their conjugates are L'+AXL"=O (Art. 43), and the CRs. of the two sets are plainly equal; hence, In any Iuinoltdion the CUs. of any four elements and of their coojizyjttes a(re equal. Coin. If six points (or rays), A, A', B, B', C, C', be in invo- lntion, then ABCA' = A'B'C'At, which tests whether the third pair be involved with the other two. Cross Ratio and Involution are of greatest import to Higher Geometry and Mechanics. Minuter treatment were out of place in this elementary work; but the foregoing, it is hoped, may excite the reader's interest, and incite him to further pursuit of thle subject. A single illustration is added in the proof of the familiar theorem:: Each, diayonal of a four-side is cut harmonically by the other two. CID Be AB, AC'. A'B, A'C the sides, AA', BB', CC'. the diagonals. Then CDC'D't = A CDC'D'Y = CI'B'A't = D CI'B'A' = C'IBA't = A1 C 'IB A' = C'DCD' t = 1: ICDC''t; .-. I CDC'D't = 1. 59 60 CO-ORDINATE GEOMETRY. But a CR. can = 1 only when two rays fall together; hence, Q.E.D. Conduct the proof for the other diagonals. Equations of Higher Degree representing Several Right Lines. 50. If L, = O, L2= O, -, Ln =O be Eqs. of n RLs.. their product L1I L2 .- L. = 0 will be of nth degree in x and y, and will be satisfied by all and only such pairs of values of x and y as reduce some factor, as Lr, to 0; i.e., the Eq. L, -- Ln. = 0 will picture all and only such points as lie on the RLs. L1=Q, L =(0, --, L.=0. If all the L's be homogeneous in x and y, so will be their pro(luct, and not otherwise; but then all the RLs. go through the origin; . . n RLs. through the origin are pictured by an Eq. of nth degree, and homogeneous in x and y. Conversely, such an Eq. pictures i RLs. through the origin. For, on division by x1, the Eq. takes the form o+ Y + 1 2 +..+ c + Y. X X- XX. This Eq. of ntlh degree in the ratio y: x has n roots, sl, s,. 3s, s, , and may be written C - - S1 "2 k =O' This Eq. is satisfied when, and only when, a factor equals 0: u as -- s =0, or y = skx. But this is the Eq. of a RL. through the origin, and there are n such factors. The RLs. are real or imaginar!, separate or coincident, according as the roots, the s's, are real or imaginary, unequal or equal. 51. If the Eq. be not homogeneous in x and y, we may test whether it be resoluble into factors of first degree in x and y PAIRS OF RIGHT LINES. by assuming such a resolution, forming the product of the assumed factors, and equating coefficients of like terms in x and y in the assumed and given expressions. For an Eq. of nth degree there must then hold i ' 2 I Eqs. of condition among LI 1.2 its coefficients. Our immediate concern is only with the Eq. of second degree, which may be written thus: kx2+2 hxy +jy2+ 2yx+2fy+c=O. (1) Instead of the tedious general method we may employ the following, esl)eeially as its incidental results are useful: IPass to 11 axes through the section-point (x,Y,) of the two RILs. which the above Eq. is supposed to represent. This is done by putting x + xl for x, y + Yi for y; on collection there results kx2I+ 2hxy+jy2+29'x+2f'y+c'=O, (2) where g'= kx1 + hey1 + a, f'= hxl +jyj +f, c' = kX12 + 2 hxjy1 + jy,2 + 2ygx + 2fy1 + c. This result is got by reasoning thus: terms not containing x, or y, are found by supposing x1 = 0, yj = 0, which gives the original expression terms not containing x or y, by supposing x = 0, y = 0, which gives the original expression with x1, y writ- ten for x, y ; terms containing a subscribed and an unsubseribed letter can result only from the original terms of second degree, appear doubly, and are symmetric as to the subscribed and unsubscribed letters. This reasoning gives the following as the result of the substitution kx + 2 hxy +jy2 + 2yx + 2fy + kxl2 + 2 hxly1+ jy112+ 2yx, + 2fy,+ 2kxx, + 2h(x1y + yj')+ 2jyy1+c = 0; and this collected gives Eq. (2). It is important that the stu- dent thoroughly master this argument. Now, if Eq. (1) is a pair of RLs. through (xl,y,), Eq. (2) is a pair through the origin; hence, (2) imust be homogeneous of second degree in x and y; hence, terms of lower degree must 61 62 CO-ORDINATE GEOM.IETRY. vanish; hence, g'= 0, f'= 0, c'= 0. But c' may be written (kx1 + hY1 + g) x + (hxl +jY, + f)yj + (gxl +fyl+c) = 0; and the coefficients of xi and y,, being g' and f', are 0; hence, so is /x1j+fyll+c; hence, between xl and Yi must consist the three Eqs., kx1 h+y1+y=, hxl+jyl+f=O, gxl+fyl+c=O. The condition of this consistence is k h g =0. h j .f g fec This Determinant is named Discriminant of the Eq. of second degree, and is denoted by A. Hence, A = 0 is the condition that thee Eq. of second degree represent two RLs. The co-factor of any element of A will be denoted by the cor- reslponding capital letter. The Cds. (x,,y,) of the section of the RLs. may be found from anV two of the above three Eqs., and are x=G: C, y,=F: C. If C, or kj - h2, be 0 x1 and Yi are finite, the RLs. meet in finity; if C= O, 1 and y, are oc, the RLs. meet in oo ; i.e., the RLs. are II. 52. It is to note that changing the origin but not the axial directions does not change the terms of highest, i.e., of second, degree. To find the direction coefficients, it suffices to factor kx2 + 2 hxy +jy2, h hk k or Y2+2xy+Ty2. J) J If the factors be y - sl, y -s2x, then sh+s2=-2 ., S kS2= whence, sl-s2 = 2 5A-kj EQUATION OF BISECTORS. S1 = h-+ VhJ :j. 82 = h -t Vt _ Aj: j- Hence, by Art. 26, if the RLs. enclose an tan(k=2/ h-/j-jsinco: 5k+j-2hcosc),, or tan 2 = 2_j U +j; the RLs. are I when k+j-2hcoso=0, orwhen k+j=O, if w=90, and are imaginary when h2 - kj 0. 53. To find the pair of RLs. halvinrg the 4s between the pair Ax2 h 2hxy+jy2=O, or y-slx=O and y-s2x=O. Brought to the N.F., these Eqs. are Y-S1X: V +8i2=0, ,J-s'x: 1 +s,22=0. Their sum resp. difference is the Eq. of the outer resp. inner bisector; and the product of these is Y - 1I +82-y - s2X: + S' = 0, 2 2 2 or 812 82 Y -2._2 2-S1 sI-s2xy-s12_-s22. x2 = O. Divide by , - s., replace SI + 82, S812 out of Art. 52 ; whence, y2 + h- x. - _2 = O, the Eq. sought. Note that this pair of halvers are alwtays real, though the pair kx2 + 2 hxy +jy2 = 0, whose Ws they halve, may be imaginary. 54. The general Eq. of second degree is not hb:nogeneous in x and y; we may make it so by multiplying the terms of lower degree by fit powers of some linear function of x and y. So we get COORDINATE GEOMETRY. kx2+2hxy+jfi + 2gx+2fy+c=0, (A) Ix +my=1, (B) kx2 + 2 kry +jy2 + (2ggx + 2fy) (7x + my) +c(lx+my)2=O. (C) By Art. 50, (C) represents two RLs. through the origin, and by Art. 30, they go through the intersections of (A) and (B). Note the method of this article, as it solves the problem of finding the. Eq. of the RLs. through the origin and the section- points of a RL. with a curve of any degree. EXERCISES. 1. What RLs. are pictured by the Eqs.: (a) x-c.x-d=o; (f) 4y'-15xy-4x'=O; (b) xy = 0; (g) y2 + 13xy _JX2 = 0; (C) X2-4y2=O; (h) y2 -2xy + 3x' = o; (d) x2-y2=0; (i) y2-2xyseco+X2=0; (e) X2-5 xy+4 y2=; (j) X2+Xy-6y2+ 7x+31y-18=0 2. Find the 4s of the above pairs, and their bisectors. 3. For what values of A do these Eqs. picture RLs.: (a) 12x2-1Oy+2y2+ 11x-5y+A =0; (b) 12X'AXY2Y'11-5112=O; (bc) 12 x2 + 6xy + A2 - 6 5 y + 2 = O (c) 12X2+86Xyq+Ay2+6)X+6.y+3=0 4. Show that the RLs. joining the origin to the section-points of 3x2+5xy-_3y2+2x+3y=O and 3x=2y1 arel. 5. N, =O, ,2 = , N3-= 0 being sides of a A, find the RLs.: N,+ N2+ 3=0; N1- N2+ N3=0; N1+N, -N3=O; -1+ N2 + N3=0. Rectilinear Loci. 55. The sum total of positions to tchich a point is astricted by .some geometric condition is called the locus of the point. 64 THE RIGHT LINE AS LOCUS. To express the condition through constants and the current Cds. of the point is to find the Eiq. of the locus. For doingr this no fixed rule can be given; each problem or class of problems must be solved for itself. In general, the expressions will be made more simple by choosing the axes with special reference to the figure of the lproblem, but more symmetric by avoiding such reference; sometimes the one advantage, sometimes the other, will out- weigh. Often the point is fixed as the junction of pairs of corresponding lines in two systems of lines; the common par- ameter of the two systems must then be eliminated. Thus, if the point lie on the curve F(x, y; p) = 0, and also on the curve O(x, y; p) =0, by giving p any value we may find a pair of values of x and y satisfying both Eqs. ; i.e., we may find one position of the point; but by eliminating p we find a relation between every pair (x, y) which satisfies the two Eqs., whatever p may be; i.e., we find a relation between the Cds. of the point in any position; i.e., we find the Eq. of its locus. There may be more than one parameter; the number of Eqs. needed is, in general, one greater than the number of parameters. EXERCISES. 1. A point moves so that the sum of its distances from the sides of an 4 is constant; what is the point's path If 11x + mly + n, = O. 12X + m2y + n2 = 0 be the sides of the 4, then 1x + ma/ + nl + 12x + 121' + n' s is the locus, a RL. + i12 V'/122 + M22 The sides being axes, at once x + y = s: sin w. Draw the RL. 2. The sum of a point's distances from n RLs. is s; find its path. 3. The ratio of a point's distances from two RLs. is 1,a u2; find its path. 4. From side to side of a given RL. are drawn 1I tracts; a point cuts them all in a fixed ratio; find its path. 5. The ends of the hypotenuse of a right A move on the rectang. Cd. axes; where does the vertex move 65 CO-ORDINATE GEOMETRY. ( 1 ' l HA x= t . cos (60-a) = b cos (600 =b(1+v'3.tana) :2 (tan a- y:x= -tan (60o- a) = (tan a-V .-.tan a= (x-3+Y) : (X - YV) ; whence (y+ s 3 ') + x 7-s) =2b, a RL. sloped 60 to the given RL. Dri 6. Given one RI. 4 fixed, another turning about a fixed point F; find the locus of P cutting the junction-line of the sections of fixed and moving sides, in the ratio X -': PT HINT. Take the fixed sides as axes, express OA resp. OB through x resp. y, and project on OF. 7. Tracts are drawn from the origin to any point of a RL., y _ sx + h; find the locus of the vertices of equi- lateral As constructed on the tracts. X Take the slope a of a tract to the X-axis as par- ameter, and proceed thus: y 1= 7,: (1-s cot a), t = yI: sin a = b: (sin a - s cos a); D a): (sin a - s cos a) - s); (I) (+ 4./3tan a) 1w it. 8. Find the locus of the intersection of Is to the sides of a A, cutting the sides at points equidistant from the ends of the base. Take the base as X-axis, x either end as origin; the Eqs. of the sides are y = sIX, y=s2x+b; take the dis- tance d as parameter; the 66 EXERCISES. 67 Eqs. of the Is are y --= +cl, y- --x C2; express cl, C2 through S12 S2S (1, so we find the IAs are sty + x = d11 , S2Y + x + =d V/1 + S2" I lence eliminate (d. 9. Find the locus of the intersection of Is to the sides of a A through the points where they eaut 11 to the base. 10. On the sides of a given 4 are laid off from its vertex tracts whose sum is s; Is to the sides through the ends of the tracts meet at P. Where is P 11. Given a vertex, the directions of the sides, and the sum of the sides, of a parallelogram; find the locus of the opposite vertex. 12. Find the locus of the intersection of RLs. joining crosswise the Joints where pairs of rays through a fixed point cut the sides of a 2 fixed 4. If the axial intercepts of the rays be a,, b1, a,, b2, the cross lines .i re B1 ( + = (t-2b P if (x,, yt) be the fixed point, el b, (12 1') Form the difference of the first, and also of the second, pair of Eqs.; their quotient gives _=__ /l, the locus sought, which is seen to be the same X/. ' for all positions of P1 on the RL. OP1. In the following problems, about a A ABC, take AB as the + X-axis of rectang. Cds. Five noteworthy points of a A are: mass-centre (intersection of niedials), orthocenttre (intersection of altitudes), centre of vertices (or of circumscribed circle), centre of sides (or of i-scrieled circle), intersection of transversals from vertices to contact-points of opposite eseribed circles. 13. Find the loci of these points for similar A with a fixed 4 A. If r be the radius of the circumscribed circle, the sides of the A are 2r . sin A, 2r . sin B, 2r. sin C; the Cds. of the points (in order) are: 68 CO-ORDINATE GEOMETRY. x=, 'r(sinlA cosB+2cosAsinB), y= 2 rsinA.sinB; x = 2r cos A sin B, y = 2r cosA cos B; x=rsinA+ B, y -rcosA-t- B; 4 B A A+B A .B AB. x = 4r cos Sl sin 2B- CosA B. y=4rsin- sin-. Cos 2 2 2' 2 2 2 B _' x = 2r (sin A-sin B + cos A. sin B), y = 8r sin- sin . 2 2 Eliminate the parameter r from each pair; it is seen that as the A swells, the five points move out on five RLs. through the centre of similitude. 14. Find locus of centre of sides; only A and its counter-side a constant. 15. Find locus of orthocentre, only B and b constant. 16. Find loci of first, second, third, fifth points, only B and a constant. 17. Find loci of the first four points, only A and b constant. 18. Find loci of the first four points, only A and c constant. 19. Find loci of third and fourth points, only C and b constant. 20. Find locus of mass-centre when c is constant in size and position, while C moves on the RL. 'y = sx + n. 21. Given an 4 of a A and the sum of the including sides; find lo(us of P cutting third side in a fixed ratio. In problems about tracts of changing length and direction, measured from a fixed point, polar Cds. are recommended. 1 ] 22. Chords through a fixed point of a circle are produced /Z \\ till the rectangle of chord and chord produced is constant; find locus of end of produced chord. Take the diameter through the fixed point as P polar axis; then OP= p, OC / =d cos 0, OC. OP=-k'; pcosO A, = V: d, a RL., as is also clear from similar A. 23. Two tracts whose D lengths are in a fixed ratio D enclose a fixed 4 at a fixed point, and the end of one moves on a RL.; how moves the end of the other FAMILIES OF RIGHT LINES THROUGH A POINT. 69 Take as polar axis the I from the fixed point on the fixed RL.; be OA and OP the tracts, and OP: OA =r. Then p =" Sec (t-a), a RlL. D)r4Iw it. 24. From a fixed point 0 is drawn a ray cutting two fixed RLs. at A and A', and on it P is taken so 4iat OP is the harmonic D mean of OA and OA'; find locus of P. OX l)eing taken as polar axis, the Elqs. of the fixed RLs. DA and DA' are 1 p= I cos 0 + mr sin 0, 1 p = /' COS 0 + ni' sin 0; these Eqs. hold for the same 0 Mhere p is OA1 resp. OA1' in the first resp. second. By hypothesis, 2: OP 1: OA + 1: OA'. Writing p for 0!', we get 2: p = (I + l1) cos0 + (m + m') sin 0, the Eq. of a RL. through I. 25. (Generalize Ex. 24 by taking n instead of 2 IlIs. through D. 26. Given base an(l (lifference of sides of a A; at either end of base is (Irawn a I to the conterminous side; find locus of its intersection with the inner lhalver of the vertical angle. Families of Right Lines through a Point. 56. Thus far in each problem have been two conditions, enabling us to determine the two parameters in the general Eq. of a RL. Had there Veen but one condition in the problem, the result would have contained one parameter undetermined, and so would have represented a family of RLs. Should a parameter appear in a result linearly, then, by Art. 29, all RLs. of the family would pass through a point. When the parame- ters appear linearly, both in the general Eq. of a RL. and in some Eq. of condition, all RLs. of the family go through a point; for, by help of the Eq. of condition, we may eliminate one parameter anld leave the other appearing linearly. Such is the case, e.g., when the parameters are the current Cds. x', y' of a RL. ; for then they fulfil some linear condition, Ix'+ my'+ n = 0. 70 CO-ORDINATE GEOMETRY. EXERCISES. 1. The vertical ; and the sum of the reciprocals of its sides are given in a A; find the Eq. of the base. Take the sides as axes; then, the reciprocals of intercepts of the base on the axes being I and m, the Eq. of the base is Ix+ n y = 1; also, / + i = c. Hence the base turns about a fixed point. Find the point.' 2. If, as the vertices of a A move on three fixed RLs. through a point, two sides turn about fixed points, the third turns about a fixed point. X3 A ' Take two of the fixed P.Ls., as OA,1, OB, for axus; tehn y = sx is the third RL. 0C. Be (x1, y), (X2, Y2), the fixed points, (x', y') the third vertex C; then y'=sx', and the Eqs. of AC and BC are (X1 - x') Y - (Y - sX') X + Xt (y - sX1) = 0, (X2 -X) Y-(2-sXr) x + X (Y2-sx2) = O. Hence, find OA, OB, and form the Eq. of the third side AB, x (Y2 --s'J ): (Y2 -SX2)- Y (XI1-X/ ): ( I - SXJ = Xt. The parameter x: enters this Eq. linearly; hence AB turns about a a point. Find the point as the intersection of two base RLs. by solving the Eq. as to xH. 3. All RLs., the sum of proper multiples of the distances of n fixed points from any one of which equals 0, form a family through a point (the centre of proportional distances of the points). Be (xk, yk) one of the points, juk the proper multiplier, x cos a + y sin a -p = 0 one of the RLs. Then, by hypothesis, L- n :r4XkCo5 a-- ,L yk ksina -p Uk.= Between this Eq. and the Eq. of the RL. eliminate p; on division by Cos a (which, and sin a, may be written before the summation sign 2) tan a appears, as parameter in the result, linearly. By proceeding as directed in Ex. 2, the fixed point is found to be ( 2 /4 k:Ok, 4Yk:yk2 k) - THE CIRCLE. CHAPTER IlI. THE CIRCLE. I3efore treating the general Eq. of second degree, it may be well to treat a special case of great importance. 57. BIy Art. 15, if r be the distance between (x, y) and ('XI YO), theen ,xz,+oxw, + 2 x .-. CO G-/leo= r2- () Iold( r and (x,, yi) fast, letting (x, y) vary; (x, y) will keep r distant from (x1, ,x); the sum total of its positions is called a Circle, with radins r, centre (x,, y,). Ilence (If) xy is the rectilinear Eq. of a circle. For - = 90 the important rectang. E1q. is X1, Y1 x xi2 + y _ 2= (I) For x1 = 0, yi = 0, i.e., for the central Eq., the simpler forms are x2+y2+2xycosw=r2 (II') atncl x2 + y2 = (II) 58. A circle is known completely when are known its radius (iii length) and its centre (in position). Since in (I') resp. (I) radius and centre are expresse(l generally, it is clear that (I') resp. (I) is a general form to which any Eq. in oblique rcsp. rectang. Cls. of any circle may be brought. We note that the coefficients of X and y2 are equal: each =1, and the coefficient of xy is 2 cos resp. 0. Divided )by k the general E' q. of second degree takes the form I 1 CO-ORDINATE GEOMNETRY. 1+2-xy+y'+2 + 2-f- + =0 (III') Lk k k k k Accordinglv, if this be the Eq. of a circle, then must j: k = 1 or j = k, and h: k = cos w or h = k cos . These, then, are conditions necessary that the Eq. of secon(d degree picture a circle. They are also sw-fficient, for where they are fulfilled, the three arbitraries x1, yl, r, may be so chosen as to satisfy any set of values of the three coefficients, 2g 2f c k' k' k Two problems may now be solved: 1. GiWen centre and radius, to.form the Eq. of the circle. Substitute in (1') resp. (I), and reduce. 2. Girea the Eq. of the circle, tofind centre and racdius. Divide the Eq. by the coefficient of x2, equate the coefficients of x and y, and the absolute term, to their correspondents in (I') resp. (I), and so firA x1, YI, r. EXERCISES. 1. The centre of a circle is (3,-4), radius 6; find its Eq. 2. Find the Eqs. of the circles whose centres and radii are (0,0), 9; (7,0), 3; (0, -2), 11; (-4,17), 1. 3. Can these Eqs. picture circles, and of what radii and centres: 15Xl2+ 15yl+ 15xy-90y-45=0; 3x- 3xy + 3g2 - 9x + 12y -10= 0 4. Find centres and radii of 7x' - 7y2 + 49x + 84y + 14 = 0 and 5x'+ 5.y2+25x- 15y+20=0. 59. As is clear on comparing (I') and (III'), the general Eqs. for determining xj, Y1, r, are 7'2 DETERIMINATION OF THE CIRCLE. x, + y cos 0) = - 9: h, Y, + Xi cos o = -f: A X12 + y12 + 2x1yj cos -_2 =C:k; or, for rectana. axes, more simply, = -- q: Ic, y1=-f: k, x12 + Y2 _ A c .2 = (g2 +f2_ kc): 2. The equivalent forms, in which the coefficient of X2 + y2 is 1, X - X12 + y - - 2 = 0, or 2+y22g+2fy+ =0, (III) may be called the 5Normal Form (N.F.) of the rectang. Eq. In the N.F. x1= -, y1=-f, r2=g2+f2_c; i.e., the C(ds. of the ceitre of a circle are the negatice half-coefficients of x and y ine the N.F. of its rectang. Eq.; the squared radius is the sumn of their squares less the absolute term. The circle is a real, a point-, or an imaginary, circle, accord- ing as Y 2+.2-kc0,. or =0, or 0. The Cds. of the centre do not contain the absolute term; hence, if this alone change, the centre does not change; i.e., circles whose Eqs. differ only in absolute terms are concentric. If the absolute term be 0, the curve goes through the origin. 60. The Eq. of a circle contains three arbitraries, x1 Yj r, or J .f, c. Hence, three conditions are needed and enough to fix a circle. To find a circle fixed by three conditions, express these through Eqs. and thence find the arbitraries. Thus, fitud the circle through the points (1, 2), (3, -5), (-2,1). Since each Cd. pair satisfies the general Eq. X2 + y2 + 2gx + 2fy + c = 0, 73 CO-ORDINATE GEOMETRY. we get 2g+4f+c+5=0, 6g- lOf+c+34=0, -4q+2f+c+5 =0; whence, finding g, f, c, and substituting, we get 23(I+y2)-2!9x+87y- 520= 0. Or, by Determinants, more neatly, thuns: If the circle x, + y2 + 2gx + 2fy + c = 0 goes through (xi, yD), (X2, Y2), (x3, y,,), then oy" + y,7 2gx, + 2fyk + c = 0 for k = 1, 2, 3. These four Eqs. consist for the same values of g, f, c when, and only when, X22 + xjXy 1 =0, XI12 + !y 2 X2 y 1 X3/2+ X3 y 1 which is therefore the Eq. sought. Clearly, it is also the con- dition that four points, P, P1, P2, P3, lie on a circle. EXERCISES. 1. Find the circles through (0,0), (a, 0), (0, b); (a, 0), (- a, 0), (0, b). 2. If (x,, y,) (x2, y.) be ends of a diameter, the Eq. of the circle is X - Xi X -x. + 1 Y -Y!z2=0, or, for oblique axes, X - X - X2 + YY - Y3Y2 + X-Xl'Y2 + y- -X - 2Y-Y cos W= 0. N.B. The following familiar propositions in the Theory of the Quadratic Equation are here recalled once for all: Be C2x2 + 2 CIx + C0, = 0 the general Eq. of second degree in x, 1 and 2 its roots; then 74 DETERMINATION OF THE CIRCLE. 75 (1) r+r2 =-2C1. C2, r, .2= Co: 02. (2) For C,=0, one root=O; for C, =0 and C, =0, b)oth roots = 0. (3) For Q2=0, one root = o; for C2=0 and C,= 0, both roots = r. (4) For C, = 0, r = - r2; i.e., the roots are equal, but unlike-signed. (5) For C12 = C, C2, the roots are equal and like-signed. 61. To find where the axes cut a curve of second degree, equate y resp. x to 0 in the general Eq. ; so we get kX2+ 2yx+c=O resp. jy2 + 2.fy + c = 0. The roots of these Eqs. are the intercepts on the X- resp. Y-axis. (They are equal, i.e., the X- resp. Y-axis meets the curve in two consecutive points, i.e., is tangent to the curve when, and only when, i2= =c resp. f2 = hcc. See Art. 64.) Conversely, given the intercepts on the axes, to find the Eq. of the circle. Suppose c =j = 1, and be a1, ac2 resp. b,, b2 the intercepts on the X- resp. Y-axis; then 2g= - (a + a2), 2f=-((b1 + b2), c = a, a2 =: b, b- Let the student interpret these Eqs. geometrically. Hence, X2 + y2 _(a1 + a.2) + 2fy + (j ua2 = 0, X2+y2+2yx-(b1+b2)y + b6b =0, resp. x2 + y2 ( a. +2)x-(b1 + bk) y + I (a1a,2+ bb2) = 0 are equations of a circle in terms of its intercepts: on the X-axis, on the Y-axis, resp. on both axes. N.B. Of course, only three intercepts can be assumed at will ; then the fourth follows froin al a. = b, b2- CO-ORDINATE GEOMETRY. EXERCISES. 1. Wheredotheaxescut X+ +y2+5x-y-6=0 2. A circle touches each axis a from the origin; find it. 3. Find the equation of a circle through (0,2), (0, -4), (5,0). 4. Find the equation of a circle referred to a tangent, and a chord through the point of touch. We have a1=b,=b2,=, a,= +2rsinw; hence, x2 + y2 + 2x1y cosw 2rx sin= 0. If the chord be a diameter, w 900; hence the important form X2 + y 2 rx = 0. Verify geometrically and explain the double sign. Polar Equation of the Circle. 62. Be d the tract from the pole to the centre, a its inclina- tion to the polar axis OD, r the radius. Then the equation sought is p2-2 dp cos (O-a) + d= i=. The product of the two roots, pl = OP1, P2 = OP2, is con- stant, and = cit - r, a familiar theorem. The Circle in Relation to the Right Line. 63. By Art. 16, a line of first decree (a RL.) cuts a line of second degree (as a circle) in two, and only two, points. CIRCLE AND 1.IGHT LINE. For, in the equation of second degree, put for y (say) its value sx + (say) taken from the equation of first degree; so we get an equation of second degree in x; its two roots are the x's of the two points common to the two lines. If 't = (mn - Y7n + f12): (12 + M2), andl v = -/ 1 (in gm+fj2)2 (2 gln-12c-n2) (12+,n2) (12+In2) and (X1,' Yi2) be the common points of the RL. and the circle lx+ my+it= 0, x2+y2+2gx+ 2.f+c= 0, the student may find (lower index going with lower sign) n Mn = __+ _ (n 1 i), 1=-(u v), )lut lhe will not find any l)leasure withal. To shun such labo- rious reckonings and suet] unmanageable formulas, we have Recourse to special forms of the equations. Thus, the I'L. y = sx + b meets the circle X2 + y2 = r2 in X12 =1s2 Vl )/( + 6`5: 1+ r ) Y1, = 5b .S j-2(I + 82u) - 24 01+ S24 The RL. x cos a + y sin a = p) meets the circle X1 = p COS Si ili -a/ r -P _)2 Yi = 1) sin.a : COS a -\r2l_ X2+y2-=2 in (B) These pairs are real and different, or equal, or imaginary; i.e., the common points are real and separate, or consecutive, or imayinary, according as in (A), r2(l+ s2)-b2 is 0, or =0, or 0; in (B), 1-2 _p2 is 0, or = 0, or 0. 64. Real and separate points present no difficulty; imaginary points have no existence in our plane, and are to be treated 77 (A) CO-ORDINATE GEOMETRY. further on; but consecutive points it is essential to his progress that the student understand clearly no0w and here. Coincident points fall together, have exactly the same posi- tion, and are distinct only in thought. Thus, we may think of the intersection of two lines L, L' as L nlmade up of two points fallen together, and we may call it P or P'according as we / think it belongiing to L or to L'. Consec- utive points become coincident and in a particular way: by nearing each other on the same definite path (curve). Thus, if C be any curve, P and PI any two points on it, we may think P as fixed and P1 as taken at will nearer and nearer P; or, what comes to the same, we may think P1 as moving nearer and nearer to P along C. Be L a RL. through P and P1. As P, nears P, L turns about P, and the position of L is fixed completely by P1. As P1 falls on P, L turns into some position TI Now P anid PF, thoulght siml)Iy as coincident, cannot fix a IbiL. for they form but one point, anl through this one point a RL. _ mav be drawn in any direction. A // \\ But P and P1, thought as consecu- / L'\\\ tive points (of C say), do fix the T' position of L; for, as consecutive \ points of C. they become coincident AtF by nearing each other along C. At every stage of this becoming- coincident, L turns into a definite position, as P1 nears P; and at the end of it, the being-coincident, as P1 falls on P, it is left turned into a definite position T. It is specially to note that the being-coincident of P and P1 has no power to fix the posi- tion of L; it is only the particular way of their becoming coinci- dent that fixes it. As P,' becomes coincident with P (or P') in some other way: by nearing it along some other curve C', the RL. L' turns into some other position T'. Looking at the algebraic side of the matter, we find the 18 CONSE CUTIVE POINTS. likeness perfect. The two pairs of common roots (Xzi,Y), (x2, 1y) of the two Eqs. y = sx + b, X2 + y2 = r2, are equal if r2 = b2 (1 +s); but not simply are they equal; they become equal in a particular way: not by b and s passing at random through any one of an infinite number of series of pairs of values ul) to that l)air which makes r2 = b2: (1 + S2) , but lV their p)assing through that 1)articular series each one of whlose pairs satisfies the two conditions, Y2 = sx2 + b and X2,2 + y= r2. The whole series of pairs of values of s and b being fixed, the last pair, which makes r2 b2 : (1 +s 2), is fixed; i.e., the last position of the RL. y=sx+b through the consecutive p)oints (x1, Y), (X21 Y2) of X2 + y2 = 2.2 is fixed. Again, we see it is not the fact of being coilcident, but the way of becoming coincident, which is significant. We further see that the concepts, coinci(lent points and consecutive points, are not in themselves complete; we think, though we do not always sav : two coincident points of two curves; two consecutive loints of one curve. 65. In the light of the above, we may now define a tangent to a curve as a RlL. through tutwo consecutive I)oiflts of the curve. Where the points fall together is called point of touch, contact, tangenicy. If for "RL." we put "curve," the definition still holds. We also see that the al(gebraic condition that a RL. and a curve (or two curves) be tangent is, that two pairs of common roots of their Eqs. be equal. Hence, if y = sx + b touch 2 + y2 = -2 b2 = 2(1s + 2) or, The ilL. y = sx b v I + .2 touch7es the circle X2 + y2 = r2 for all real values of s. This so-called magic equation of thle tangent determines it by its direction (s), not by its point of touch, and is useful in problems not involving this lpoint. So, too, if x cos a + y sin a = p touch X2 + y2 = r2, r2=p2. Hence, the Cds. of the point of touch are XI =rCOSa, yi=rsinfa. 79 CO-ORDINATE GEOMETRY. Substituting for cos a, sin a, p in x cos a + y sin a =p, we get T7he RL. xx1 + yyj = r2 touches the circle x2 + y2 = r2 at thepoint (x1,y1) of the circle. This equation determines the tangent hy its contact-point (x1,y,), and is useful in problems involving that point. 66. The doctrine of Chords is so of a piece for all curves of second degree, that it is deemed best to state it here at once in full generality. We know bow to find the intersections of a given RL. with a given curve; the converse would be to find the RL. through given intersections with the curce. The general method is this: Combine the general Eq. of a RL. through two points with two Eqs. which say the given points lie on the given curve; the result will be the Eq. sought. This tedious general method we may replace by special methods in special cases. Such is the following method of Burnside for the curve of second degree: Form an Eq. whose terms of degree higher than the first shall cancel (hence, it will picture a RL.), and which. shall be satisfied by the Cds. of the points (x1,yl), (x2,y2), only when these points lie on the curve. Such an Eq. formed after this prescription is kx2+ 2hxy+jy2+ 2gx+ 2fy+c = k(x-x1) (x-x2) +2h(x-xx) (y-Y2) +j(y-y1) (Y-Y2). This, therefore, is the Eq. of a secant through (x,,y1), (x9,y2) of the curve of second degree. The condition that this secant become a tangent is, that (x,,y,), (x2,y2) fall together. ..x2 + 2 hxy +jy2 + 2gx + 2fy + c 80 EQUATION OF THE TANGENT. or, after expanding, cancelling, transposing, remembering that kx12 + 2 hxl y1 + jgy2 + 2gxl + 2fy1 + c = 0, finally, kx1x+h(xly+ylx) +jyly+g(x,+x) +f(y1+y) +c=O (D') is the Eq. of a RL. tangent at (X1, Yi) to kxx+h(.ry+yx) +jyy+g(x+x) +f(y+y) +c = . (D) The Eq. of second degree being written thus, the Cds. appear in pairs, and we get the Eq. of the tangent by substituting for the first current Cd. in each pair the corresponding Cd. of the point of touch. These general Eqs. of secant and tangent include all special cases, and are here deduced once for all. EXERCISE. Form by these methods, then simplify, the Eqs. of chords and tangents of the curves Xy = VI y2 = 4px X2 +y2 =r2; ( - a)2 + (y _b)2 =r2. 67. From any point (xe, y') may be drawn two, and only twvo, tangents to a curve of second degree. For, write F(x, y; x, y) = 0 for Eq. (D) of Art. 66; then the Eq. (1)') of the tangent is F(x,, ye; x, y) = 0; and clearly F(xr. Yi; x. ) = F(x,y; 1, ye). If the tangent at (x1, y1) go through (x', y'), then F(x1, y1; x', y') = 0; and, since (x1, y1) is on the curve, F(x1, yi; xA, Yi) = 0. In the first of these Eqs. xa, y, x', y' all appear linearly; solving as to y1 (say), we get y, = an expression linear in x1, x'. Substituting this in F(x,, y1; x1, y,) = 0, which is of second degree in x1 and y,i we get an expression of second degree in x1. Solving this, we get two values of xl, to each of which, since F(xj, y1; x'. y') = 0 is linear in xl, Y1, corresponds one, and only one, value of y,; hence, there are two, an(l only two, pairs of values (xl, Y,); i.e., two, and only two, points of tangency. 81 82 CO-ORDINATE GEOMETRY. Of course these pairs may he real and unequal, or equal, or imaginary ; accordlingly, the tangents will be real and separate, or coincident, or imaginary. T1'le second case, of coincidence, arises when the point (x'., y') is on the curve; for, since the Eq. of a tangent through a point of the curve has been found uni- versallv (see Art. 6'), there is hut one such tangent, which may, however, be thought as two fallen together. To tell when they are real, when imaginary, we may reason thus: 68. .Any curve, F(x, y) = 0, bounds off all points of the plane for which F(x, y) 0 from all points for which F(x, y) 0. Y x For, assign y any value, say y,; then, as x varies, F(x, y) will become 0 only where the RL. y = y, cuts the curve F(x.y) =0; but, as, and only as, x passes through a root x, of F(x, yj) 0, F(x, y,) changes sign. We may call the sides of the curve plus resp. minus. In the Eqs. of the curve and the tanglent, (D) and (D') , let us relplace 1 by v resp. v,, so as to make them homogeneous in x, y, x1, yl, v, v,. They then take the forms k-xx + h (xy + yx) + jyy + g(xv +vx) +f(yv+ Vy) + "V= 0, (E) xl xx h(x.y + yj x) +4-jy y + g(x v + vlx) +fAyiv + ray) + cviv = 0. (El) 82 INSIDE AND OUTSIDE. If, now, we proceed as sketched in Art. 67 to find x1, we shall get a result of the form AX1+Br1= V+, where A, B, C are functions of x', y', v', k, h, j, U, f, c. Since it=- (ix'+ hyl'+gv'), + (fi'.x'+NfcvYv') , (hx'+jY fV') 2 we readily see that C will be homogeneous of tenth degree in all the arguments: of fourth degree in x', y', v', and of sixth degree in k, h, j, A, f, c. [Consider that in replacing Y, the parentheses ( ' ( ) w, ( )3 will be squared, which squares will be again squared in completing the square (in x, y,), as will the coefficients k, h, etc.] Now, if (x', y') be on F(x,y; x,y) = 0, the two values of x1 are equal; but then C= 0; hence, F(x', y'; x', y') = 0 makes C= 0; i.e., F(x', y'; X', y') is a factor of C. 'Now, since .' enters F(x', y'; x', y') in sec- ond degree, but enters C onlv in fourth degree, it follows that F(xv', y'; x', y') cannot appear in C in higher than second degree. If it appear in second degree, we may extract the second root, and write V C= F(x', y'; x', y') A-/. Here 1 cannot contain x', y', or v', since each enters F in second degree, and each entered C in only fourth degree; hence R is a function of k. It, j, g, f c only, and that of second degree. Hence, whether C be real or imaginary will depend only on the sign of R, that is, only on k-, h, j, f, g, c, not at all on ', y ; that is, not at all on the position of the point (x', y') from which the tangents are drawn. Hence, tangents from all points will be either all real or all imaginary. This is so for any curve of second degree, hence for the special curve, circle. But now we know that some tangents to the circle (from points outside the circle) are real, while some (from points inside) are imagi- nary. Hence. F(x',y'; x', a') cannot enter C in second degree, but only in first degree. F(x', y'; x', y') changes sign at every point or the curve F(x, y ; x, y) = 0, and no other 83 CO-ORDINATE GEOMETRY. function of x and y does. Hence, C changes sign along the same curve, is plus for all points (x', y') on one side, minus for all points (x', y') on the other; hence, tangents from all points on one side of the curve are real, from all points on the other side are imaginary. The side on which the tangents are real, we may call the outside; the other, the inside. 69. The Eq. F(xl, yi; x, y)--O, of the tangent to F(x, y; x, y) = 0 is symmetric as to x and xl, as to y and yl, a fact of highest import to the whole theory of curves of second degree. This import we shall now in part develop. Thus far, the point (x1, yj) has been taken on the curve; the query is natural, What does F(x1, yi; x. y) = 0 picture when (x1, y,) is not on the curve To answer it, suppose tangents drawn from (xl, Yi) touching the curve at (x2, Y2), (x3. y3). The Eq. of one is F(x2, y2; x, y) = 0. Since it goes through (xl, Y,), F(xa. /12; x1 Yi) = 0. But this Eq. also says that F(x, y; x1, y,) = 0 goes through (x2, y2); by like reasoning, we show that it goes through (x3, yS); hence, F(x y; x1,i Y1) =O or F(xl, V1; X. Y) = is the RL. through the tangent points of tangents from (xS, q1) to F(x,y; x,y)=0. Such a RL. is named polar of the pole (x,, Yj) as to F(x,y; x,y)=0. Since the equation of condition F(x1, yj; x2, y2) = 0 may be read either (x1, yj) is on the polar of (x2, y:), i.e., on F(x, y; Y:, Y2) = 0 or (x2, y2) is on the polar of (x1, y,), i.e., on F(x1, y,; x, y) = 0, if one pole be on the polar of a second, the second is on the polar of the first: or, if one polar pass throughli the pole of a second, the second passes through the pole of the first. 84 POLE AND POLAR. 85 Two poles, each on the polar of the other, or two polars, each through the pole of the other, are called conjugate. Hence, if a point be on each of a system of polars, i.e., be their intersection, the poles of each polar will be on the polar of the point; or, as a RL. turns around a point, its pole glides along the polar of the point; or, as a point glides along a IRL., its p)Qlar turns about the pole of the ILL. If we convert the definition of the polar of a point, we shall get a definition of the pole of a RL.: as the intersection of the tangents to F(x, y; x, y) = 0 through the intersection of the RL. and the cutve. If, now, this IlL. turn about a point, its pole will glide along the polar of the point; hence, the polar of a point is the locus of the intersection of the pair of tangents to the curve through the intersection with the curve of a RL. through the point. 70. These two definitions of polar are equivalent. The first yields a geometric construction only when the pole is outside the curve, for only then are the tangent points real. If the point be inside the curve, we may still draw through it two chords of the curve, and draw the two pairs of tangents through their ends; the RL. through the two intersections of the tangent pairs will then be the polar sought. When the pole falls on the curve, the Eq. shows the polar becomes a tangent through its pole as point of tangency. Hence, the tangent is to be thought as a polar through its own pole. It is carefully to note that the terms pole and polar are mean- ingless without reference, express or implied; to a curve of second degree, which we may call the referee. The notions of pole and polar are still deeper inwrapped in the notion of curve of second degree. as what follows may show. 71. Be P : P-2 the ratio in which F(x, y; x, y) = 0 cuts the tract (xi, Yl) (x2, y2) ; then the Cds. / P1I2 + P2X1 IA'Y2 + P2Y1 /-Li + /42 gII1+1L2 CO-ORDINATE GEOMETRY. of the section-point must satisfy F(x, y; x, y) = 0. Substi- tuting herein and arranging terms, we get 112 F(x2, y2.; x2, Y2) + 2IMo2F(xi, yi ; x2. y2) + 2F(xl,,y; x, y,) = 0. (G) [N.B. Reason as in Art. 51; the result must be homoge- neous in all and P2, and symmetric as to indices 1, 2; the eocffiCient of ,u2 resp. 2u we get by supposing P2 respt. pi to be 0; the coefficient of 1p2 must be double and symmetric as to the indices 1, 2.] This quadratic yields two values of the ratio Pil: P2' say r' and r". If (x1, Yl) and (xby2) be conjugate (each on the other's polar). then F(x1. il; y2,y9) = 0, anD(l then r'= - r" ; i.e., the tract (X1. Yi) (x., 12), from pole to polar, is cut by F(x, y; x. y) = 0 innerly and outerly in like ratio; i.e., is cut harmonically. Hence, any tract from a Poole to its polar is cut harnionically by the referee. Hence, once again, we may define polar thus: The locus of the harmonic conjugate to a fixed point, the other pair being section-points with the referee of chords through the point, is called the polar of thefixed point (as to the referee). Thus is justified the use of conjugate in Art. 69. P.P X If, now, there be given five points of the referee, we may construct it with the ruler only, thus: Through P1 and P2, P, 86 POLAR AS LOCUS OF An FOURTH HARtMONIC. and P4 draw secants meeting in I; on them find fourth har- Jnonics 1I, III conjugate to I; HIP1 is lpolar of I. Draw 1P2 cuttilng JIll' at HI". Then is 1'P, the fourth harmonic to P5, the other pair being I, II", a point of the Referee. As we may choose four out of five in five ways, and join each four by twos in three ways. we may thus construct fifteen sixth-points. Recombining these tweutv. we may r o onl to construct any number of points of the Referee. 72. In Eq. (G) drop the subscript 2; then the roots r', r" are the ratios in which F(x, y; x, y) = 0 cuts the tract from (xI, yj) to (x, y). When these ratios are equal, the tract is cut bv the curve in two consecutive l)oints; i.e., the RL. through (x,1.. /) and (x, y) touches the curve; but when the roots are equ al, F (xl, y, ; xj, yj - F(, J; x, Y) =F(xl, yj ; . 1)2 In this Eq. (x, y) is any point on a RL. through (x1. Y,) tan- gent to F(x, y; x, y) =0; hence, this Eq., being of second degree in (x, y), pictures the Pair of tangents through (x1, Yl) to F(x,y; x,y)=0. The righlt member being a square, is always plus; hence, the factors on the left are like-signed; hence all points (x, y) of a tan- gent lie on the same side as any one point (xy, yi) ; i.e., the tangent does not cut the curve; i.e., the curve is throughout convex or concave on the same side; not like this figure. s This we may (do l)y think- ing [P ')2 a diagonal of a T four-side cut by another di- agional at 1. D)raw ASPI, 5P2, _ I AS!; draw PIVR at will; p1 H (Iraw 1.1 T; (raw (Q T; it Q cuts P1P2 at II, by Art. 49. 87 CO-ORDINATE GEOMETRY. Thus far all geometric representation has been purposely avoided, to show more clearly how the notions and prop)erties of pole and polar all lie enfolded in ttie algebraic fact that in the Eq. of the tangent to the curve of second degree current Cds. and Cds. of the point of touch appear symmetrically. The figure illustrates the definitions given. Poles and polars are marked by the letters P and L, witi corresponding indices. It is not necessary to know aught of the curve except that it is of second degree. We mav now return to the special properties of the circle. 73. By Art. 63, the section-points of x cos a + y sin a =p and x2 + y2= =2 are xi. =p cos a sin a dp2 Y,2 =p sin a Tc os a i-P2. The half-sums of these pairs are the Cds. of the mid-point of the intercepted chord: xm =p cos a, y, = p sin a. Bv eliminating p we get a relation holdina between the Cds. of the mid-points of all chords having the same a, i.e., all Ii chords. Hence, .m:xr=tana or y=tana X 88 NORMALS. is the locus of the mid-points of a system of 11 chords. Such a locus is called a diameter, and in this case is clearly a Rl. through the centre I to the chords. A RL. through any point (x,, Yi) of a curve I to the tangent at that point is named Normal to the curve at that point. The tangent to X2 + y2 = r2 at (X1, y') is XX1 + yyj = Thene, en t hence, (X-xj)y-(y-yj)x1=O, or x:y=xI:y, is normal to the circle x2 + y2 = -2 at (ZI, Yi). The absolute in this Eq. is O; hence, Normals to a circle pass through (envelop) a point, the centre. Also, all normals are diameters of a circle. 74. In the expression Xi-i' +y-y -.2, or X2 + y2 + 2gx + 2fy + c, either of which equivalents equated to 0 is the N.F. of the rectang,. Eq. of the circle whose centre is (x, Y) and radius r, (X,) - XI + Y - Yi is the squared distance of any point (x,y) from the centre ; and since the radiusi- is I to the tangent at its end (Art. 73), the difference x -XI2 + y _ -82_ r- is the squared tan- Xb Y. gent-length1 from (x, y) to the cicea._,,2 + -- y2 _ 9.2= circle X -XI Y -2 r2=O. So, too, is x2 + y2 + 2gx + 2fy + c the squared tangent-length from (x, y) to the circle XI + y2 + 2x q+ 2f, + c = O. This squared tangeit-length is called the power of the point (x, y) as to the circle. To find it, replace the current Cds. in the N.F. of the Eq. of the circle by the Cds. of the point. Like reasoning and results hold for oblique axes. The power of the origin (0, 0) is the absolute in the N.F. of the Eq. 75. The centre being origin, the Eq. of the circle is x2+ y2 - r2, however the axes be turned. Turn them till the X-axis 89 90 CO-ORDINATE GEOiEtrzY. passes through any assumed pole. Then is Y, = O, and the tDr polar is xxc,= r or x XI But this is a RL. I to X-axis, distant x from origin. Hence: (1) The circle being referee, the polar is I to the radius through the pole; (2) the radiuis is the geometric mean of the distances of pole and polar from the centre. L. We may now trace the movement of pole and polar thus: When the pole P is at P., the polar L is the tangent L,; when P falls on P2, L falls on L.,; as P nears the centre, L retires to x ; as P passes through the centre, L passes through o; as P nears Ps', L nears L,'; as P retires to x. L nears the centre; as P passes through x, L passes through the centre; as P nears PF, L nears L,. So long as P stays on a diameter, L stays iI in all positions; i.e., L turns around its point oo, also P and L move counter. As P glides along a tangent, L turns about the point of tangency; as P glides along any RL., L turns about the pole of that RL.; as P glides around any circle concentric with the referee, radius a, L turns around a second concentric circle, radius '; if P glide around this second circle, L will turn around the first. All these are special EXEPRCISES. cases of the general proposition: as the pole glides along (traces) any curve of nth degree, the polar turns around (en- velops, enwraps) a curve of nth class; and conversely. Such pairs of curves are called reciprocal. N.B. A curve cut by a RL. in n points is of nth degree; a curve touched bv n RLs. through a point is of nth class. Curves of second degree and curves of second class are the same; but in general curves do not rank alike in degree and class. See Art. 160. EXERCISES. 1. Find tangents to X2 + y2 - 6 x - 14 y -3 = 0 at the points whose x is 9. From the Eq. of the circle we find the corresponding value of y: 12, 2. The Eq. of the circle in the form I'(x, y; x, y) = 0 is xx +yy -3 (x+ x) -7 (y + y)-3 = 0. Hence, the tangents are 9x+12 y-3(9+x) -7 (12+y)-3=0, 9x+ 2y-3(9+x)-7( 2+y)-3=0; or, reduced, 6 x + 5y = 114, 6 Ox-y =44. 2. Similarly, find the tangents thus defined: X2 +y2- 4x+22y+25=0, x,=3; (X - 5)2 + (y + 8)2 = 113, xl = 13. 3. Find the tangents to X2 + y2 + 10 x-6 y-2 = 0 11 toy =2x-7. The Eq. of the circle may be written (x + 5)2 + (y - 3)2 = 36; or, if x' = x + 5, y'=_y - 3, X/2 + g'2 = 36. The Eq. of the RL. becomes 2x' -20. A RL. II must be y' =2x' +b- This is tangent to x'2 + ye = 36 when, and only when, 36 (1 + 22) = b2; i.e., when l = (iV5. Therefore the tangents are y'=2x't6v'o/; or, y =2x+ 13 6v'1. 4. Draw tangents to x2 + y2 = 58 inclined 603 to 4 x-3 q= 12. 5. Through (xl=9,y 0) on X2+ y2-12x+2y+3=0 draw RLs. inclined 45' to the circle. HINT. The RLs. sought halve the angles between tangent and normal. 6. Find the angle between two tangents to a circle. 7. Find the power of (-11,-9) as to (x-3)2+ (y-7)2=25; of (4, 1) as to4x2+4y2-3x-y-7=0. 8. Find the circle tangent to y= 3x- 5, centre at origin. 9. Find the tangents from (16, 11) to A2 + y2 = 169, and where they touch. 91 CO-ORDINATE GEOMETRY. 10. Find and draw the polars of: (11, 17) as to (x - 3)2 + (y + 5)2=81; (8,-5) as to X2+y2+14x+6Oy+22=0; (-2,-7) as to x2+y2 -18x+2y+57 =0. 11. Find the polar of (X,, y) as to the point (-circle) (a, l). The Eq. of the point regarded as circle of vanishing radius is (X-a)2 + (.y-b)2 = 0; ..the polar of (XZe yl) is (X,-a) (X-a) + (y1-b) (y -b) = O. This RL. goes through (a, b) 1 to the junction-line of (xl, ye) and (a, b). Show that the polar of (11, 3) as to (4,- 2) is 7 X + 5 y-18 = 0. 12. Find the polar of (Xi, yl) as to a RL. Regard the RL. as circle of infinite radius, and write its Eq. (x2+y2)(1+k) +2(A,+kA2)X+ 2 (B1+kB2) y+ C,+kC2=0. For k =- 1 this circle passes over into a RL. The polar of (x1, yJ) is, for k=-1, (Al-A2) (XI + x) + (B1-B2) (Y1 + ) + C1-C2 = 0, the RL. is 2(A1-A2)x+2(B1-B2)y + C1-C'2=0. Hence, the polar is If to the RL., midway between it and the pole. Show that the polar of (8,20) as to 5 1-3 y + 7 = 0 is 5 x-3 3-6 = 0. 13. What is the pole: of y = mX + b as to X2 + y2 = r2 of Ax+By+C=0 as to (X-a)2+(y-b)2=r2 14. Find the pole (x2, y2) conjugate to (xI, Yi) as to X2 + y2 = r2 and on the junction-line of (xl, yi) and the centre of the circle. Since (X2, Y2), as conjugate to (xl, YI), lies on the polar of (Xi Y1i), X112 + Y1Y2 = r2; since it lies on x: y = XZ :y, X2 Y2 = xI: 911 Hence, X. = r21X: (XZ2 + yl2), y = r2y : (X_2 + 1y2)_ Hence, xl = r2x2: (Z22 + y22), y3 = r'y2: (x22 + Y22). Hence, (xiI, y) and (12,1Y2) are called related as to x2 + y2 =r2 15. Show that each of two related points is the pole of a RL. through the other 11 to the polar of the other. Systems of Circles. 76. Be Cl c0, C2=0 Eqs. of two circles in normal form. Then is C1-XC2 = 0 the Eq. of a circle (by Art. 58) through the common points of 0, =0 and C2=0 (by Art. 30). 92 SYSTEMS OF CIRCLES. 93 Or, Cl- X2=0 is the Eq. of a system or family of circles through two fixed points: the section-points of 1 = 0 and 2 = 0; X is the parameter of the system, and ranges from -o to + c. Since Cl and C2 are the powers of any point as to Cl= 0 respectively C2 = 0, and since X is clearly the ratio of these powers, we see that The ratio of the powers of any point on any circle of the system as to any two circles of the system is constant. The ratio is of course different for different circles or as to different lairs of circles. For I= 1 the terms of second degree vanish, the circle passes over into a RL., i.e., a circle of infinite radius. This RL. is always real, though the section-points be imaginary. Clearly the powers of its every point as to the two circles are equal, since its Eq. is Cl - C2 = 0. Hence it might be named Equtipotential Line or simply Power-Line of the circles (called Radical Axis by Gaultier, 1813). 77. The power-lines of three circles taken in sets of two are C1-0C2 = 01 C2-03 = 0 Q3-01 = 0; added, these Eqs. vanish identically; i.e., the three power-lines meet in a point-the power-centre of the three circles. Hence, to construct the power-line of two circles, C0, and C2, draw 03 and C4 cutting Cl and 0,. The power-centres of C1, C2, 03 and C1, C2, C4 fix two points on the power-line of C0 and Co. CO-ORDINATE GEOMETRY. 78 Two points determine a RL. as common power-line of a system of circles: C1 -,kC . 0, through the points. Thte power-lines of each circle of such a system and a fixed circle C pass through a point. For the power-line of C and any circle C' of the system cuts the given power-line of the system say at I; which then is the power-centre of C, C' and any second circle C" of the system; hence the power-line of C and C" also passes through L EXERCISES. 1. Find in co-ordinates the power-centre of (x - 7)2 + (y9)2 = 36, (+ 3)2 + (y - 2)2 = 16, (x + 4)2 + (y + 5)2 = 9, and draw the figure. 2. Show that the power-line of two circles is I to the junction-line of their centres (or centre-line, as it may be called). 79. The form C, - AC2 =0 is not convenient for study- ing a system of circles. The power-line and junction-line of centres, being I, naturally suggest themselves as axes. The latter being taken for X-axis, the term in y falls away; also, for x= O the values of y are equal and unlike-signed for all the circles; hence the parameter A can enter only the term in x, and we can write the Eq. of the system, x2 + y2 - 2x +62= 0. Here A is the changing distance of the centre from the origin; 6 is the fixed distance to the section-points from the origin; these points are real or imaginary, according as 62 is - or +. 80. The Eq. of the system of polars of any point (x1, Y,) as to this system of circles is X1X+y y-X(xl+x)+62=0; X appears in first degree only, hence this system of polars pass through a point, the section of x1x + y1y +82=0 and X1 X-=0. In general, then, the polar of a point changes with the circle of the system, turning about a point; but if the two RLs. which fix this point, xJx +yy1 + 62 = O. xs + x = 0, be the same 94 ORTHOGONAL CIRCLES. RL., then are all RLs. of the system xx+yy -A(X- +x) + 62 = 0 the same RL. This is so when, and only when, Yi= 0, x1= 3, for otherwise the two Eqs. are not the same. Hence each of these two and only these points (8, 0), (-8, 0) has the same polar as to all circles of the system, namely, a RL. through the other I to the line of centres. These points are real or imaginary according as 82 is + or-, i.e., according as the section-points of the circles are imaginary or real. Writing the Eq. of the system thus, y +(x-x)2 =2 -82, we see if 82 be +, and so the above critical points real, then the circle is imaginary, for every X 8. For A very large the centre retires toward o along the X-axis, the circle flattens toward the Y-axis; as A nears 8, the centre nears the critical point (8, 0), the circle shrinks toward and around that point; and as A equals 8, the circle vanishes in that point. Hence the critical points (8, 0), (-8, 0) are themselves circles of the system, point-circles, and are hence named by Poncelet limiting points of the system. 81. The powers of any point of the common power-line as to these circles, i.e., the squared tangent-lengths from any point of the power-line to the circles, are all equal ; i.e., the ends of all such tangent-tracts, the points of tangency, lie on a circle with centre on the common power-line. The radii of this circle, as tangent to the other circles, are I to the radii of those circles; i.e., each circle with centre a point of the poicer-line and squared radius the power of that point cuts orthogonally the whole system of circles. As the limiting points are circles of the system, each orthogo- nal circle passes through the limiting points. Hence the Eq. of the system of orthogonal circles is x2+y2- 2Xy-82=0=x2+ (y- x)21_2_82. Deduce this directly as the Eq. of the orthogonal system. 95 CO-ORDINATE GEOMETRY. Note that these two mutually orthogonal systems are comple- mentary: in the one 82 is + in the other - ; the power-line of one is the centre-line of the other; the section-points of one are the limiting ppints of the other; of the one the section- points are imaginary, the limiting points real, - of the other, vice versa. To construct this double system. Draw any number of cir- cles through two points. To any one draw any number of tangents. About the points where these cut the RL. of the two fixed points describe circles with the tangent-tracts as radii. Do not fail to carrv out this construction. EXERCISES. 1. When are C = 0 and Cl = 0 1 The squared distance between their centres must equal the sum of their squared radii; i.e., 96 EXERCISES. (g -g )+ (f-f)2 = g2 +f2 - C + g2 +fA2 - C, or, 2 gg+ 2ff1-c--l = 0. (A) Hence, if C = 0 cut both C1 = 0 and C2 = 0 at right angles, we Itave 2gyg+2ff-c-c,=0 and 2g2g+2f2f-c-c2=O. From these we may express any two of the three symbols g, f, c through the other linearly; substituting in C= 0 we get an Eq. of a ecircle containing one parameter linearly; hence all circles cutting two circles orthogonally form a system through two points, i.e., with common power-line, as already proved. If C= 0 cut three circles orthogonally, then we have three Eqs. of the form (A), which with C= 0 give, through elimination of g,f, c, as the equivalent of C = 0: X2 Xas-x -y 1 =0. C2 .91 f2 1 C3 93 fA 1 Geometrically this is clearly a circle about the power-centre, its squared radius the power of that centre. 2. The circle orthogonal to Cl =0, C2= 0, C8 =0 is also orthog- onal to Al1e + A2C2 + A3 C3 = O. Use condition (A). 3. The polar of one end of a diameter of a circle as to any orthogonal eircle passes through the other end. Be X2 + y2 = 7.2 the given circle; then by (A) an orthogonal circle is X2 + y2 + 2gx + 2fy + r2 = 0. As to this the polar of (r, 0) is rx+g(r+x)+r2=0, which always goes through (- r, 0). 4. Powers of points of one circle as to another vary as their distances from the power-line. 5. To find the angle al, under which C= 0 and Cl = 0 intersect. a, equals the angle between the radii to a section-point. If r, r, be the radii, d the distance between the centres, then 2 rr1 cos al = r2 + r12-d2. If we hold C, = 0 fixed and a, constant, then since d2 - r2 is the squared tangent-length to C1 = 0 from the centre of C= 0, we have this relation between the Cds. of the centres of all circles, a= 0, which cut Cl = 0, under the angle a,: 97 r2 - 2 rlr Cos a, = Cl. (H) 98 CO-ORDINATE GEOMETRY. Since x and y enter Cl, this Eq. contains three arbitraries, x, y, r. Impos- ing the further condition that C= 0 cut C2= 0 under c2, we get a second Eq.: r-2 - 2 r2 cos a, = C2. These two Eqs. (1o not yet fix the centre (x, y) and the radius r of C =0; but they determine its family. For it will cut any circle of the system Cl - AC2 = 0 under a constant angle. We have, namely, r2_2- 2r r1 cos a, -Ax2 cos a2= C_-A 1-A 1-A which declares that the varying circle C = 0 cuts the circle C'= C.- 2=O' 1-A radius rt, under an angle y such that r' cos -y = (r1 cos a, - Ar2 cos aj): (1-A). We may express rr through the constants of CG and C2 thus: r'2=(1- A))(-,2 Ar22)-Ad21: (1-A)2. Hence y is determined univocally, i.e., is constant. If we assign 'y at will and substitute the value of r', we get a quadratic for determining A; i.e., there are two circles of the system C1 - XC' = 0 which the varying circle C= 0 cuts under any given angle. As a special case, for y = 0, cos y =1, we see that there are two circles of the system which the varying circle always touches, on which it rolls. 6. Through the section-points of X2 + y2 + 4 x - 14 y -68 = 0 and X2 + y2 -6 x-22 y + 30= 0 draw a circle tangent to X-axis. 7. Under what angle do X2+y2= 16 and (X - 5)2 + y2 = 9 intersect 8. Find the power-centre of X2+y2-2x+6y-15=0, 2+y2+ 14x+ 12y+81-=0 and the point (3,-7). 9. Find a circle through (-5, -4) and cutting orthogonally x2+y2-4x-6y+9=0, x2+y2+6x-4y+4=O. 10. Find a circle through (-4, 3), (-2, -3) cutting X2 + y2 - 6 x - 7 = 0 orthogonally. 11. Show that the point-circles cut a diameter of every circle of the system harmonically. 12. What circle of x2+.y-2Ax+ 52=o cuts (x-a)2+ (y-b)2=r2 orthogonally CENTRES AND AXES OF SIMILITUDE. 99 13. It has been shown (Art. 80) that the polars of a point as to a sys- tem of circles pass through a point, -the centre of the polar family. These two points are called poles harmonic as to the system of circles. Show that the power-line halves the distance between any two harmonic poles, and that the circles cut the junction-line of the poles in an involu- tion of points. 14. Find a circle whose power-lines with two given circles go through their centres. 15. Show that the centres of all such circles lie on a RL. 11 to the power-line of the two circles, named secondary power-line. 16. Show that the circles halving the circumferences of two given circles form a system, and find the power-line. 17. Show that the secondary power-lines of three circles go through a point. 18. Find a circle halving the circumferences of the three circles: X2+y2 -2x-9=0, X2+y2+3x-9=0, X2+y2_6y-l0X+18=0. Centres and Axes of Similitude. 82. Congruent figures are alike in shape and size, differ only in place; they may be thought fitted one on another. Similar figures are alike in shape, but not in size. Such figures may be supposed made thus: From any point draw rays in any directions; on each ray take two tracts in a fixed ratio; the ends form two similar figures. The fixed point is called centre of similitude; the fixed ratio, ratio of similitude. Clearly the one figure may be thought as the other swollen or shrunk in like measure throughout. Bv pushing or turning either figure the shape is not changed. When simply pushed, or when turned through a flat angle, cor- responding tracts in the two figures keep 11, and the figures keep similarly placed. By the above construction of similar figures, space is doubled in thought: the space of the one figure, and the space of the other. To each point in one space corresponds a point in the CO-ORDINATE GEOMETRY. other. If these spaces be now thought pushed or turned out from each other, corresponding points remain centres of simi- larity for the two figures and spaces. For, by similar triangles, corresponding tracts, tracts between corresponding points in pairs, are proportional andl include equal angles. In what follows we shall keep the original construction, with- out pushing or turning, unless through a flat angle. a'O C 83. Clearly, tangents at corresponding points, being drawn through pairs of corresponding points, are 11- To find the figure c' similar to a circle c, radius r, ratio of similitude r: r'. Take any point 0 as a centre of similitude, lay off OC so that r:r'= OC: OC'. Let P correspond to P, then OP: OP' = r: r'= OC: OC'= CP: C'P = r: r'. Since CP is constant, so is C'P'; i.e., c' is a circle, radius r'; i.e., afigure similar to a circle is a circle. Conversely, all circles are similar. For be c and c' any two circles, radii r and r'; cut the tract between their circles innerly respectively outerly in the ratio r : r'; then, by the above, the figure similar to c is a circle with centre C', radius r'; i.e., it is the circle c'. 100 AXES OF SIMILITUDE. 101 COR. 1. Any two circles in a plane have two centres of simil- itude: inner and outer, cutting the tract between their centres: innerly and outerly, in the ratio of their radii. This property is peculiar to the circle, since it alone of plane figures, being homogeneous, may be turned around in itself. Con. 2. The two pairs of common tangents to two circles cross in the inner resp. outer centre of similitude. Or thus: A tangent to one of two circles through a centre of simili- tude is tangent to the other; for the radii to the points of touch are Jj. 84. Be C = (X _ -X)2+ (y _ y")2 - r,, = 0 any circle, and let I(C'C0) resp. E(C,02) denote the inner resp. outer centre of similitude of C, and Q. Since I(C102) resp. E(1Ct2) cts the tract between the centres in the ratio of the radii, its Cds. are rlx2 + rix, rTY2+ WI 1, + r. V rl +r2 resp. :2:= ,l X2 = 2"IY r 2-12 rl-+r2 ' r-- 2 (x', y') resp. (x", y") is the pole of the chord of contact T' resp. T"T" of the inner resp. outer common tangents. Hence substituting for X', y' resp. x", y'" in the Eq. of the polar, we get as the Eqs. of these four chords, after easy reduction: (X2-XI) (x-xI) + (Y2-YI) (Y-Y1)=r1 (rl T r2), resp. (X1-Xn) (X-x2) + (Y -Y2) (1-12) =r2(r2 w rrl)- The centre-line is (y2 - Y) (x - x1) -(x2 - x1) (y -Yi) = 0, whence we see the chords of contact are 1 to the centre-line. 85. Three circles, C1, 0A, (Q, combining three ways in sets of two, have six centres of similitude: three inner, three outer. Form the Eq. of the RL. through E(C0C0) and E(0207), multi- ply it by (r, - r2) (r - ra), and divide it by r2; the result is CO-ORDINATE GEOMETRY. Ir r2 r7 X - rl r2 r3 + 2l r2 r3 =0. (HI) Y1 Y2 Y3 XX2 X3 Y1 Y2 Y3 1 1 1 X1 X2 X3 To find the RL. through E(C203) and E(C3C1) permute the indices; this will not change H', it being symmetric as to the3 indices; i.e., the RLs. are the same; i.e., the three outer centres of similitude, E(C1C2), E(C203), E(C3Cl) lie oan a RL. By changing the signs of th2 r's properly we show that E(C/C2), I(C2Q3), 1(03Q1); E(0203), I(0303), I(C0C2); resp. E(C301), I(ClC2), I(C2C3) lie on a RL. These four RLs. are named axes of similitude of the three circles: one outer, three inner. COR. If two circles touch innerly resp. outerly, the point of touch is an outer resp. inner centre of similitude of the two; hence, from the above, if one circle touch two, the junction- line of the points of touch passes through a centre of similitude of the two: outer resp. inner according as the circles are touched alike (both outerly or both innerly) resp. not alike (one innerly, one outerly). EXERCISE. Find the centres and axes of x2 + y2 = 16, (X - 5)2 + y2 = 81, (x _ y)2 + (y -10)-2 = 4. Draw the figure. 86. If a varying circle cut the circle C = 0 under the angle a, by Art. 81, its radius R and its centre-Cds. satisfy the Eq. R- - 2 r1R cos a = C. Treating R and a, or, what is the same, Bs and R cos a as parameters, we may impose two more such conditions: R2- 2r2Rcosa= C2, B_ -2r3Rcosa = C3,; eliminating the parameters from these three Eqs., we get a relation between the centre-Cds. and constants, i.e., we find 102 THE TACTION PROBLEM. the locus of the centre of the circle cutting the three circles C, = 0 C2= 0, 3 =0 under the same angle a. So we get (rl- ri) (C - 02)- (r, - 22) (C,- 3) = 0, a RL. through the intersection of C1-C2=0 and C1-C0 = 0, i.e., th ough the power-centre of C0 = 0, 02 = 0, C0 = 0. Writing for Cl - 0, C0 - 0, their values, we find the coefficients of x resp. y are the coefficients of y resp. - x in the Eq. of an axis of similitude; i.e., the RL. is I to such an axis, and in fact the outer one. For thus far we have taken a as the inner angle between the circles, i.e., the angle between the radii to a section-point. To take a as the outer angle in case of either of the circles, it suffices: to change the sign of Cos a, since the inner and the outer angle are supplementary and cosines of supplementary angles are equal and unlike-signed; or to change the sign of r1, r2 resp. r3. Taking all the angles as outer changes all the r's, which does not affect the Eq. of the ilL. ; changing one of the r's and leaving two unchanged is clearly tantamount to changing the two and leaving the one unchanged; this can be done in three ways; so we get three other RLs. through the power-centre I each to an inner axis of similitude. Hence the whole locus of the centre of a circle cutting three circles under the same (varying) angle is a pencil of four RLs. through the power-centre, 1 each to an axis of similitude. 87. If two circles touch innerly, their inner angle is 0; if outerly, it is 1800. Hence, the I from the power-centre on the outer axis of similitude contains the centres of two circles: one touching the three circles all innerly; the other, outerly: the r's are all + or all -. Changing the sign of one r we get a I on an inner axis of similitude containing the centres of two circles: one touching two circles innerly, the third outerly; the other touching the two outerly, the third innerly. Changing the sign of each r in turn we find in all eight circles touching each of three circles; a pair of centres lie on each I through the power-centre on an axis of similitude. 103s CO-ORDINATE GEOMETRY. We might now determine another line on which the centre of a tangent circle must lie, by eliminating R between two Eqs. of condition, as C1 = , e + 2Br1, = = 12 + 2Rr2. But the line would not turn out to be a RL. or a circle, hence would not admit of elementary construction: with compasses and ruler. But the doctrines of poles and polars, power-centres and power-lines, centres and axes of similitude, enable us to solve the general Taction-Problem by use of the ruler alone. 88. Suppose the circle 0 resp. 0' touches the given circles Ci, C2, C, outerly resp. innerly. (1) Then by Art. 85, Cor., the chords of contact T1T11, T2T21, T3T3' go through P, the inner centre of similitude of 0 and 0'. (2) Hence, PT,: PS,'= PS,: PT,', PT1. PT1'= PS, . PS1,. But PT1 PS1 PT1' PS1'=q2 q'2, q2 and q12 being powers of Pas to 0 and 0'. Hence, PT,. PT,'= qq' (a constant for all directions) = PT. PT1'= PT3 PT31; hence Pis the power- centre of C,, C.,, C3. (N.B. T. and Tl' are anti-correspondent points of 0 and 0'.) 104 THE TACTION PROBLEM. (03) Since C0 and C. touch 0 resp. O' each outerly resp. innerly, as in (1), the contact-chords T1T2 resp. TjtT2' go through the same outer centre of similitude of C, and C2, say K'. Hence, as in (2), KT1 KT, = KTI' _ IT2'; i.e., the powers of K as to 0 and O' are equal; i.e., K is on the power-line of 0 and O'. So too likewise are the outer centres of similitude of C, and Q, CQ and C,; i.e., the outer axis of similitude of C0, C, ( ,, is the power-line of 0 and 0. (4) The power-lines of 0 and C, resp. O' and C,, i.e., the tangents at T, resp. T11, meet say at X,, which is then the pole of the contact-chord T1T,' as to C,; but by Art. 77 the power- line of 0 and O' also goes through X1 the power-centre of 0, 0', C,; i.e., through the pole of chord T1Tl' as to C,; hence, by Art. 69 the chord T1T,' goes through the pole of the power-line as to 0Q. Likewise the chord TT2' resp. T3T goes through the pole of the same power-line as to C, resp. C3. If, of the two circles 0 and O', 0 touch C, and 0. outerly resp. innerly and C0 innerly resp. outerly, but O' touch C, and C0 innerly resp. outerly and C3 outerly resp. innerly, then the power-line of 0 and O' passes through the outer centre of simil- itude of C, and 2, through the inner centres of similitude of 09 and C,, C3 and 0Q; i.e., it is the inner axis of similitude E31j,; the other relations hold unchanged. Hence the follow- ing rule: Determine the power-centre and axes of similitude of the three given circles; determine the pole of each axis as to each circle: each three RLs. through the power-centre and the three loles of each axis cut the three circles in contact-points of two tangent circles. REMARK. This classic problem, in which the geometry of the circle seems to culminate, was proposed and solved by Apollo- nius of Pergaw (B.c. 220). His solution was. lost, but was 105 CO-ORDINATE GEOMETRY. restored by Vieta (t1603). The first solution of the analogous problem for space: to find a sphere touching 4 given spheres, was given by Fermat (t1665). Both solutions were indirect, reducing the problem to simpler and simpler problems. Gaul- tier (1813) and Gergonne (1814) first gave direct solutions of the first problem. In the above rule replace 3, circle, axis by 4, sphere, plane, to solve the problem for space. Note care- fullv on what the solution turns: on determining the chords of contact in the given circles (spheres) by two points: the power- centre and a pole of an axis (plane) of similitude. Circular Loci. 89. 1. Given any A cut by a transversal through a fixed point of the base; through the fixed point and the intersections of the transversal with I C each side and the adjacent vertex at the base are drawn circles; find the locus of their intersection. Be ABA' the A. Take the fixed point 0 as origin, the base as one rectangular axis, say X-axis. Be OA =a, OA' = a'; then the sides AB, A'B are y=c (x-a), y = c'(x - at); the transversal is y = sx; it cuts the sides at (c'-f c!-) or C, C'- The circles through 0, A, C resp. 0, A', Cl are X2 + y2_ax - a(cs + 1) y=0, resp. x2+y2-afxa(c1s +1)3y = 0. C -S c'-s Eliminating s, we get the locus of their intersection, c(xA + y2 - ax) -ay: x2 + y _-ax + cay = df(X2 + y2 - aX) - aly : x2 + y2 - a Ix + c la'y; or, (c-e') (x2 + y2)2_ (a + aI)x(x2 + y2) + aa(x2 + y2) + (1 +cc') (a'-a)y(x2+.y2) =0; 0 106 CC , S"a ), (e-S C-3 CIRCULAR LOCI. 107 or, (+ y2) iT+y2_(a+at)x_ + ccl 1(a a') y + aa This Eq. is the product of two: the first, of a point-circle, the origin; the second, of the circle circumscribing the A. 2. A right angle turns about a fixed point; find the locus of the foot of the 1 from the vertex on the chord of the intercept of its sides on a fixed circle. The Eq. of the circle, referred to rectang. axes through the centre, is X2 + y2 = r2. (1) B Take the diameter through the fixed point P as X-axis. The chord is y= sx+ b. (2) Eliminate y between (1) and (2), whence, (1+S2)X2+ 2sbx+ P2-r2=0. (3) The roots of this Eq., X1, x2, are the x's of A, B; the y's are sxl + b, sx2 + b. If OP = c, the direction-coefficients of the angle's sides PA, PB are sx +b sx2 + b. X1 C XI1 C since the angle is right, their product is - 1; which, cleared of fractions, gives (1 +s2) XX2 + (ab -c)(x, + x2) + 62 + C2 = 0. From (3) we have Xi x= (b2-r2) ( +s2), X1+X2= -2sb : (1+s2) - ..(1+s2)(c2-r2)+2b(sc+b)=-, (4) an Eq. of condition between the parameters s and b. The I from P on the chord AB is Y=- (x -c).(6 Eliminating s and b, by means of (2), (4), (5), we get the locus of the intersection of (2) and (5), the locus sought: y2+ (X-C)2 x2+yl-CX+ C2 - 2 0. This breaks up into the Eq. of the point-circle P, and the circle about the mid-point of OP, radius - 2_ . 2 4 108 CO-ORDINIATE GEOMETRY. Clearly P is not part of the locus sought; how, then, does it appear as part in the result The pair of values, x= c, y = 0, satisfies (5) for all values, real and imaginary, of s and b; then, from (2), 0= sc + b; hence, from (4), for c r, (1 + s2) = 0, s= i. Now the problem as proposed implied only real values of s and b, but the analytic statement held not only for real, but also for imaginary, values; i.e., for the problem in question and for more; accordingly, the result yields the locus sought, for real values of s and b, and another locus not sought, for imaginary values of s and b. 3. Find the locus of the foot of the I from the centre on the chord. 4. Find the locus of the intersection of tangents at the ends of the chord. Be (x',y') the intersection; then x'x+ y'y = r2 is the chord. Combine this with x2+y2=r2; the pairs of roots so obtained (x, yJ), (x2,y2) picture the ends A, B of the chord; the coefficients of direction of PA, PB are ____ Y2 , and their product is - 1. X1- C X2-C Hence, (12+ y12)(1.2 c2) + 2r2cx'-2r4= 0; or, dropping the primes, x + r c 2+ y2= r (2 r _c2) I r2 - C2 W( - CT) If R be the radius of this circle, d the distance between the centres, then (El - d_)2= 2 r2 (R2 + dr) the condition that a quadrilateral inscribed in one circle may be circum- scribed about another, any tangent to the inner being taken as a side. 5. Find the locus of a point the feet of perpendiculars from which, on the sides of a A, lie on a RL. Be x Cos a, + y sin al-p = 0 = N1, N2 = 0, N3 = 0 the sides of the A; (X1, YM), (X2, Y2), (X3,3) the feet of the Is; (:r, y) the point. Then are N1, N2, N3 the lengths of these Is. Their projections on the axes are x-x1z=N1Vcosa1, y- =1y=N1sina1, and so with indices 2, 3; hence, Sk = X-A k cos auk, y = y - NkAsin ak, for k=1, 2, 3. CIRCULAR LOCI. 109 The feet (xl, y1), (x., Y2), (X3, y3) lie on a RL. when, and only when, 2-Yl: X2- X = Y3 - Y: X3- X; or, after reduction and substitution, N1N2 sin (a-a2) + N21Nx3 sin (a2-a2) + N3N, sin (a3-a) = 0. (1) N1, N2, N3 are linear in x, y; hence, (1) is quadratic in x, y. The first term yields as coefficients of x2, y2, resp. :y, cos a, cos a2 sin (a, - a2), sin a, sin a2 sin (a, - a.), resp. sin (a, + a2) sin (a, - a2). The difference of the first two is cos (a, + a2) sin (a, - a.2), or, I (sin 2 al-sin 2 a2). The third is - ', (cos 2 al-cos 2 a2). Permuting the indices to get the contributions of the other terms, and summing, we see that the difference of the coefficients of x2 and ,2 as well as the coefficient of xy, vanishes; i.e., the locus is a circle; the Eq. is also satisfied by putting any two of the N's =0; i.e., the circle goes through the vertices. Hence, the feet of Is on the sides of a A from any point of the circum- scribed circle, and from no other, lie on a RL. This problem deserves further notice. Suppose the RLs. N1= 0, N22=0, N =0 tangent to a circle 0, radius r, centre at origin; then, P1=P2 =P3 = r. Developing (1), we have f (X2 + y12) -Px - Qy + F= 0, where 1 = sin (a -a2) sin (a2 - aj) sin (a3- aj) If R be the radius of this circle, D the distance between the centres, RI = (p2 + Q_) 4 112 - F. D2 = (P + Q2) :4M2, whence, D2 - F2 = F. ifl If A1, A2, A3 be the angles of the A, then a2-al1=7r-A2, a3-a2= r-Aj, al-a3=v-A2; hence, if S = area of A, A1 =-sin Al . sin A2. sin A = -S: 2 R2; F = r2 (sin A, + sin A, + sin A,). and from (2), 110 CO-ORDI-NATE GEOMETRY. If s, s.,, sb le sides of the A, 2 S=r (s + s.,+ s) =2Rr (sin A,+ sinA2+sinA3); whence, F= rS: X, and hence, F: O=-2Er; 2- I)= P 2- . (2) Such, then, is the relation connecting the radii and distance eween the centres of the inscribed and circumscribed circles of a A. Now, holding the circles fixed, let us see how we can vary the A. Taking the one centre as origin, the other on the X-axis, we have Q=0, 2 _ F _2 4M12 A1' 4 J1J2 By virtue of (I), these three relations are satisfied if these two are: Q=O, F:,1M=-2Rr. Choosing one of the angles al, a2, a, at pleasure, we can still find values of the other two to satisfy these two equations. HIence, When the relation D2 = lg- 2 RIr holds between the radii of two circles and the distance between their centres, a A can be drawn in the one about the other, the direction of one side being taken at pleasure. Theorems analogous to these two for triangles and quadrilaterals hold for polygons generally. We got (1) by imposing the condition that the feet of the Is lie on a RL. Let us see how it expresses this condition. Suppose the origin inside of the A and a, a2 a. Be t4 P(x, y) any point within the A, lIt 11 / and join the feet of the Is from it .' \t/ on the sides of the A, to form a A F1, F2, F,. Then -N1, N.2, .x3 are lthe engths of these Is, all have '/O the same sign -, and are like- / 1 YF1 \ signed with the Is from the origin inclined al, of, at, as both P and the origin are within the A. Hence the terms of the left side of (1) are in order the double areas of the As FIPF,, FPF, PF-P1, the whole left side is the douLle area of FFF2,. If P'(.r, y/) be without the A, then one of the Is, say N1, becomes +, and the left side of (1) is the difference between the double area of the A F2'P'F2' and the sum of the double areas of FI PI F,' and F2'P FI', i.e., again, the whole left side is the double area of F1' F.2' FPI. Now this double CIRCULAR LOCI. 111 area is 0 when, and only when, F1, F2, F3 are on a RL. Accordingly we may generalize our problem by requiring that the area of the A F1F, be not 0 but some constant C2. Then the left side of (1) is this area doubled; also since (1) is the Eq. of a circle whose radius we call R, the left side can be written K(dl2 J- 2), where K is constant and d2 stands for the general expression, X2 + y2 + 2 gx + 2fqV. Hence N1N, sin (32 - a,) + , sin (a3 - a,) + -N1NV1 sin (a, -a) = 2 cK = (d2 -_ ) is the locus of a point the feet of Is from which on the sides NT = 0, 2 = 0, N13 = 0 of a A are vertices of a A of constant area c2, + or-. Tile locus is two circles concentric to the circle circumscribing the given A. The outer is always real; the inner, only when c2 KR2. The given area c2 changes sign for d = R, i.e., as P goes through the circle. 6. Find the locus of a point, the sum of whose squared distances from n points, multiplied resp. by given constants, shall be a given constant. 7. Find the locus of the centre of a circle seen from two given roints under given angles. 8. Find the locus of the centre of a circle that cuts two given circles at ends of diameters of each. 9. From a fixed point P are drawn tangents to a system of circles through two fixed points; find the locus of the intersection of the chord of contact with the diameter through P. 10. Be 1 2 1' 2' 8'f a regular hexagon; draw 1 3, 1 3', also any RL. through the centre, cutting 1 ., 1 3' at 4, 4'; find the locus of the intersec- tion of 2 4 and 2' 4'. 11. Find the locus of a point whose polars as to three given circles meet in a point. 12. A constant angle turns about its vertex fixed on the bisector of a fixed angle; find where the I from the vertex on the junction-lirle of the intersection of the sides of the angles meets it. 13. Find the locus of a point from which two given circles seem of like size. 14. Find the locus of a point whence two consecutive tracts L.ll, MN of a RL. seem of like size. 15. Find the locus of the mass-centre of a A inscribed in a given circle, on a given chord as base. 112 CO-ORDINATE GEOMETRY. 16. Under the same conditions as in (15) find the locus of the ortho- centre and of the centre of sides of the L. 17. Of two related poles, as to a given circle, one glides along a RL.; how does the other glide 18. How does one of two related poles as to x+ y2 = r2 glide when the other glides along (x )1 + i ) 19. Find the locus of the mid-point of a chord of a given circle, which subtends a right angle at a given point. 20. Find the locus of the foot of the I from the origin on a chord of a given circle, which subtends a right angle at the origin. 21. Two variable circles touch each other and two fixed circles; find the locus of their point of touch. 22. Find the locus of a point whose polars as to three fixed circles meet in a point. GENERAL PROPERTIES OF CONICS. CHAPTER IV. GENERAL PROPERTIES OF CONICS. 90. The general Eq. of curves of second degree, called Conics (Art. 152), is kxS +2 hxy+jy2+ 2gx+ 2fy +c = 0, or F(x, y; , y) = O. Among many ways of treating conies that seems most natu- ral which proves itself the best in the study of Quadrics (sur- faces of second degree), namely, to develop the relations of the locus to the RL. Let the student recall that The Eq. of the polar as to F(x, y; x, y) = 0 of the pole (xa, Yi) is F(xl, y1; x, y) = 0, or F(x, y; x1, yj) = 0 (Art. 69), or (kxl+hy,+g)x+(7thx1+jyf)y+gx1+fylf+c=O. (J) If the pole be on the curve, the polar is a tangent at the pole. The Eq. of the pair of tangents to F(x, y; x, y) = 0 through (xl, yl) is F(x,, Yi; xi, yl) - F(x, y; x, y) = F(xl , Y1; x, y) 2 (Art. 72). By passing to 11 axes through a new origin, O'(x',y'), k, h, j are not changed, but g, f, c are changed into g'= kx'+ hy'+ g, f'= hx'+jy'+f, c' F(x', y' ; x', y') (Art. 51). If A k h g = 0, the conic breaks up into two RLs. h j f gf C If, also, C'_ kAj - h 2 = 0, the RLs. are 113 CO-ORDINATE GEOMETRY. 91 If (x', y') and (x, y) be a fixed and a variable point on a RL. sloped B to the X-axis, the I to which is sloped a resp. P8 to the X- resp. Y-axis, and 8 be the distance between the points, then either of these two equivalent sets of relations states the Law of Sines of the A PDP: x-x' y- = 8 x-x'y-y' 8 sin ai- sin O sinas' costs Cosa sinas sin - cos a sin f-t cos/8 sin t sin w Sin sinai sin 0 q whence s _ =T.- sinas-B q Then x =x' + q8, y = y'+ q'8. y o X I To find the distances from (x', y') at which the conic, substitute in F(x, y; x, y) = 0, hence the RL. meets as in Art. 51; (kq2 + 2hqqt+jq2)82 + 21 (kx'+ hy'+ y)q + (hx'+jy'+ f)q' 8 + F(x', y'; x', y') = 0. A geometric interpretation of this Eq., according to Art. 60, lays bare the form and general properties of the conic. 92. The roots of this quadratic, 81, 82, are counter (equal and uinlikezsigned), i.e., (x', y') is the mid-point of a chord of the conic, directed by B, when, and only when, 114 DIAMETERS AND CENTRE. (kx'+ hy'+ y) q + (h x'+jy'+f)q'= 0, or (kx'+ hyy'+ g) + s (hx'+ jy'+f) = 0. (K) For 0 constant, s, q, and q' are constant, and the Cds. (x', y') of the mid-point of any chord directed by 0 are connected by an Eq. of first degree; i.e., the mid-points of all 11 chords of a conic lie on a AL. Such a RL. is named a Diameter. By changing 0 we change the direction of the 11 chords, change s, and change the diameter; s, then, is the parameter of the system of diameters, and since it enters (A-) linearly, all diameters pass through a point, called the centre of the conic. This centre is the intersection of the two diameters 7x' +hy' + g-0 and hy+jy'-Ff=; i.e., the point (G: C, F: C). It is in finitv or in o according as C 0 or C= 0. Hence conies are named conveniently, b)ut not quite correctly, centric and non-centric, accordino' as the centre lies in finity or not in finity. 93. We have got the notion of centre from that of diameter, 1)ut we may get this from that, thus: The coefficients of x and y in Eq. (J) of the p)olar of (xl, yj) vanish when kxl + hy1 + g = 0 and hx, +jy, +f = 0, i.e., when the pole is the point (G: C, F: C); but when the coeffi- cients vanish, the intercepts on the axes are both oc, i.e., the AL. lies wholly in x. Hence (G: C, F: C) is the pole of the polar at o. Hence the polar of every point at x goes through (G : C, F: C). Now a pole and any point on its polar are a pair of points to which the section-points with the referee of a RL. through the pole and point on the polar are an harmonic pair; and since one point of the first pair is at so, the other halves the tract between the second pair (Art. 41) ; i.e., the point (G: C, F: C) halves every chord of the referee (conic) through it. Such a point is named centre. See Note, page 190. 15 116 CO-ORDINATE GEOMETRY. Now take any point on a RL. or polar through this centre by Art. 71 it is the fourth harmonic to the pole at x, and hence halves the tract between the other pair, namnely, the section- points with the referee (conic) of the RL. through it and the pole. But all such RLs. are 11, since, as the point glides along the polar, they turn about the same point at ox, the pole of that polar; hence, too, they are all conjugate to that polar and no other RLs. are; hence, a RL. through the centre of a conic halves a system of 11 chords which are conjugate to it as a polar. Such a Rb. is called a Diameter conjugate to the chords it halves. Among all the chords conjugate to any diameter D there is one and onlv one through the centre, which is accordingly D's conjugate diameter D'. The chords which D' halves are 11 to D. For D passes through the pole of D' at x0, and hence all I's to D pass through that pole at A; hence all chords 11 to D are conjugate to D', and hence halved byD'. Hence, conju- gate diameters, D and D', of a conic halve each all chords 11 to the other. This, indeed, is clear from the fact that the conju- gate relation is mutual (Art. 69). 94. From (K) we see that the direction-coefficient s' of the diameter halving chords whose direction-coefficient is s is s'= _k7 h As.; whence s=- +hs- h +js h-j-js' If C hj - h2 = 0, then S =-- V=Ik(V\k + VJ. s') V!i VY (Vk+Vj s') Hence, s' is constant, i.e., diameters of a non-centric conic are 11. We may not cancel V/k+ VTs' in the numerator and denominator of s, since it is 0, and to divide by 0 has no sense; but s takes the undetermined form . To find what this means, 0 put s5 or its value for s in (IC) ; so we get the Eq. of the diam- eter conjugate to the chords directed by s, i.e., the 11 diame- CONJUGATE DIAMETERS AS CO-ORDINATE AXES. 117 ters. Reducing, we get Ox+ Oy +gI;+fVk=0; but this, by Art. 92, is the Eq. of a RL. at o. Now this RL. at oo goes through the centre at ao, and so halves all chords through that centre, i.e., halves all the 11 diameters, i.e., is conjugate to them all; and this it does whatever its direction may be. Hence, the common conjugate to all diameters of a non- centric conic is the RL. at A, and may be thought 11 to them all. 95. For a = 90, q' is 0, and the diameter is k1x + hy' + g = ; this, then, is the diameter halving chords 11 to the X-axis. So, too, hx' +jy'+f = 0 is the diameter of chords 11 to the Y-axis. These diameters are themselves 11 to the Y- resp. X-axis when and only when h = 0; but then they are con- jugate, as each halves chords 11 to the other; hence, the condi- tion necessary ansd sifficient that the axes be 11 to a pair of co01JU- gate diameters is h = 0; i.e., the term in xy must vanish. If the centre be taken as origin (which can be done always and only in centric conies), the new coefficients of x and y, kxj + hy, + g and hx, +jy1 +f, vanish, and the central Eq. becomes kx2 +2 hxy +jy2 + c' =0. If, besides, a pair of conjugate diameters be taken as axes, the term in xy vanishes, and the Eq. takes the form k'x2 +jy2 + C = 0. This Eq. is a pure quadratic both in x and in y: to any value of either correspond two counter values of the other, each axis halving all chords l1 to the other. 96. In general, diameters are oblique to their conjugate chords; are they ever I Choose rectang. axes; then s and 8s become tan 0 and tan 0', and tano'= - k + htanO h + j tan0 CO-ORDINATE GEONIETRY. When chords and diameter are I, tan 0. tan 0' - 1; or, on reduction h7tan &2 + k -j tan 0- h 0. This quadratic in tan 0 has two roots: tan 0I, tan 02, both always real. They vield each an X of values of 0, but as they differ among themselves only by some multiple of -r, the two o 's determine but two different directions of chords I to their diameters; also, since tan 0, tan 0=-1, these direction.s are I to each other: hence, if either be thought as the direc- tion of the chords, the other will be the direction of their diameter. Again, by Art. 52, these two I- directions are the ones that halve the angles between the directions fixed by the pair of RLs. kx2 + 2 hXy +jy2= 0. Hence, there is one and only one pair of I conjilUate diameters: the pair halving the angles between the Asymptotes (see Art. 97). They are named Axes of the conic. N.B. Of course, in the non-centric conic only one diameter I to its chords is in finity; it is called the Axis of the conic. 97. If the coefficient of the second power of 6 be 0, one root, 8, of the Eq. is x ; i.e., one distance from (x', y/') to the conic in a direction fixed by kq2+21uqj'+jq'2=0, or kD+ 2hs+j-= 0 is x. This Eq. is quadratic in s; hence there are two such directions., which are real and separate, real and coincident, or imaginary according as h2.j(or -C) is 0, =0, or 0. The conic is named accordingly Hyperbola (excess), Parabola (likeness), or Ellipse (lack). Denote them by H, P, E. H has points at xz in two directions, P in one, E in xlone (real). The co-factor C is called the criterion of the conic. The direction in which lies the point at X in P is fixed by the value of s: s = Vk: Vj, since 7jj- h2 = 0; but this is the direction-coefficient of the 11 diameters; hence all diameters of 118 ASYMPTOTIC DIRECTIONS. 119 t1le non-centric P meet it (it x, and hence can mect in only one point in finity. 98. If the coefficients of both powvers of 3 vanish, then 1)oth roots, 8, 8., are c ; the coefficient of 8 vanishes when :Lnd only when both xr' + hy' + g = O and I' +jy' +f = O; i.e., only when the origin is at the centre ; i.e., in H and E, not in P. Hence, two RLs. (Irawn throwugh the centre in directions fixed by k + 2 hs +jS2 = 0 meet the centric conic each in twopoints at . N ow thle points at co onl a RL. are consecu- tive; hence, these ILs. meet the centric conic at two consecu- tive points at oc ; i.e., they tocch it at oo. These RLs. throuigh the centre tangent to the centric conic at cc are narned Asymptotes ; they are real in H, imaginary in E. Their directions may be named asymptotic. All RLs. drawn in asymptotic directions, except the asymptotes, meet the conic in one finite, and one infinite, point; 1)oth points are real in H, imaginary in E. In the non-centric conic, P. the one diarnetral direction rep- resents two coincident asymptotic directions; all diameters of P meet the curve in one finite, and one infinite, point. 99. If E, E', C be ends and centre of a diameter, P any pole on it, P' thle section of the diameter with P's polar, then, by Art. 71, E, P, E', P' form an harmonic range; EP -E'P' _ -1, or EP: PE'= EP : E'PP. C E On compounding and dividing, results CP: CE- CE: CP'; i.e., The geometric mean of the central distances of anypole and its polar, measured on any diameter, is half that diameter. CO-ORDINATE GEOMETRY. In P the centre C and one end of the diameter, say E', retire to co; hence E' halves PP' outerly, and hence E halves PP' innerly. By definition the poles of a system of 11 chords lie on the diameter, D, conjugate to the chords, halving them; hence the tangents at the ends of any one of these chords meet on the diameter, D, in the pole of that chord; if the ehord be a diame- ter, D', its pole is the point at X on D, and the tangents through its ends are accordingly 11 to each other and to D; i.e., tangents at the ends of a diameter are 11 to its conjugate. In P this conjugate is at oc, but the tangent is still 11 to the diameter's conjugate chords. 100. We have reduced (Art. 94) the Eq. of centric conies, H and E; to reduce that of the non-centric, P, take as X-axis any diameter, as YBaxis the tangent through its end. Then the absolute vanishes, the origin being on the curve; the Eq. becomes a pure quadratic in y, since to any value of x must correspond two counter values of y, the chords 11 to the Y-axis being halved by the X-axis; the term in x2 vanishes, since, for any value of y, one of the x-roots of the Eq. must be x, one finite, the Il's to the X-axis, i.e., the diameters, meeting the curve in one point in finity, one in x ; there remain only the terms in x and y2, which may be written conveniently thus: y2 = 4 q'x, the Eq. of the P referred to a diameter and the tangent at its end. The figures on page 121 illustrate fully the foregoing articles. 10L1 Let us resume the study of the quadratic in S. The coefficient of 82 contains k, h, j, 9. an(l may be written 0(k, hj; 0); then the product of the roots 3,, 82 i.e., of the distances of (x', y') from the conic in the direction 0', is F(x', y'; x', y') : 0 (k, h, j ; Ad') 120 FIGURES. 121 122 CO-ORDINNATE GEOMETRY. The product of the distances of (x", y") in the same direc- tion is F(x" y" x", y") : -O(k, hj; O') - The ratio of these products F(xl, y'; xs', y') : F(XI.yit y; x"1, yF) is independent of 0', i.e., the same for all directions; i.e., the ratio of tihe products of the distances of any two points from a conic is constant for cal 11 directions of the distances. By taking two fixed directions 6', 6" and one arbitrary point (x', y'), instead of two fixed points (x', y'), (x", y") and one arbitrary direction 6', we get as ratio of the product of the distances O(k, hAj; 6'): O(k, h,j; 6"), -a result independent of the point (x', y'), the same for all points; i.e., the ratio of the product of the distances of a point from a conic, measured in two fixed directions, is constant for all points. The interest of these theorems lies mainly in the special cases: (1) Take the centre as the point; then the two distances in any direction are equal, being halves of a diameter; hence the ratio of the prodicts of the distances froin any point to a conic is the ratio of the squared diamieters 11 to the distances. Ff" \' (2) Take the directions tangent to the conic; then the dis- tances are again equal; taking the second root of the ratio of RATIO OF DISTANCE-PRODUCTS. the products, we see that the ratio of two tangent-lengths from a point to a conic equals the ratio of the 11 diameters. (3) Take as directions those of the diameter through the point and its conjugate chordl; then the distances on the chord are equal, and those on the diameter are its segments; hence the square of any chord varies ais the product of the segments into which it cuts its conjitgate diameter. See the figures. 102. By Art. 51 k, h, j are not changed by a change of origin; to find how they are changed by a chance of axial directions, we might use the general formulae of transformation (Art. 21); much neater, however, is this method of Boole: The transformation formulae being homogeneous in Cds., in passing from axes X, Y to axes X', I, inclined w resp. a', the expression kx + 2 hxy +jy2 changes into kFx2 + 2 hI XIy +jyI2, so that kx2 + 2 hxy +jy2 = k'a92 + 2 h Ix7yr +y'12. Also x2 + 2 xy cos w + y2 xX2 + 2 xty' cos WI + yl2, since each is the squared distance of the same point (x, y) (x', y') from the common origin. Add this Eq., multiplied by an arbitrary A, to the first; there results (k+,)x2 2(h+Cosw)xy +(j+ )y2 = (I'9+ A,)Xe2 + 2 (h'+ ,u cos Wr)Xlyt + (i'+ ,U)y'2. Each side of this Eq., equated to 0, represents the same locus: a pair of RILs. through the origin (Art. 44) ; if ,u be chosen so 123 CO-ORDINATE GEOMETRY. that the RLs. fall together, each side becomes a perfect square; i.e., the same values of .t make both sides perfect squares; i.e., the roots, p, p2, of the two Eqs.: (k+p)(j +P)=(h+ cosw)2 and (kl + p) (j'+ (h' + C WI) 2 are the same; i.e., corresponding ratios of the coefficients of the powers of P in the two Eqs. are equal; i.e., (k +j - 2h cos): sing = (k'+j'-2 h'coso,) : Sin V2, (7j _ h2): sin wB (kd- he) sin :m'; i.e., the ratios (k + j-2 h cos o) Sin: 2 and (kj-h2) : sin W are unchanged by any change of axes. Geometric Interpretation. Suppose the central Eq. of a centric conic brought to the form: ki + 2 hxy +jy2 = 1; then are _ _ the intercepts on the axes, and are half-diam- eters. 1. For w= 90, from the above, k+j is constant; i.e., the sum of the squared reciprocals of two rectang. diameters is constant. 2. For conjugate diameters taken as axes, h = O; hence ician ktjj 1 1 ___2 and __2_ are constant; hence their quotient - + -, or sin. sin k J the sum of two squared conjugate diameters, is constant. sinlw Also, by inverting and taking the second root, is con- stant; i.e., the area of the parallelogram of two conjugate half- diameters is constant. 124L THE FOUR CENTRICS. CHAPTER V. SPECIAL PROPERTIES OF CONICS. Centric Conics: Ellipse and Hyperbola. 103. The Eq. of the centric conic referred to conjugate diameters is k'x2 +j'y2 + ce = 0 (Art. 94). Two general cases present themselves: I. k' and j' like-signed, say both +; then the criterion C-= j - V = k' 0- 0; hence the curve is an ellipse. Under this head are two special cases: (1) c' 0; then the ellipse is real, denote it by E. (2) c' 0; then the ellipse is imaginary, denote it by E'. For clearly no real values of x and y satisfy its Eq. II. k' and j' unlike-signed, say k' + and j'- ; then the cri- terion C_ kj - 7t2 = kho - 0 0; hence the curve is an hyperbola. Under this head are two special cases: (1) cl0; then the hyperbola is primary, denote it by H. (2) c' 0; then the hyperbola is secondary, denote it by H'. Now write I for -I,' - on observing signs there a12 b f2 7 result these Eqs. of E, E', H, H', referred to conjugate diameters: X2 + y2 ____y2 a'2 b12 a b ' -2 2 2 1, 2 a'2 b12 a'2 b12 125 CO-ORDINATE GEOMETRY. To denote that the pair of rectang. conjugate-diameters, or the axes of the conic, are taken as Cd. axes, drop the primes fromn cc and b. 104. Thus far little reference has been made to figures, for the shapes of the curves were supposed unknown, and it was deemed important to illustrate how the properties of curves mav be deduced while their forms are +et unknown. Reason far outruns imagination. We may reason colTectly about forms we cannot imagine at all. But we may now find out the shapes and draw the figures of three of the above curves. Only the imaginary El is unrepresentable in our plane. Putting y = 0 in the Eqs. of E, H, H', there results x= a, x= +a, x= ia; i.e., all three cut the X-axis on each side a from the origin (centre): E and H in real points, H' in imaginary points. So, if x= 0, then y = b, y= ib, y= b; i.e., all three cut the Y-axis on each side b from the origin (centre): E and H' in real points, H in imaginary points. It is common to assume a b in the E; then 2 a and 2 b are called axes major atnd minor of the E. 2 a resp. 2 b is the real (commonly called trannserse) axis of H resp. H'; 2 ib resp. 2 iI is the imaginary axis of H resp. H'. The real axis 2 b of H' is often called, though hardly properly, the conjtigate axis of H. Plainly, like results hold when a and b are primed; i.e., when any pair of conjugate diameters are taken as Cd. axes: E cuts both in real points like-distant from the centre; H cuts only one of two conjugate diameters in real points, while IH cuts the other. Hence, while all the real ends of one system of diame- ters lie on H, all the real ends of their conjugates lie on H'; and conversely. Hence H' is commonly called the conjugate of H; strictly each is the conjugate of the other. Since the rectang. Eqs. are pure quadratics in both x and I. each curve is symmetric as to each of its axes. 126G THE CENTRICS TRACED. 127 Clearly a and b are the greatest values of x and y in E; a is the least value of x in H, b the least value of y in H. 105. If we pass to polar cds., putting p cos 0, p sin 0 for x, y, there results on reduction and inversion: 2 b2 l- b2 1-e2 cos 0 av,2 f or H. p2= -b2 where e- 2 1-e-cos t a 2 -b2 a2+b 2 for H'. P2 _ 7 , where e- 2+b 1-2eOS 02 a2 as central polar Eqs. of E, H, H', one side of the real axis of H being polar axis. That of H resp. H' is got from that of E by simply changing the sign of b2 resp. a2. The geometric mean- ing, of e2, used here for shortness, wilt be seen later. These Eqs. are pure quadratics both in p and in cosO, also OS02= cos - =0COS(7r - 0) 2, and two counter p's make a diameter; therefore, Diameters like-sloped to an axis are equal, and equal diameters (ire like-sloped to an ax is. 106. Let us trace E. For 0= 0, p = a; as 0 increases to 2-, p decreases to b; as 0 increases to 7r, p increases to a; as 0 increases to 3, p decreases to b; as 0 increases to 2 v, p increases to a. The greatest resp. least diameter is 2 a resp. 2b. In H, for 0 = 0, p =a; as 0 increases to cos- ,p e increases to o, all values of p in H' being meanwhile imnagi- nary; as 0 increases from cos-! to 7r and thence to 7r - cos-, c 2 e p in H' (lecreases from cc to b. and thenice increases to ao, all values of p in H being meanwhile imaginary; as 0 increases CO-ORDINATE GEOMETRY. from 7r- o--1os to 7r and thence to c + cos -1, p in H e e decreases from X to a, and thence increases to a, all values of p in /H being meantime imaginary; as 0 increases from vr + cos`- to and thence to 2 7r - cos1, p in AH decreases e 2 e from oc to b, and thence increases to a, all values of p in H being meantime imaginary; as 6 increases from 27r -cos 1 e to 2 7r, p in H decreases from X to a, all values of p in Hf being meantime imaginary. Y 0 a, x T'he two directions 0 = cos-1 with their counters corresuond to e and 0=7r-Cos-' -I e 128 I MAJOR AND MINOR CIRCLES. tanOa = + Ib and tan = which are the direction-coefficients of the asymptotes: X" = 0. e2 b2 Hence the two H's have common asymptotes, and along these asymptotes they close in upon each other at ao. 107. Solved as to y resp. x the Eq. of E is g - Vaa -H X rest). s,, = v _Y. a b Now Y, Vaa2_x2 resp. x,= Vb- y2 is the Eq. of a circle about the centre (origin), radius a resp. b, which may be called the major resp. minor circle of the E. For any value of x the corresponding values of y in the E and the major circle are in the ratio y,: ye = b: a. Hence the E is the orthogo- nal projection of its major circle under the W cos- 1b- a So the minor circle is a like projection of the E. Think the surfaces of E and the major circle made up of elementary trap- ezoids, or covered with threads I to the common diameter, cor- 129 130 CO-ORDINATE GEOMETRY. responding elements of the two surfaces will have the fixed ratio b: a; hence the whole areas will have that ratio; i.e., area of E= b ra2 = 7rab = V\ra2 _ rb2 a equwls the geometric meant of the areas of major and minor circles. The student can easily convince himself that the ratio of the projection of any plane area to the area projected is the cosine of the angle of projection. b The Eqs. of E and H solved as to y: y = - a .r- and y Vx( a-' declare that any ordinate is the fith part of the geometric mean of the segments into which it cuts the major axis; for the segments are a + x, a - x in the E, and x + a, x - a in the H. In the E the section is inner; in the H it is outer. For a = b, the E reduces to a circle or equiaxial E. The equiaxial H is x2 _ y2 =a2. It corresponds to the circle, and is called also equilateral or rectangular, since its asymptotes are I. It is congruent with its A': -_ =-a2 and falls on it when turned through 900. As in the circle, so in the equiaxial H, any two cnju gate diameters are equal. 108. Be Art. 93 the direction-coefficients of two conjugate diameters are connected by the relation, s' = __ k + s or, h +js o CONJ UGATE DIAMETERS. 131 if conjulgate dciameters be axes, and so h =0, ss=- _k i.e., their _prodeuct equcdls the negative ratio of the coefcients of X2 and y23; hence, in the present form of the Eqs. of E resp. H, 7J ,2 1_ f2 ss - ,- resp. ,ss' = - - or tan0-tan6'=-2 resp. = +-I . (t2( Hence tan 0 and tan 0' are unlike-signed in E, like-siglned in H i.e., 0 and 0' lie in udjjacedt quadraifts in! E, in the scmoe quad- rant in H; i.e., of two conjugate diameters of an E, one lies in first and third, one in second and fourth, quadrants ; but of an H, both lie in first and third or 1)oth iii second and fourth. Ju E, if taiO= b tan O= ;: i.e., when one of a' a two conjugate diameters of an E is one diagonal of the rectangle of the tanigents 1 to the axes, the other is the other diafgonal. This pair of diameters of the E are named eqgi-conjitgate. In H, if tan 0= bC, tan0' = t b; i.c., tzo conjugate diameters fall together on each diagonal of the rectangle of the tangent 11 to the axis ; i.e., on each of the asymptotes; hence each asymnptote is a self-conjgy(ate (liamfleter. In E, as 0 increases, 0' increases; in H, as 0 increases, 6' decreases. 109. The Eq. of the ta2gent to E or Hf is (the up)per sign going with E) 1 =1 or I2 fI 1 F b'2 (2 b2 The Eq. of the diameter through (.x, ) is - = -. Its conl- jugate is 11 to the tangent, and goes through the centre; hence its Eq. is X1X /l 0. aC h 132 CO-ORDINATE GEOMETRY. To find the Cds. xa, y, of an end of this conjugate, combine its Eq. with the Eq. of the curve, thus: -y2y,,a ir=1,o 22 a 2 b /2 i X 2y2Y2 or x12 b4 62 62x212tb2 a 2 For E resp. H the parenthesis t is +1 resp. -1; hence, for E, X2 = TF-Yls Y2= iXI b a for A, 2=iby1, Y2=i-bz . b a Again we see only one of two conjugate diameters has real ends on an H. 310. Plainly, if the x (or y) of one end of a diameter is known, the diameter itself is known as one of two equal diame- ters like-sloped to the X-axis; hence we can express any squared half-diameter through the x of its end. If (x, y,) be the end of a' in an E, then a2 = 2 + y 2 = 1 + b- (a2 _ X2) 2 + a bx2= 2 +e22. a 2a By Art. 102 a'2 + 612 = a2 + b2, hence bF2 = a2 - e2X2. Now b6=x4+y22; hence, by Art. 109, a2+ 6,2 a2- 2 = a' - e2x-2. 62 a-1 Changing the signs of b2 and b62, we get for the H, a'2 = _b2 + e2x12 -612 = a2 - e2X12 = 2 +b 2) 111. Bv Art. 102 the area of the parallelogram of two conju- gate half-diameters. and therefore its fourfold: the area of the parallelogram of the tangents through the ends of two conju- CONJUGATE DIAMETERS. gate diameters, is constant. If 4, be one 4 between these diam- eters. then this area is 4 a'b' sin 4, and = 4 ab; whence ab sin i = ,a'b' Hence 4 is least when atb' is greatest; and a'b' is greatest when a2 + b2 + 2a'b', which = a"2 + be2 - 2 a'b', which = (a' - bW)2, is least; i.e., when a' = b'; i.e., when the diameters are the equi-conjugates. For equi-conjugates, i 4 = ab 2 ab a'b' a2 +b2 Of course this last reasoning is necessary and applicable only in case of the E; in the H the 4 be- tween conjugates is least when they fall together in an asymptote. If 2 a', 2 b' be two conjugate diam- eters, 2p the distance between the tangents 1i to 2a', then the constant area is 2a'. 2p = 4 ab, or p = ab; i.e., the distance from the centre to a tangent is a fourth propor- tlonal to the half-diameter 11 to the tangent and the half-axes. 133 134 CO-ORDINATE GEOMETRY. 112. The definition of the normal (Art. 73) yields as its Eq. a2 b2 a(2b2 aX ( _xi) bF (y - y) = O ; or, - -T - - ry = a2 T: b2. XI Y, XIYi The intercepts of the tangent on the axes are - and -; those of the normal are 12 T b- a2 XI a n da and __ be The product of corresponding constant: (a2 F b2). intercepts of T and N is a The intercepts on the tanyent between the point of touch and the X- resp. Y-axis may be named X- resp. Y-tanfgent (lengths). To like intercepts on the normal like names are given. The projections of these tangent and normal-lengths, each on it. own axis, are named sub-tangents and sub-normals. The following table needs no explanation: X-subtangent - _X - a=X= X1 b2Yl; b2_ b-"-y = b2X, Y-subtangent =b Y1 = .xi _; YI yI aAy, (1) (2) TANGENT AND NORMAL. I -AT S Ti a2T b2 bx X-subnormal = x,2- X1 = -l a2 a2 (3) Y-subnormal = Yi aL2 b2 a2 - Tb2 Yi= i ; L--tangent = y 2 + !4 2 y2 y a I'=1 __ t " 2 X-tangent 1 b4XI bx1 a2 + a2 Y12 =- 671 . b'; bx1 2I + b4 XI2I=bx 5 a22 2 P Y-tanogent = 12 + _4 2 +-2 12 I. ay1 I ay, b' a I = ayl . b'; ay, X-normal Y-normal = Iy2 + b4 2j ti- = . b ; Cl = a2+64 a 2 ,2 L+ ,2 = 1 X,2+ y j I =I X 2 - lf b (4) (5) (6) (7) (8) 135 bj a 2y2+ b 2X 2 I'll - - I - I a b2 W CO-ORDINATE GEOMETRY. Hence these results are evident: (1) Product of ST's = product of SN's = product of Cds. of point of touch. (2) Product of T's = product of N's = squared half-diam- eter 11 to tangent. (3) Product of X- resp. Y-normal by central distance of tan- gent = b2 resp. a:. 113. As the tangent is but a special case of the polar, so the normal may be subsumed under the more general concept of a I through the pole to the polar. In lieu of a better, give this I the name Perpolar. Since the Eq. of the polar has the same form as that of tfie tangent, the Eq. of the per- polar has the same form as that of the normal. As the pole glides along a RL., the polar turns about (envelopes, en- wraps) a point; but in the Eq. of the perpolar, which may be written a2ylx w: b-.xy = (a' 2 b2 )Xyl, the parameters r,, Yi do not appear linearly; hence, when con- nected by some linear relation, it will not in general be possible to eliminate one from the Eq. of the perpolar and leave the other in first degree only; i.e., as the pole (x1. Yi) glides along a RL., the perpolar will not in general turn about a point, but about some curve. But in three cases it is possible: when xl is constant, when yi is constant, when y4 :xl is constant; i.e., when the pole mores on a RL. 11 to either axis or through the centre, the perpolar turns about a point, the perpole of the RL. If the RL. be 11 to the Y-axis, x1 is constant, and the pole of the RL is on the X-axis, distant a2: x1 from the centre; then the perpole is also on the X-axis, distant e2x1 from the centre. (This is seen at once on writing the Eq. of the perpolar thus: 136 FOCI OF THE CURVE. 137 b2xiy + y1(a2X -a 2aT b-'x). For x, constant, this is the Eq. of a pencil of RLs. whose base-lines are y = 0, i.e., the X-axis, and a2x-(a2 T b2)x O, i.e., X=-2 x= e2x1; a2 . . the perpole is [e2'x, 0]). The product of these two central distances is the constant a2e2; hence, poles and peipoles of RLs. 11 to the AY-axis form an involution of points on the X-axis, whose centre is the centre of the conic, whose foci are distant ae from the centre. Likewise it is proved that poles and perpoles of RLs. 11 to the X-axis are in involution on the Y-axis, but the constant product of distances of a pair from the centre is - ae-. Hence if either pair of foci are real, the other are imaginary. The student will readily see that poles and perpoles of all RLs. through the centre lie on the RL. at x. 114. The foci of the involutions on the axes are called foci of the curve; hence a centric conic has four foci: two real, two irnayinary. They enjoy important properties. The central distance, ae, of a real focus is called the linear eccentricity of the conic; e itself is the eccentricity proper. It is the ratio of the central distance of a focus to the half-axis on which the focus lies. Now in case of the imaginary focus the central distance of the focus is imaginary in both E and H; but the half-axis is real in E and imaginary in H; hence their quo- tient is ima-inarv in Ebut real in H; i.e., one eccenbtricy is real, one iniayinary in E. both are real in H. The polar of a focus is called a Directrix. Suppose the real foci (ae, 0), (- ae, 0) on the X-axis; the directrices are a e By Art. 113 a focus is a double pOimt in which have fallen together I)ole and perpole of a certain RlL., the directrix. As a pole glides aloIg this directrix, both its 1)olah and its perpolar turn about the focus always I to each other. CO-ORDINATE GEOMETRY. Call the tract from a focus to a point a focal radius of that point, and ally RL. through a focus a focal chord. From any point P of the directrix draw a chord cutting the conic at I an(l I' to the pole Q of this chord, andI Q from the focus F, draw FQ cut- ting the chord at P. By Art. 71, since the polar of P' is FQ, P', 1', P, I form an harmonic range; Ihence FIP'J'PJ is an harmonic pencil. But FP is dlearly the per- polar of Pt, as this perpolar must go through P' aimd through F. Hence FP and FP alre I; hence they halve the 4s of Fl and F' (Art. 41). Hence, the focal radius of thee pole of a chord halves the 4 at the focus subtended by the chord. 115. By Art. 113 polar and perpolar cut the axis of involu- tion in a pair of conjugate points, harmonic with the foci; hence the focal radii of the intersection of polar and perpolar form with these two an harmonic pencil ; and since these two are I, they halue the angles beticeen the focal radii. When polar and perpolar are tangent and normal, their inter- section is the pole, the point of tangence on the conic; hence tangent and normal halve the 4s of the focal radii of the point of touch. The normal halves the inner resp. outer 4 in the E resp. H; hence an E and an H with the same foci, i.e., confocal, are I to each other. 116. These relations of position imply several relations of size: the intercept between the foci being 2 ae, the X-intercept of the normal being e2m (Art. 112), the segments of the focal intercept are ae + e2x,, ae - e2x, in E, where the normal cuts it innerly, and e2x1 + ae, eax1 - ae in H, where the normal cuts it outerlv. 138 PERPOLES AND PERPOLARS. 139 Hence, as the focal radii r, r' are proportional to these segiments, r _a+ ex, r ex, +a r- 1 in the E, and r, in the H; r, C- ex, rr ex, -a or r+r__ in the E, and r+t = ex, in the Af. r-r' 2ex1 r-r' 2a But, plainly, r- r_2 = (ce + x 2 - (ae -_) 4 aex, or (r+r')(r-r')= 2a.2ex,; hence r + r'= 2 a, r r'= 2 ex1, r =a + ex,, r'= a- ex1, rr'= a2 _ e2XJ2 _ V resp. r + r'= 2ex, r -r'= 2a, = ex, + a, r'- ex, - a, rrf= e2x12 - a2 = - Or, in the E resp. H, the sum resp. diference of the focal radii of a point is a constant, namely, the cnajor resp. real axis; and in both E and H the product of the focal radii of Ca pol.nt is the squcared half-diameter conjaysate to the diameter through the point. The distance of (x1, y,) on the E from the directrix x = a e a is clearlV- x_ , and the distance of the same point from the focus is a - ex,; i.e., in the E the ratio ofthe distacces of c point froiTnfocus candl directrix is Ca constmit, the eccentricity. Plainly the like holds for the H. In E this ratio is 1, in H it is 1. Let the student show that the locus of a 1)oiut the sum resp. diffei-ence of whose distances from two fixed points is constant is an E resp. H; also, the locus of a point the ratio of whose distances from a fixed point and a fixed RL. is a constant 1 resp. 1 is an E resp. H. The sum resp. difference of the focal Is on a tangent is twice b the central I on the tangent; i.e., it is 2a - b; the sum resp. difference of the focal radii is 2 a; also, the ratio of focal I CO-ORDINATE GEOMETRY. to focal radius is the same for the two foci, being the sine of the slope of radius to tangent; hence this ratio, this sine is by Dividing the central 1 on the tangent by this ratio, we get the central distance to a tangent, meas tred 11 to a foeal raditus to the point of touch, namely, a. Hence the locus of the section 140 ASYMPTOTIC PROPERTIES. of a tangent and a diameter 11 to a focal radius to the point of touch is the major circle. The same major circle is the locus of the foot of the focal I on the tangent; for the As PF'N, MOFare similar, since OF = e and FIX= ae_ e2xi OM F' a-ex Asymptotic Properties. 117. Thus far the properties of E and H have corresponded; but the asymptotic properties of the H have no real correspon- dents in the E, as the asymptotes of the E are imaginary. Accordingly, in what follows, reference is to the H alone. If XI- , = 1 be the H, -2 = are its asymptotes. aF2 bf2 a1 b2 One of these, ,- = 0, or = is clearly the a b' y b' central diagonal of the parallelogram of the conjugate half- diameters a', b'; the other is the central 11 to the other diagonal, x + Y = 1. a b' Hence, given a pair of conjugate diameters, we can find the asymptotes; or, given the asymptotes, we can find the conjugate to any given diameter. x 141 142 CO-ORDINATE GEOMETRY. By Art. 42 the RLs. x=O, y=O; y= b,x, y= ba i.e., the asymptotes and any pair of conjugate diameters form an harmon ic pencil. Hence the asymptotic intercept of a 1[ to any diameter is haled by its conjugate diameter, and the inter- cept between two conjugate diameters of a 11 to one asymptote is halved by the other. As a special case, the asymptotic inter- cept of a tanigent is halved at the point of tan gence. Since the same conjugate diameter that halves the intercept between the asymptotes also halves the chord of the curve, clearly the intercepts between the asymptotes and curve are =, or, CS= C'S'. From the Eqs. of the curve and the asymptotes there follows: b' M is = Mns' =bFX YA = inC= inC'= a'x 1- 2 a '. y- ,= CS = CIS'= It'gc-If2 Ya +yA=CS'= C'S = b' + X ; CS - CS'= C'S' C'S = b'2; i.e., the product of the distances of any point of an H to the asymp- totes, in any direction, equals the squared 11 half-diameters of tihe H. Clearly CS can be made large, and so CS' small, at will. 118. The tangent-intercept between the asymptotes TT' being halved at the point of touch A', the A TOT' = 2 TOA'= the parallelogram A'OB'T of the conjugate half-diameters, a' Hence coojiqate diametersformn an Involution of WNich the asylmptotes are the double or focarlays. Like may be easily proved of the conjugate diameters and imaginary asymptotes of the E by noting Art. 107 and form- ing the Determinant of Art. 47. POLARD EQUATIONS. and b', i.e., = the constant ab. From A1' draw to each asymp- tote a 11 to the other, and call them u and v ; they are the CUs. of A', the asymptotes being axes. With the asymptotes they form a parallelogram which is clearly half the A TOT'; hence, if 0 be the : of the asymptotes, we have uv sin + = ab: 2. But by Art. Il1, since the asymptotes of H fall on the equi- conjugate diameters of E, sin= _2 ab UV cla2+b2 a2+ b 4 This, the Eq. of the H referred to its asymptotes, says the parallelogram of asymptotic Cds. of a point is of constant area. Focal Is on the asymptotes are clearly equal; the asymptotes being tangents, their product is - 62(Art. 112); hence each is b in length, but they are counter-directed. This is also plain at once from trigonometric considerations. Polar Equation of Centric Conic. 119. Take the right focus as pole, the riaht X-direction as polar axis; then, by Art. 11;6, p = a-exc, resp. p = ex-a in E resp. H; x being here reckoned from the centre, x=ae+pcos9; a(l -el) _bliees hence P= 1-2 = b l+eeosO 1+ecosO a resp. p= e-l) = :1-ecos9. 1-ecosO a The expressions for the right focal radius being unlike in E and H, we might have expected these Eqs. of E andA H to turn out unlike. This makes the Eqs. just found unhandy. But the one expression for the left focal radius is p = a + ex. 143 CO-ORDINATE GEOMETRY. The left X-direction being taken as polar axis and 9 reckoned clockwise being taken as positire, we hav e x =-(ae + p cos 9); 1- = - b2 1+ e cos9 in E resp. H. 1 +ecosO a These Eqs. of E and H, like the central Eqs., differ only in thesignof b2. For 0= - or -, p= . Hence 2 is 2 2 a a the focal chord I to the axis; it is called the parameter or latus rectum. It is interesting to trace the curve from the polar Eq. The corresponding values of p are unlike-signed in E and H; in E p is measured bounding the :s 0, in H it is measured counter as long as p falls out negative. The value of e is not the same in the two Eqs. ; in E it is 1, hence the divisor 1 + e cos9 remains throughout + and finite; but not so in H. For 9=0, p=a-ae; in Ethis is +, but in H it is hence it is reckoned leftward in E, rightward in H. As 9 increases to cos(, _), p traces out (by its end) the left lower e branch of hA. For 9= cos '(-Dl p =- , and is drawn in the left lower asymptotic direction. Just here p changes sign, and beginning at + x, traces out the right upper branch of H till 9 = ir, when p sinks to a +ae at the right vertex; thence staying +, it rises to + x, tracing out the lower right branch of H till 9 reaches 7r + cos- l-. Here again p changes sign, becoming - oc, and as 0 increases to 2 7r, it traces out the left upper branch of H, reaching the left vertex as it reaches 2 7r. The student himself may readily follow the course in the E. It is noteworthy that the right upper branch of H is thus seen to be continuous (in cc ) with the left lower, and the right lower with the left upper. We are forced to think the H thus by this reasoning also: The asymptote touches the H at c in two counter directions; hence unless it meets each branch in the same two consecutive points at xc. it must meet H in four points, 144 NON-CENTRIIC CONIC. which is impossible. While, then, to imagination the H con- sists of two distinct branches, to reason it consists of one branch closed in two directions (the asymptotic) at x. Non-Centric Conic: Parabola. 120. By referring P to a diameter and the tangent through its end as X- and Y-asxes, its Eq. is brought to the simplest form, i , oflfl\ - - 4 -I- IAn A, -Iv AM Y U 1 K kAAALL. AL',J'IJ) One and only one diameter is -l to its conjugates (Art. 94); it is called principal dianeter or axis of P. The Eq. of P referred to these Is is called the vertical Eq. of P, the origin being the vertex, and is written y2 4 qxu; 4 9' is called par- amneter of the corresponding diameter, 4 g is )rincipal parameter, or simply parameter. From these Eqs. we may now draw out all the properties of the P, as is done in most texts; but another method seems directer. According as 17 - 7j is 0, _ 0, 0, the conic is E, P, H. Hence, any two of the symbols k, h, j being held fast, as the other changes, the conic becomes in turn an E of this or that 145 146 CO-ORDINATE GEOMETRY. shape, a P, and an H of this or that shape. P is .thus seen to be a critical curve between E's and H's, a border or limit of the two. What kind of a limit, we shall sec. For this investigation of the relation of P to E on one hand and H on the other, the central Eq. of E and H is ill suited, as the central Eq. of P is unmanageable, the centre being at a. Since the vertical is the simplest Eq. of P, let us move the origin to (say) the left vertex in E and H. The Eq. becomes x - d y2i y2 b2 x- iY =1 or y2 l (2 (i x -2). a2 V o (12 be Here =- A=O, j= 1; hence, if either E or H 62 h is to pass over into P, bW must vanish to make -Aj = 0; 26b2 and a must stay finite to make the Eq. y2 =4q7x. Now 26b2 b2 or 4 q is the focal chord I to the axis; hence - or 2 q is a a b2 the ordinate at the focus, yr, and - or q is the focal abscissa A, 2 a or distance of the focus from the vertex. Accordingly, we keep finite by holding the focus and vertex fixed ; to make - a a vanish, we must let a increase toward x ; i.e., let the centre, and with it the otherfocus, retire to a:. But then a2 4:b2 b2 o e2=. a2Fb= I -T b = 1, or en_1 a2 a Hence we may treat P as an E (or H) whose parameter has kept constant while its centre andl one focus have retired to x, or as an. E (or H) whose eccentricity has increased (or decreased) to ]. The properties of P are the properties of E (or H) at this limit. viz.: 121. P is symmetric as to its axis (Art. 104). P is the locus of a point equidistantfrom focus and directrix. P AS THE LIMIT OF E AND H. The vertex (origin) is distant q from the directrix; hence any point (Xa, Yi) of P is distant x1 + q fronm focus and from direc- trix. The poles of all 11 chords lie on the diameter of those chords (Art. 98). Also a pole P, the intersection 31 of its polar and the diameter through it, and the intersections I, I' of the diameter and the curve, form an harmonic range; and as ' is at A, I is midwcay between P and M; i.e., the intercept on a diameter between a pole and its polar is halved by the curve P. If the diameter be the axis, I is the vertex, and the inter- cept is the subtangent; i.e., the subtangent is halved at the vertex. Hence the subtangent is 2 x' long; also it is cut by the focus into segments xA + q and x1 - q. Hence the focal distance of the intersection of tangent and X-axis (axis of P) = the focal dis- tance of the point of tangence= x + q. Hence the focal I on the tangent halves the tangent-length; so too does the vertical tangent; hence the locus of the foot of the focal 1 on the tangent is the vertical tangent. Further, clearly the focus halves the distance between the intersections of tangent and normal with the axis; hence the 14L7 CO-ORDINATE GEOMETRY. whole intercept is 2 (x, + q); on taking away the subtangent 2 x1 there remains the subnormal = 2 q; i.e., the subnormal in P is the constant half-parameter. Plainly a circle about the focus and through the point of touch goes also through the intersections of tangent and normal with the axis. Polar and perpolar and focal rays through their intersection form an harmonic pencil (Art. 114). The second focus being at, x, the second focal ray is the diameter through the intersec- tion. Hence any polar and iperpolcar halve the 4s between the focal ray and the diameter through their intersection. As a special case, the tangent and normal halve the :fs between the focal ray and the diameter through anypoint of a P. Hence, too, since the diameter meets the axis at A, the focus halves the axial intercept between the conju gates: polar and perpolar, or specially, tangent and normal. This has already been proved geometrically in the special case. 122. All these relations are readily drawn out from the Eqs. of taucent and normal: yy, = 2 q(x + x1) and (y-y1) 2 q + 1y(x --x) = 0. a useful exercise left for the student. Eliminating x1 by the relation y12 =4 axe, we get yyi = 2 qx + and (y-y9)2 q + yj - 0- 2 '\( 4q Solved as to Yi' x and y being treated as known, these Eqs. vield two resp. three roots, values of y, i.e., ordinates of the points where tangents resp. normals drawn through (x, y) meet the P; hence, through any point may be drawn two tangents and three normals to P. The tangents are real and separate, real and coincident, or imaginary, according as y2-4qxis O, = O, or O; i.e., according as the point from which they are drawn be with- out, upon, or within the P. The sum of the roots is 2 y; i.e., 148 NORMALS TO P. the ordinate of the point through which the tangents are drawn is the half-sum of the ordinates of the points of touch; i.e., the point is on the diameter of the chord of contact, as already known. The reduced Eq. of the normal is Y1z + 4q (2 q-x)yl-8q2y = 0. The absence of the term in y'2 shows that its coefficient, the sum of the roots, vanishes; or, Y1/ + Y1" + Ultrl = 0; i.e., the sum of the ordinates of the points where three norrmals to a P. througyh a point, meet the P is 0; or each is the negative sum of the other two. The sum of tile ordinates of the ends of 11 chords is constant, namely, the double ordinate of their diam- eter; hence all third nornmals through the intersections of pairs of normals at the ends of 11 chords cut the P at points hav- ing the same y1, i.e., at the same point; i.e., are the same normal; i.e., pairs of normals at the ends of 11 chords meet on a third normal. To find this normal, draw one of the 11 chords through the vertex:; then the point symmetric as to the axis with the other end of the chord is the point of P through which the third normal goes. One of the normals is always real, wherever (x, y) he taken; the other two are real and sel)arate, real and coincident, or imaginary, according as y'- 4qx0, =0, or 0; i.e., according as the point they are dr-awn from is within, upon, or without the P. This is also clear geometrically, since plainly real intersections of normals take place only within the curve. 123. In the vertical Eq. of P, y2 = 4qx, the principal parameter 4 q is the double focal ordinate, or focal chord. In the Eq., y2 = 4 q'x, of P referred to any diameter and the tangent at its end, on putting x = q', the parameter 4 q' also appears as a double ordinate; but is it also a focal chord Be p the focal ray to the origin or point of tangenec, 0, o the tangent's slope to the axis, p' the focal ray 11 to the tangent 149 CO-ORDINATE (;EOMETRY. and sloped w, p1" the counter-ray sloped 7r + o ; then p = x' + q, p'=2q:l-coso, p"=2q:1 +cosw, p'+p"=4q:sin . The second Eq. is got from the polar Eq. of the E by putting e = 1, 0 = 7- ; the third from the second by putting + X for o). Clearly p' + p" or 4 q: sin (,) is the focal chord 11 to the tangent ; the x of this chord is the focal distance of the inter- section of tangent and axis, i.e., the focal distance of the point of tangence, i.e.; p. Now project p on the focal I to the tangent, and project this projection on the axis; by Art. 121 the last pro- jection is q; i.e., p = q: sin . 4 p =p' + p =focal chord. Hence the abscissa to the focal ordinate is half that ordinate; but this is the property of the abscissa q'. Hence 4 q' is the . 2 double focal ordinate, or focal chord, and 4 q: sin oj, where w is the axial angle or slope of the tangent. 150 MAGIC EQUATIONS. CHAPTER VI. SPECIAL METHODS AND PROBLEMS. Magic Equations of Tangents and Normals. 124. Thus far the Eq. of the tanigent has been expressed through the Cds. of the point of touch. This Eq. does not in itself determine the tangent, but only by help of an understood Eq. of cowlition declaring the point of touch to be on the curve; without this latter, it were the more general Eq. of a polar. Thus, YiY = 2q(xl + x) touches the P y2 = 4 qx only in case (xj, yi) be on P, i.e., only in case Yi2 = 4 qxl; other- wise, it is but the polar of (x1, yl) as to the P y2 = 4 qx. This implied Eq. greatly cumbers operations about the tangent. Where not the point of tangence but only the direction of the tangent is involved, this cumbranice may be avoided by express- ing the Eq. of the tangent through the direction-coefficient s as the parameter of the Eq. This form of the Eq. of the tangent is called the magic equation of the tangent. It may be got by putting for y its value sx + d in the general Eq. of the conic, and expressing the condition of equal roots of the quadratic in x, whence may be found d in terms of s. But this is tedious. Better is it to get the special forms for E, H, P separately. Thus, after the above substitution in y2 = 4 qx, the roots are equal when s2d2= (sd - 2 q)2, or when d = q: s; q ..y=sxc+- is the magic Eq. sought; s being thought as changing, it is the Eq. of a family of RLs. touching the P y2 = 4 qx. In finding the magic Eq. for the E, we may exemplify another method. Solved as to y the ordinary Eq. is 151 CO-ORDINATE GEOMETRY. b2 x, b2 Y=- . .+ _. a' Yi Yi Here s _b2x a Yi Y, square s, multiply by a2, add b2; results d2 = s2a2 + W2. Hence y = Sx Vs\a& + b2 resp. y = sx V/s2a2-b2 is the magic Eq. for E resp. H. N.B. The steps in this elimination are suggested by the reflection that ayl2 + b2x12 =a2b2. 125. Magic Eqs. of normals are easy to find. Thus, in P, 2q 2q q Y=-I= XI q.- Yi 8 8 Substituting in the Eq. of the normal Y-Y,= -s(x-x), and writing s for--, we get as Eq. sought qs3 + (2q-x)s + y = O. In E, y, =b 2: V8a2 + b2, whence x= - s2a2: \/s2a2 + b2; whence, on substituting as in case of P, there results (y- SX)2 = s2(b2 - a2)2 (2b2 +a2) as Eq. of normal to E. Changing 62 to - b6, we get a like Eq. of normal to H. 152 THE ECCENTRIC ANGLE. These Eqs. are of fourth degree in s; hence may De drawn from any point four normals to an E or all H. 126. The use of the magic Eq. may be illustrated in finding the locus of the intersection of a pair of 1 tangents to an E: ysx+Vsa2 + b2 and y=_-1 b2. Clear, transpose, square, sum. and divide by 1 + S2; results xc2+y2=a2 +b2; or, in case of H, b2+y2=a2_b2 These are named director-circles of E resp. H. If E and H be co-axial, the D.C. of each goes through the foci of the other. Show that the D.C. of P is the directrix. The Eccentric Angle. X2 Y2 127. By Art. 107 the E + 12=1 is the vertical a2 b2 shadow or 11 projection of the circle x2 + y2 = a2. In this projection chords II to the plane of the E are projected at full length, chords 1L to these are shortened in the ratio b: a, every other system of 11 chords are shortened in some ratio between 1 and b :a. Clearly the centre of the circle is projected into the centre of the E, hence the diameters of the circle into the diameters of the E. A pair of I diameters. in the circle are conjugate, each halving all chordls 11 to the other; hence their projections are conjugate diameters of the E, each halving all chords 11 to the other. Call the diameter 11 to the plane of the E, which is projected into the axis major of the E, the axis of the circle; then the 4 which any diameter makes with this axis is called the eccentric magle of the projection of that diameter. The eccentric 4 of a point of the E is the eccentric 4 of the diameter throtugh it. Hence the eccentric 4s of two conjugate diameters differ by 90. 153 154 CO-ORDINATE GEOMETRY. If the projected circle be turned round its axis through Cost- at it will fall on the major circle of the E. If - be the eccen- p tric 4 of (x, y) on the E, then the Q/ Cds. (x', y') of the corresponding e point of the circle are xS=aCosc, y'=asinl; hence x =a cosE, y =bsinc are the Eqs. of the E in terms of E. 128. The eccentric 4 is especially useful in dealing with chords and tangents. Thus the chord through l, E, is x y 1 =0, a cos cl bsin5 1 1 a Cos c2 b sinq9 1 which on reduction takes the form -cos C, + S2 + - sin '+= cOS 1 C2 a 2 b 2 2 Putting c = c2= e, we get the Eq. of the tangent at E: -COsE+ sinfe = 1. (l b Replacing x, and y1 in the Eq. of the normal, we get its Eq. a x _by = a2 b 2. COS cl sin cl The advantage of these Eqs. lies in the fact that the arbi- trarv E's are free from condition. We may illustrate their use in findingo the locus of the intersection of the pair of tangents at the ends of conjugate diameters. Such a pair are -cosE+-sine= 1 and --sinE + -COS= 1. a b a b QUASI-ECCENTRIC EQUATION OF H. 155 Finding hence cos e and sin (, an(l placing the sum of their squares equal to 1, we get, on reducing, + y2- 2au2 262- a co-axial E with axes multiplied by v'2. 129. We have seen that equiaxial H is to H in general as the circle (eqtiaxial E) is to E in general. Any H may be thought as a vertical shadow or parallel projection of an cquiaxial H under cos-b (a b). If a b, we may think the relation a reversed: the equiaxial H the projection of H in general. Now since see,- tan 2_= 1, if we put x=asecq. as we may, we must have y = a tan y in equiaxial H. or y = btan, in H in general. These, then, are the Eqs. of H in terms of a. -- P To construct 7 we have but to form a right A with base a and hlypoteuause x; the 4 at the base will be vq. This is done by drawinig froml the end of the abscissa x a tangent to the major circle of the H. The tangent-length is the corresponding y in the equiaxial H and the (-th part of the y in the H in general. b 156 CO-ORDINATE GEOMETRY. The equiaxial H' y2 - x 1 is got by exchanging x and y in the equialialA xH -y2= 1; hence its Eqs. are x=atanq, y=asece. In the generil H' a22 -1, y is changed in the ratio Z a2 b2 b while x is unchanged; hence the Eqs. of it are a x= a tany, y = bsec q. In the equiaxial H's conjugate diameters are equal and like- sloped, the one to the X-, the other to the Y-axis; hence the points corresponding to like values of v in the two pairs of Eqs. are ends of conjugate diameters; after projection conjugate diameters remain conjugate, each still halving all chords 11 to the other; hence like values of v yield ends of conjugate diam- eters in the pairs of general Eqs. Noting the signs of the trigonometric functions, we see that zr ranging from 0 to - yields all points in the right upper 2 branches of H and H'; -q from 7 to 7r yields all points on the 2 left lower branches of both; -q from 7r to 3 -yields all on the left upper branch of H and the right lower branch of H'; z from 3 to 2 -r yields all on the right lower branch of H and the left upper branch of H'. Hence the ends of conjugate diame- ters, answering to like values of -q, will be in the same quadrant for v, between 0 and 7r, but in counter-quadrants for -q between and 2 7r. Observing this, we find as Eqs. of tangents through ends of conjugate diameters in first and third quadrants x y x - see v-q tan v = 1, -- tany - _- seecq -1 a b a b whence, on addition to eliminate -, we get x__IY o a b HYPERBOLIC FUNCTIONS. 157 i.e., one asymptote, as locus of the intersection of the tan- gents. Putting X- for r, in the second Eq., we get - tan,+ sec 1=-1 as Eq. of the tangent through the other end of the second diam- eter, whence +M -0, a b i.e., the second asymptote, results as locus of the intersection. 130. There is another noteworthy way of expressing the Cds. of a point on an H through a third variable. As the student may know, sine and cosine may be defined analytically, without any geometric reference, through exponentials, thus: cos =i.(eio + e-) , sin = 1 (ei0-e-io), 2 i where i.i=-1. All properties of sine and cosine may be drawn out from these definitions with greater ease and generality than from any other. If instead of i be written 1, the resulting expressions -(eO +e-0) and i-(e - e-) are named resp. hyperbolic cosine and sine of 0, and may be written -c0 and WsO. We see at once that hc0 - 1tS = 1, and hence we may write in H X = -co, Y' = hsO, or x=ahcO, y =bhsO, at b and in H' x = a hs, y = biC0; the analogy of which to the eccentric Eqs. of the E is plain. Hyperbolic functions are of some use in higher analysis, and these Eqs. are of interest in Kinematic. CO-ORDINATE GEOMETRY. Supplemental Chords. 131. Two chords through a point of an E or H and the ends of a diameter are called supplemental. They are 11 to a pair of conjugate diameters, for a diameter halving one is clearly 11 to the other. Hence, if on any diameter of the conic as a chord be described a circle-segment containing a given angle, and a point where this circle-segment cuts the conic be joined to the ends of the diameter, the diameters 11 to these chords will be conju- gate and inclined at the given angle. The problem of drawing conjugate diameters making a given W with each other is thus solved and soluble only when the circle-segment meets the conic in real points. As these points will in general be two, there are in general two pairs of conjugate diameters havin(g a given slope to each other. Auxiliary Circles. 132. Of these have already been found several, as: (1) and (2), the major and minor circles (Art. 107); the major is the locus of the foot of the focal I on the tangent / \ (Art. 116). (3) The director-circle X2 + f = LL2 b2 (+ in E, - in H), being the r locus of the intersection of pairs of I tangents (Art. 125). (4) To these we may now add the two counter-circles. If either focal radius of any point of an E resp. N be lengthened resp. shortened by the length of the other, the point thus reached will clearly lie on a circle about the first focus, radius 2 a. Since the tangent halves 158 AUXILIARY CIRCLES. the angle of the focal radii, the point and the other will be symmetric as to the tangent, or the point will be the counter- point of that focus as to the tangent. H-ence the locus of the counter-point of either focus is the counter-circle about the other focus. Also, the counter-point of either focus (as to any tan- gent), the )oint of tangence, and the otherfocus lie on a RL. (5) The system of focal circles. Of these the X-axis is the common power-line, the Y-axis is the centre-line. Any one meets the tangents through its centre on the vertical tangents (at the ends of the major or real axis). For in the E the radius of such a circle is Va2 + b2 cotE2, E being the eccentric W of the point of tangence of a tangent through its centre; also the intercept of such a tangent between the vertical tangents is 2Va 2+ b2 cotE2. Like reasoning holds for the H, cot E chang- ing to cosed. These circles are helpful in problems of construction. 133. In P the minor circle lies wholly in co; the major reduces to the vertical tangent, the locus of the foot of the focal I on the tangent (Art. 121) ; the director-circle becomes the directrix, the locus of the intersection of pairs of I tangents (Art. 106) ; so, too, does the counter-circle of the focus, since plainly the counterpoints of the focus lie on the directrix (Art. 121). All this the student may also prove analytically by pass- ing to the vertex as origin and reducing the Eqs., remembering that in P, a= _ X, e= 1 = 42= . The focal circles all a reduce to the axis of P, since the other focus is at x, and so are little useful in construction. Their place is filled in a measure by the circles about As circumscribed about P, all of which pass through the focus, as may thus be proved. Let three tangents touch at P1, P2, Pa, and meet by twos at 13, 11, by Art. 1 14, P P1FI3= =I3FP2, and : P2FIA= I1FP3; 159J CO-ORDINATE GEOMETRY. whence 13F11 ,F1-P2 + W P2F11 - I P1FI3 + 13FP2 + 172F1 + I1FP -J+PFP3; i.e., the intercept of any tangent between two fixed tangents to P subtends a fixed angle at the focus: half the : subtended by the chord through the fixed points of tangence. Now as the focal I on the tangent meets it on the vertical tangent, the slope of the L to the axis = the slope of the tangent to the vertical tangent; but by Art. 121 the former is half the slope of focal radius to the point of tangence; hence the difference of the slopes of two tangents to the vertical tangent, i.e., the W between two tangents, is half the angle between the focal radii to the points of tangence. On appleying this to the case in hand, it appears that the Xs 1113 and ITF11 are supplementary; hence the circle about the A 1j1,2 goes through F; Q.F..V. Circles circumscrib- ing circumscribed As we may namefocal. Vertical Equation of the Conic. 234. If x-a resp. x + a be put for x in the central Eq. of the E resp. H, the reduced Eq. takes the form 160 LENGTHS OF TANGENTS. 161 2 2b2 . _ .x2 resp. y2 =- b. + b. 2, a a 2 a a2 which is therefore the Eq. of the E resp. H referred to the axis and the tangent through the left resp. right vertex. If in either 2 b2 we put 2 =4 q, a= x, we get the vertical Eq. of P a 2 2b2 y = 4 qx. Now 4 q or - - is the parameter or double ordinate a throuah the focus; lhence these Eqs. state the geometric fact that the squeare of the ordinate, as compared with the rectangle of parameter and abscissa, shows: in the Ellipse, lack; in the Parabola, likeness; in the Hyperbola, excess. From this fact the curves seem to have been named. Hence the Eq. of any conic may be brought into the form y2= Rx + Sx2, where S is 0, = 0, 0 resp. for E, P, H resp. Lengths of Tangents from a Point to a Conic. 135. These have been found to varv as the [I diameters in the centric conic (Art. 101). In the P be P1(x1, y,), P2(x2, Y2) the points of tangence, tj, t2 the tangent-lengths. Then the Cds. of the intersection of the tangents are: X = l= 'VXX2 y = Y1 + Y2. i.e., are the geometric resp. arithmetic means of the Cds. of the points of tangence. Hence, p, and P2 being the focal radii, ___ 2__ ./-/ J ;.1=Y2-Y1 (1+- 22 2 t y=Y2 f1 (21Yq)=Y2_Y1 p+; 4q 4q -2 so t2 2=Y2Y1 . tj2: t2 =PI: p2i i.e., the squared tangent-lengths from a point to a P vary as the focal radii to the points of tangence. CO-ORDINATE GEOMETRY. Areas of Segments and Sectors of a Conic. 136. Bv Art. 107 the area of an E is to the area of its major circle as b to a, chords I to the axis major being all in this ratio. It is also clear that any segment of the E is to the corre- sponding segment of the circle as b to a, and the same holds equally of half-segments reckoned from the axis major. Two corresponding sectors AOP, AOP' are made up of two half-segments in the ratio b: a, and two A in the same ratio; /F hence are themselves in that / A ; ratio. So, too, are any other corresponding sectors AOQ, o o A AOQ'; hence so, too, are their " FA differences POQ, P'OQ'. Since the centric 4 AOP or \/ and the eccentric E or AOP' are connected by the relation y: x=tan =-. tanE, if the a sector be given by its centric 4, we may still use the eccentric. The area of any focal sector PFQ is the difference of the focal sectors AFQ and AFP, each of which is, again, the dif- ference of a central sector and a A. 137. In P lbe P1, P1' any two points, 1 the pole of P1PI, ,f1 the mid-point. Then the RL. I-27IL is a diameter, the tract IAll, is halved by P at T1, the tangent at T1 is 11 to PIP,', and halves the tangents from I, at I., I:. The As 1112h' and T1P1Pl' have equal altitudes; the base and therefore the area of the second is twice that of the first. We may proceed with the As T1I2Pl, TlII'Pl' exactly as with P111P1', cutting off by mid-tangents at T2, 727 outer areas and by chords to T2, T2' inner areas twice as large; and so on without end. The limit of the sum of the outer areas is the outer sector P1I1P1', and the limit of the sum 162 AREAS OF SEGMENTS AND SECTORS OF A CONIC. 163 of the inner areas is the inner parabolic segment P1 TI PI', cut off by tile chord P1P1'; hence this latter is twice the former, or 2 A or 39 of the A PI 1P1', or of the parallelogram P1 Q1 of the chord an(1 the 11 tangent. The focal sector from the vertex V to the point P(x, y) is2 -Z j.--- (x - 17)Y' , or -X-j + -2- gy. / If the axis and the diameter1 through P cut the directrix at I)' and D, the area VD'DP is xy + qy, i.e., twice the h area of the focal sector. Hence, any focal sector PFP' is half the area of the corre- sponding outer segment be- tween the curve and the directrix and the diameters through P, P'. 138. If through any two points P., P2 on an H be drawn 1Is to either asymptote, meeting the other at Y1, XX2, then PIX1X2P2 is called an hyperbolic segment cor- responding to the hyperbolic sector PIOP2. The A P1OX1, P2OXY2 are equal, being halves of equal parallel- .1 ograms (Art. 118) ; taking each in turn from the fioure P1OX2P2, we get 0 ma m1 in turn PIXX2P2 and P1,0X; hence, 2 corresponding hyperbolic sector and segment are equna. If P1, Ql and P2, Q2 be ends of two 11 ehords, the sectors P10P2 and Q10Q2 are equal; for the con- jugate diameter of the chords halves both the triangular and CO-ORDINATE GEOMETRY. hyperbolic areas, halving every element of each: taking away equals from equals, we have left the sectors, and therefore their corresponding segments, equal. If y = sx + b be any chord referred to the asymptotes, then the roots xl, x2 of the Eq. xy _ sx2 + bx KI are the x's 2 of the ends of the chordl; their product, - , is independent of S b, i.e., is the same for all 11 chords. When the product of the x's is constant, so is the product of the y's by virtue of the rela- tion xy = K 2 Hence either set of asymptotic Cds. of the ends of two Ii chords form a geometric progression. Plainly the converse holds: if any number of like asymptotic Uds., x's or y's, be taken in geometric progression, they will belong to the ends of 11 chords, and hence the area of sector or seg- ment determined by any two consecutive ones will be constant. Take now the hyperbolic segment between the ordinates MY= K an(1 y= a in theH Xy=K2, and cut it into n equal sub-segments by ordinates in geometric progression. If r be the ratio, then a _ rAK, and r = The end-abscissue are x = K, X= -. The a /I sides of the first segment are K and rK; a . K 0 _ _ its base is - - K; its area, if 4) be the r asymptotic angle, is K -K) sin . 'and rK (- K) sin 4. Hence the area S of the whole segment lies between Ai IK2 sin4)n(--1) and K2 sin 0,t( 1-r). The ratio of these extremes is r; if n be taken ever greater and greater, r nears 1, the extremes near each other, keeping S always between them; hence S is the common limit which they 164 HYPERBOLIC AREAS. 165 both near as n increases without limit. We can readily evalu- ate n(1-r) or n 1 for n nearing ao, by expanding To do this, we write it as a binomial, thus: (i + I _ as a is Km --1 is 1; accordingly, the binomial expansion K is ap)plicable, being convergent. On expanding, taking from 1, and multiplying by n, there results the series, ( __- --1 -2 n t I a - n a 1 t K h K 1.2 K As n rises above all limit, 1 sinks below all limit, nears 0, U and the numerators become -1, -I -2 , -1. -2 -3, and the series becomes - ---.--+ a-1a a 4,K 2 AC 3 X 4 X K which, from Algebra, we know to be the negative of the expan- of a whe 0 a sion of the natural logarithm of when 2, as is the K K case here. Hence, at last, S=-K2 sin nat. log-, or S =K2sin nat.log- K a The area of a segment whose end-ordinates are a and b, b a K, S(a, b) = K2 sin ( nat. logy--nat. logK K2 sin 4t-nat. log a 166 CO-ORDILNATE GEOMETRY. If K be taken as linear unit, and if 4 = 900, i.e., if the H be equilateral, then S= nat. log!, S(a,b)=nat. logb; i.e., the area of any segment, reckoned from the vertex, is the natural logarithm of the end-abscissa, and the area of any seg- ment wholly on one side of the vertex is the natural logarithm of the ratio of its end-ordinates. Hence natural logarithms have been called hyperbolic logarithms. Thus far, segments and sectors lie outside of H; but problems about inner ones can now present no difficulty. Varieties of the Conic. 139. We have found three species of conic: E, P, H, accord- ing as -C or h2-kj is 0, = 0, 0; but of each there are several varieties, which are now to classify. By Art. 51, on passing to 11 axes through a new origin (xl, Yi), the new abso- lute term becomes c'= (kAx1+ hy,+g)xi+(hx1+jy1+f)y,+gxl+fy1+c. If the new origin (x,, Yj) be the centre of the conic, the coeffi- cients of xl, yj vanish, and the values of xi, yj are G C; hence, T ' et = gxl fyl + c=g 9 + f F '+ c. C = A. C C CAC Hence the Eq. of the centric conic referred to its centre is kx2+2hxy+j+ AC =0. C Now, in the E, C or kj - h2 is 0, hence k and j are like- signed, hence when A and they (k and j) are unlike-signed the Eq. is clearly satisfied by real values of x and y, hence the E is real; but when a and they are like-signed the Eq. is satisfied VARIETIES OF THE CONIC. 167 by no real values of x and y, hence the E is imaginary; also when A = 0 no real values satisfy it but the pair (0, 0), hence the Eq. pictures a pair of imaginary RLs. intersecting in the origin (xl, yj). In the H, k and j are unlike-signed and C 0; plainly the Eq. is satisfied by real values of x and y in all cases, but for A = 0 it pictures a pair of RLs. through the origin (x1, y,). In case of the non-centric, P, C= 0, hence the first three terms of the general Eq. form a perfect square, and we have (kx -Vy)2+2gx+2fy+c=-. Solved as to the parenthesis, this Eq. takes the form Vk7-xJ-y+ f = V2Gx-K. Vj vj Accordingly, the Eq. pictures a real P save when G = 0. Then it pictures two RLs., which are always 11, and are real and separate, coincident, or imaginary, according as K O, = 0, or 0. If j = 0, like conclusions hold on changing G to F aud K to J. Hence the following table ( kA 0 . . . . real ellipse. C0 A =O . . . . two imaginary RLs. l7A 0 . . . . imaginary ellipse. A 0. . . . real parabola. C=O A = . . : . two parallel RLs. A 0 . . . . real parabola. f kA 0 . . . . real hyperbola. C0 A=O . . . . two real RLs. kA 0 . . . . real hyperbola. 1CO-ORDINATE GEOMETRY. CHAPTER VII. SPECIAL METHODS AND PROBLEMS (Continued). Determination and Construction of the Conic. 140. The Eq. of the conic has six constants, but division by any one reduces the number to five, which are therefore the independent arbitraries of the Eq. Hence five independent simple conditions are needed and enough to determine a conic, since they determine the five arbitraries. Conditions are inde- pendent when no one can be drawn out from the others, simple when each fixes but one relation among the arbitraries. Such are that the conic shall go through certain points or touch cer- tain RlLs. A mdltiple condition fixes more than one relation among the arbitraries. Thus, that the conic touch a certain RL. at a certain point is a double condition, namely, that the conic pass through two given consecutive points; that the centre be (x,, Yi) is a double condition, for it makes kx,+hy,+g=O and hx,+jy,f=O; that a given direction be asymptotic is simple, since it fixes one point at so, but that a given RL. be an asymptote is double, since it fixes two points at o ; that a given point be focus is double, since three tangents besides would determine the conic by determining three points of the major circle ; to give the direction of an axis is to give a point (at o ) in P, or to give the relation 2 h = a constant in E and H, a simple condition; k -j hence, to give both axes in position, since this fixes the centre besides, is to impose a triple condition. A given eccentricity is in general one simple condition, but if e= 0, the conic is a 168 SYSTEMS OF COIICS. circle, which implies the two conditions of Art. 58. For exer- cises, see pl. 210-213. 141. The case of a conic fixed by five points merits special attention. If the five lie on a RL., that RL. and any other are the conic; there is a two-fold so of solutions in pairs of RLs. If only four lie on a RL., that RL. and any other through the fifth are the conic; there is a one-fold X of solutions in pairs of RLs. If only three lie on a RL., that RL. and the RL. through the other two are the conic; there is but one solution. Thus far the solutions have been pairs of RLs., since no curve-conic has three points on a RL. If only two points lie on a RL., the Eq. of the conic is got by assuming kx2 + 2 hxy +jy2 + 2 gx + 2fy + c =0, which with five like Eqs. got by replacing the current Cds. by the Cds. of the five points will form a system of six Eqs. homo- geneous of first degree in the six unknowns, k, h, j, g,f, c; since the absolutes are all 0, the condition of consistence is that the determinant of the coefficients of the unknowns vanish; hence that determinant equated to 0 is the Eq. sought. But to avoid the tedium of reducing this determinant, we may proceed better thus Be LI2= 0, L2 3 = 0, 0 L4 1 = 00, L4 = 0, L24 the six RLs. fixed by four of the points; to denote that in any L the current Cds. are replaced by the Cds. of the fifth point, prefix the superscript 5. Then L12.L34=0 and L23.L41=0 are a pair of pairs of RLs., i.e., a pair of conies through the four points; hence L12'L34-XL23.L41 = 0 is a conic, being of second degree, through the four points. If this conic goes through the fifth point, then 5L12. 5L34 - 54L23'5L41 = 0 whence finding the value of A\ and putting it for X in the other Eq. we get the Eq. of the conic sought. Since we can find a fit value of X for any fifth point, it follows that L12-L34 - XL2.- L41 = 0 represents a conic passing 169 170 CO-ORDINATE GEOMETRY. through the four points and any other point. Also a conic through four points of which no three are on a RL. can sustain but one more condition, and X may be taken to satisfy any one condition; hence the above Eq. represents the whole system of conies through four points ; and since for any fifth point there is found but one value of X, it is seen that thron h five points, no three of which lie on a RL., one, and only one, conic passes. 142. Two special forms of this Eq. of a system of conies are specially useful. (1 ) Take two of the L's for axes; then their Eqs. are x = 0, y = 0; the Eqs. of the other two are lx+my+1=0 and l'x+m'y+1=0, and the Eq. of the system is (lX+my+1)(l'x+m'y+ 1)- Xry=O. The terms of second degree are lI'x2 + (In' + I'm - X)xy + mn je'y2. They form a square, i.e., the conic is a P, when, and only when, (jbtn + Int- _ - )2 = 4 ll'nmv'. This Eq. is of second degree in X; hence two, and only two, of a system of conics throuh fJourpoinits are P's. The student may easily investigate the conditions under which the P's are real or imaginary, separate or coincident. (2) Take the Eqs. of the RLs. in the N. F., and drop the sec- ond subscript; then h- X-A X.- V4= 0 is a conic about the 4-side whose counter-sides are AN = 0, 1X3= 0 and 32 = O. X4 = 0. Call the lengths of the sides, i.e., the chords of the conic, cl, c2, C3, c4. From any point P draw rays 9 1a, ar, r4, to the vertices of the 4-side. Then the double areas of the As are CONIC THROUGH FIVE POINTS. 171 X1 - cl = r, r2- sin r,2,2, N2 - c2 = r2 r-3 sin r2r3, N3 c3 = r3 r4 sin r3r4, N4 c4 = r4 r, sin r4r1; whence N1.N1 c c C sin r s n r-r Si -23Sin r4rl 3 The right member of this Eq. is the cross-ratio of the four rays,. and the left is constant so long as N1 N8: N2. N4 is; i.e., so long as the centre of the pencil, P, is on a conic through the four points. Hence the cross-ratio of a pencil from any point of a conic through four fixed points of the conic is con- stant. 143. If now we take any sixth point on the conic, we shall have an inscribed 6-side. The sides and vertices numbered 1 and 4, 2 and 5, 3 and 6, i.e., whose numbers differ by 3, are called counter. Let the Eqs. of the sides of the primary 4-side stand as in Art. 142; then the Eq. of the fourth side of the 6-side, i.e., the side from 4 to 5, will be N3-KN4=0, since it is a ray of the pencil (_73, N4) ; so, too, the sixth side, from 6 to 1, will have for its Eq. N1 -KN4 = 0. Since 5 is on the conic N1X3 - XN2N4 = 0, we have =_ = ix A2, or A N4 N1 N CO-ORDINATE GEOMETRY. hence the ray from 2 to 5 is soning, the ray from 3 to 6 is fifth side is the common ray of KN1--X = 0 te 3t-o pen2 = - the two pencils By like rea- Hence the 1Y3- xN. + (KN1- X2) = 0, These Eqs. are the same when ju = 1: Kt. ptl - ; hence the fifth side is KN X- KN2 + K1X3 -V K'4 = 0. The RL. through the intersections of counter sides 1, = 0 and N3 - K-N = 0, N 3= 0 and N1, - K'N 4= 0 is the common ray of the pencils N,-r(N83- KN4) = 0, N, -K'NL,-r'1N = 0, i.e., it is the through the NX2=0 and RL. K , - KK'N4+ K N = 0. intersection of the third pair K-JCL - KK'NV + KN3 - X2 i' 0. BLt this RE. goes of counter sides Hence we have Pascal's Theorem. The three pairs of counter sides of a 6-side inscribed in a conic intersect on a RL. called a Pascal (RL.). Since six points may be thought arranged circularly in 3! or 120 ways, and since a ray joining two points may b)e counted in two ways, it follows that there are sixty really diflf'rent orders, and so sixty really different Pascal RLs. X - 4 + /1" ("-V;l - -YD = 0 - PASCAL'S THEOREM. Pascal discovered this beautiful relation, and built upon it a theory of the conic. 144. We have seen (Art. 141) how to form the Eq. of a conic through five points; Pascal's Theorem enables us to con- struct it without knowing its Eq., and more rapidly than by Art. 71, thus Draw through 5 any RL.; it will cut the conic in 6; find the intersections of sides 1 and 4, 2 and 5 (the ray just drawn); through them draw the Pascal P RL.; it will pass through the intersection of sides 3 and 6; 2 from 1 draw side 6 through the intersection of the Pascal 5 and side 3; it will meet side 5 in point 6. So we may find any number of points of the 1/ conic. If side 5 be drawn 1I to side 1, the RL. halving the tracts 1 2 and 5 6 will be a diameter of the conic; a second diameter can be got by drawing side 5 11 to side 3; these two diameters fix the conic's centre. If side 5 is to touch the conic at 5, point 6 must fall on 5 hence, draw side 6 from 1 through 5; through the intersections of the counter-sides 1 and 4, 3 and 6, draw the Pascal; it will cut side 2 in a point of side 5; the RL. through this point and 5 will touch the conic at 5. So we may draw any number of tangents to the conic. 145. We have learned to construct the conic by points and form its Eq. from given conditions; it remains to determine its elements, centre, axes, foci, directrices, asymptotes, from its Eq. By Art. 93 the centre is the point (G: 0, F: C). If C be 0 resp. 0, the Cds. are finite, the conic is an E resp. H. 1T3 174 CO-ORDINATE GEOMETRY. Transformed to 11 axes through the centre, the Eq. becomes kx2+21 xy+jy2+ A= o. C The asymptotes are now k2 + 2 hxy +jy2 = 0; hence, if we pass back to the old origin and axes, the Eq. of the asymptotes becomes kxl 2 hxy +jy2+ 2 x + 2fy + c 0; C- i.e., the Eqs. of the conic and its asymj)totes differ only by the constant A The axes halve the 4s between the asymptotes; hence they are the RLs. X2 _ y2= k 1-x y by Art. 53. As tan 2 0 2 h h byirt 0 being the slope of an axis to the X-axis, we may now find the directions of the axes by drawing the RL. y = 2 h x and halv- =1 7, -j ing its slope to the _Xaxis; but to determine the lengths of the axes is still tedious. A guide to a simpler solution of both problems in one is the reflection that the diameters of a conic through its intersections with a concentric circle are equal, being diameters of the circle, hence are like-sloped to either axis of the conic; accordingly, when these diameters fall together, it is on an axis of the conic. Now by Arts. 30, 50 X2 A a2 kx2+ 2xy +jy2 + - . H = 0 C is a pair of RLs. through the intersections of the conic with the concentric circle x2 +,2 = r2, and the origin (centre). They fall together when the Eq. is a square, i.e., when (k+ )(J+ ) h2. ECCENTRICITIES. The roots .12, r22 of this Eq. are therefore the squared half- axes of the conic; putting them in turn for r2 in the Eq. of the pair and taking the second root, we get the Eqs. of the axes. The axes thus found in size and position, the foci are found by laying off on the proper axis the proper focal distance, ae = Va2 T b2. The directrices are easily constructed as polars of the foci. To find the eccentricity, suppose the Eq. of the curve referred to its axes to be Ax2 + By2 = 1; then, as A and B are the reciprocals of the squared half-axes, we have A= B(1 - e2). Now pass back to the original central Eq. kx2 + 2 hxy +jy2+ C=. By Art. 102, since o = 900, A+B=k+j and AB=kj-h2. On elimination of A and B from the three Eqs., there results e4 + (k-_j)2+ 4 72(e2_ 1)=O e+ lj - h2 This Eq. teaches that there are two pairs of counter-eccen- tricities of a centric conic, corresponding to the two pairs of foci, one real, one imaginary; if e12, e22 be the squares, e 2 + e2 = e2 . e22 =_ (k _)2 + 4h2 1 21 2 kj -h2 Hence 1 + 1 = 1 , the sum of the reciprocals of the squared el2 e22 eccentricities is 1. Also, if kj - h2 0, the product of el2 and e 2 is -; i.e., the squared eccentricities are one +, one -; i.e.. in E one pair of eccentricities are real, one imaginary. If kj - h2 0, both the sum and the product of the squared eccentricities are + ; hence both squares are +, hence in H both pairs of eccentricities are real. The real foci lie on the real axis, which latter falls on the X- resp. Y-axis when A resp. B is +, and the corresponding squared eccentricity is 175 CO-ORDINATE GEOMETRY. B-A A-B B resp. A 146. The Eq. of P, as the highest terms form a square, may be written (KX +,y)2 + 2 gx + 2fy + C= O. The RL. KX + ty = 0 is a diameter (Art. 97), and the RL. 2 gx + 2fI + C = O is a tangent at its end, since on combining the last Eq. with the Eq. of the P there result two equal pairs of values of x and y, which also satisfy KX + ty= 0. This diameter and tangent are not in general I, since not in general is -7K =-1. But by adding and subtracting 2 KX +2 tX + X2 we may write the Eq. of the P thus: (KX + y + X) 2(g - KX)X + 2(f- tX)y + c-A = 0. The RL. KX + Ly + X = O is still a diameter, being 11 to KX+ty=O, and 2(g-KX)x+2(f-IX)y+c-X2=0 is the tangent at its end for the same reason as before; they are 1 if __ i.e., when, and only when, A- Kf ai L 9- KX K2 + t Hence when X has this value, the two RLs. are resp. axis and vertical tanfqeit of the P, and their intersection is the vertex. The parameter 4 q is the ratio of the squared distance of a point of P from the axis to its distance from the vertical tan- gent; i.e., 4q- (Kx+ty+X)2 2(g-KX)X+2( f-(X)!y+CX2 lC + L2 2V (_ + (f_ A)2 _2 J(qKX)2+(f-LA)2 2 +L K2 + el whence on substitution for A, 4q= 2(Kq-f ) = 2(V2 -Vjf (K2 + t2) 3 (k +j)3 176 TRACING CONICS. The exponent a leaves the sign of 4q at will, as it should be, since the + direction of the X-axis is yet at will. The P clearly lies on the - side of 2(g - KX)X+ 2(f- tX)y + c-X2= 0. The focus is on the axis q distant from the vertex; or we may determine its Cds. by finding the intersection of the tangent 2 gx + 2fy + c = 0 and the vertical tangent, and through it drawing a I to the former, which will cut the axis at the focus. 147. When the elements are found, the curve may be actually traced by the methods of points, of tangents, or of continuous motion. (1) Cut the major axis of an E anywhere within the foci; with the two parts as radii draw circles about the foci; each pair of circles, the sum of whose radii is the axis major, inter- sect in two points of the E (Art. 101). Like holds for the H on changing within to without and sum to difference. (2) From the end of any radius of the major circle of an E, drop a I on the axis major; from the intersection of the radius with the minor circle draw a 11 to the axis major; it will cut the I in a point of the E; for it cuts it in the ratio b a. From any point on the real axis of an H draw a tangent- tract to the major circle; also draw a 1I tangent-tract to the minor circle; at the point lay off the equal of the second tan- gent-tract I to the axis; its end is a point of the H; for the first tangent-tract is clearly a tanq, and the second is b tan d or y. (3) Through a focus draw chords of the major circle of the E resp. H; through their ends draw Is to them; the Is touch E resp. H; for the feet of focal Is on the tangents lie on the major circle. (4) Join cross-wise the intersections of a focal circle with the vertical tangents to an E resp. H; the junction-lines will touch the E resp. H, by Art. 132. 177 CO-ORDINATE GEOMETRY. By either (3) or (4), more easily by (4), enough tangents may soon be drawn to shadow forth the curve quite clearly. (5) If the ends of a string 2 a long be fastened less than 2 a apart, and it be stretched by a sliding pencil, this will trace an E whose axis major is the leinyth, whose foci are the ends, of the string. (5') If the ends of a string he fastened, one at a point, the other at the end of a ruler 2 a lon-er than the string, and the string be kept stretched against the ruler by a )encil while the ruler turns about its other end fastened at a second point more than 2 a apart from the first, the pencil-point will move on an H whose foci are the fixed points and whose real axis is 2 a. These constructions rest on the same properties as those of (1) . (6) If one end N of a ruler a long slide on a fixed bar BB' while a fixed point N' of the ruler slides aloni a I bar AA', the other end P of the ruler will trace an E whose axis major is 2 I-UP= 2 a, whose axis minor is A 2 N'P= 2 b. For draw a circle with radius a about the cross- p)oint C, and a radits U,-IP' 11 to NP. Then COP: MIP'= b: a. The three bars form a pair of elliptic comilpasses. (6') If a ruler RBR' bent at B slide along a straiglht-edge DD' while a pencil- l)oint P keeps a string RB long(J stretched against the ruler, one end of the string being F fastened at R, the other at a fixed point F, then P will trace an H, of which F is a focus, D' a directrix. BR an asymptotic direction. For the distances of P from F and DII are clearly in a fixed ratio, namely, sec0, where 178 CONFOCAL CONICS. 0=41RBR'-900; hence e=sec0, c =cos0, and cos- e is the slope of an asymptote to the X-axis. Among constructions of P by points this seems simplest: (7) About the focus with any radius q draw a circle; from where it cuts the axis lay off 2 q toward the focus; through the point thus reached draw a I to the axis; it will cut the circle in points of /'; also the junction-lines of the points of P and the other end of the diameter will touch the P (Art. 121). (8) From the focus draw any ray to the vertical tangent; through their intersection draw a I to the ray; it will touch P; also a second focal ray having twice the slope of the first to the axis will meet the tangent at the point of tancgence (Art. 121). Or, the point of touch may be found by remembering that the focal I on the tangent halves the tangent-tract from the axis. Constructions of P by points and by tangents involve each other. (9) Construction (6') of H will yield P when the ruler is bent at right angles; for then 0 = 0, e = 1. Confocal Conics. 148. Confocal conies are clearly also co-axial, for the foci fix the axes in position; if 2 a, 2 b and 2 a,, 2b, be their axes, then a2 - b2 = a2 -b = squared central distance of focus. Conversely, if two conics be co-axial and a 2 _ 2= a2 -bl 2, they are c6nfocal, the foci clearly falling together. On transposing, we get a2 - a 2 = b2 - 6b2; accordingly we may put a 2 + A for a 2, at the same time putting b2 + A for b12. Hence +f x =1 is a family of confocal conies, a 2+X b2 +X' and all conies confocal with the base conic + = 1 are a2 62 got by letting A range from - X to + oo. For - a A - a2, the conies are imaginary ; for A a2, the conic is a pair of RLs. fallen together on the Y-axis; for 1,19 CO-ORDINATE GEOMETRY. - a2X - b2, the conics are H's sinking down to the X-axis; for X = b- - 0, the conic is the doubly-laid X-axis witho t the foci; for X = - b2 + 0, it is the doubly-laid X-axis within the foci; for - b2X x, the conies are E's swelling out from the X-axis toward the concentric circle with x radius. The Eq. + u- - 1 = 0 is of second degree in A; a2+XA b2-j-X hence to any pair (xy) correspond two values of A; i.e., through any point in the plane pass tbo, and only twvo, confocals. For X very great and +, the quadratic is-; for A nearing - b2, it is + ; for X = V-b, it springs from + X to - c ; for A nearing -a2, it is again + ; for A =-a2, it again springs from + x to - (, and thence stays -. Accordingly, the quadratic changes sign by passing through 0 only for X between + so and - IA, and for A between _b2 and - a2; i.e., the two confocals which pass through any point are the one an E, the other an H. The tangent and normal of the E halve the outer resp. inner 4s of the focal radii to the point, the normal and tangent of the H halve the same 4s; hence they are the same pair of RLs., the normal of one curve is the tangent of the other; hence conIfoccfds cut each other, wherever they cut, orthogonally. 149. Cojfocal curves (and especially confocal su)faces) form a system of rectcang. Cd. lines (and stwjaces) of great iumport to Higher Geometry and Mechanics. It is in place here only to show how ordinary Cds. may be expressed through con/bcal Cds.: pairs of values of A yielding confocals through the point. 2 X "/ Be -+ =1 and + =1 b 1 C2 + XI 62A' a 2 + b2 Ax the confocals. Multiply by the divisors of y2 and subtract; there results x2 = bA X2I+ XI X -- (a2 +Al a 2+ A2. = (Al- A2) (a2 + ,\(a2 a2 A2)_ (A2-)A) (a2 - IA) (,,2 + j (a2 + A) a2 _,2 180 SIMILAR CONICS. and on exchanging a2 and V2, 2= b+A) (b2 + X2) a -b Hence x2+y2=a2+b2+AI+A2=a2+ X+b2+A2 = a + A2 + b2 + AI = . This r being regarded as a. half-diameter of the confocal AX, its conjugate rl' is given by the Eq. r= ,2 + A1 + b2 + Al - a2 + AX +b2+ A2 = 1- A2; and so r2 2 = A2-A1X whence r' 2 +r2f 2 =-; i.e., the conjugates to a common diameter of two confocals are alike in size, one real, one imaginary. Also, by Art. 111, if p resp. P2 be the central I on the tangent to Al resp. X2 11 to r1' resp. r2', or through the end of r, then 2= a2 + A1. b2+ A a2__________ X Pi = +t - , and p22 -= a XI - 2A- k Thus may all geometric elements of the system of confocals be expressed through the parameters X, X2, and their relations studied. Similar Conics. 150. From Art. 82 it is clear that any conic KR similar to the central conic K must be congruent with some conic K2 concen- tric with K and similar to it; Kl and K2 differ only in that 14 has been pushed and turned as to K. Since the centres of K and Is fall together in the centre of similitude, the centres of K and K, are corresponding points or centres of similarity. As 1, and 14 are congruent, in dealing with their metric relations we may put tile one for the other. Now the eccen- tricity of K is Ia2 - b2: a't&, that of K2 or K1 is Ial2 - b,2: alltk; 181 CO-ORDINATE GEOMETRY. in K and K2, a and a, (or a,) correspond, so do b and bl; hence, a b a a, -=-, or -_ a, b, b b' hence the eccentricities of similar conies are equal. Conversely, if e=e2=e,, orif 2 2 2 2 a al by decomposing, we get can b _2 = b' or the conics are similar. Hence eccentricities equal is the necessary and sufficient condi- tion of similarity in central conies. As P is simply a central conic whose centre has retired to so while the parameter has stayed finite, we may at once infer that all P's are similar, having the same eccentricity e = 1, and all conies similar to a P are P's. Or we may place the P's vertex on vertex, axis on axis; then their Eqs. will be y2 = 4 qx, y = 4 qux; or p sin =4 qp cos 0, Pl sin 0= 4 q1p, cos A; hence for 0 = 01, p = q: q1; or the P's are similar, the ratio of similitude being the ratio of their parameters. Since in similar conics the eccentricities are equal, the expres- sion (2 - e2) : - e-2 must be the same in them; hence, by Art. 145, (k +J)2 kj - h2 must le constant and equal 22 k1 +J : kj1 - hl2. This, then, is the condition that K and C,, given by their Eqs., be similar. For oblique axes we must write k+j- 2hcoswo for k +j, as is plain from Art. 102. If now K and Kl be like-placed, i.e., have corresponding tracts 11, the Eq. of K2 becomes that of KX on change of origin 182 CENTRAL PROJECTION. only; by such change kl, hl, j' are not changed, hence they are the same in the Eqs. of Kl1 and K2; but in K and K2 the inter- cepts on the axes, origin being at centre, vary inversely as \'k, V/kC (or -Vk), Vj, v'ji; and since these tracts correspond, k :k= j:j=r2; and since (k +j) is constant, we have also kj -h 2 In two P's like-placed, or with 11 axes, it is plain that k _ --, since each is the squared direction-coefficient of the c h_ axis; hence in all cases the triple equality - = -- = - shows k, hl j that the two coniCs are similar and similarly placed. The Conic as the Projection of a Circle. 151. By Art. 107 the E is a 11 projection of a circle, the only kind of projection yet spoken of. The notion may be widened thus: The point P' where any plane HI cuts any ray from S is called the central projection of any point P of the ray, on the plane. S is called the centre of projection, H' the plane of projection. Only when S is at oo does central pass over into 11 projec- tion. Rays from S through points on a RL. lie in a plane; hence their intersections with H' lie on a RL. ; i.e., the projection of a RL. is a RL. If this RL. cut a plane curve in n points, its projection will cut the projection of the curve in n points, the projections of the original n points; since n fixes the degree of the curve, that degree is unchanged by projection; e.g., the projection of a conic is a conic. If the points in which the RL. meets the curve be consecu- tive, so will be their projections; hence, the projection of a tan- gent to a curve is tangent to the projection of the curve. 183 CO-ORDINATE GEOMETRY. Hence, too, the projections of pole and polar as to a conic are pole and pohir as to the projection of the conic; also, projections of conjugates, points or RLs., are conja gate. All points of a plane through the centre 11 to II', and no others, are projected into x ; hence, to project a RL. into oo, rroject it on a plane 1I to the plane through it and the centre. A pencil of rays is clearly projected into a pencil of rays, whose centre is the projection of the centre of the pencil. This latter projection will be in fin ity unless the centre lie on a RL., the intersection of the projected plane H and the plane H" through the centre S, 11 to HIr; then it is in oo. Hence any pencil whose centre is on this RL., IJ", is projected into a penl- cil of 11 RLs. The centres of all such pencils lay on a RL. before projection; hence 'they lie on a RL. after projection, namely, the RL. at x. The plane I."' through the centre S 11 to IH meets H in a RL. at x, and meets DI in a RL. in finity, IP'L'; hence all pencils of 11 RLs. in H are projected into pencils of intersecting RLs. in H', whose centres lie on the RYL. IPL'. 152. Since 'I planes meet a third plane in 11 RLs., and since 4s between 11 pairs of (like-directed) RLs. are equal, clearly any 4 in H is protected into equal 4s on 11 planes. Hence, if the sides of any 4 at A cut IL" at B and C, then the projection 4B'A'C' of 4BAC on H' equals 4 ]SC, the projec- tion of the same 4 on 11". Hence, to project any 4 BAC into an 4 al, draw BS and CS making BSC= 4al, then project on any plane 11 to the plane BSC. This may be done in an x of ways. If some RL. of the plane of BA1C is to be projected into c:, let it cut the sides of the 4 at -A, in B,, C,; on BC, as chord draw in any plane anl are to contain the given 4 cl; the centre S may be taken anywhere on this arc, and H' mav be anv plane 11 to the plane BISCJ. If two 4s in a plane, at -A, and A,, are to be projected into two grien 4fs, a, and ar while a given RL. in the plane is to be projected into oc, con- struct as before in any plane on BCG as chord an arc to contain 184 ORIGIN OF THE NAME CONIC. a,, and in the same plane on BQC, as chord an arc to contain a2: the intersection of these arcs determines the centre ', which, however, may still lie anywhere on the circle of intersection of the surfaces generated by revolving the arcs about their chords. The plane of projection I' may be any plane 11 to the plane through the centre S and the given RL. On this theorem, that any two angles in a plane may be pro- jected into angles given in size and at the same time a given BLI. of the plane projected into o, is based the theory of projections. An immediate deduction is: 153. Any conic may be projected into a circle and at the same time any point into the centre of the circle. -P2 Q Q Be C the point to be projected into the centre of the circle. Let the polars of any two points P1, P2 on the polar of C as to the conic cut that polar in Q1, Q2; then are P1, Q1 and P2, Q2 pairs of' conjugate points. By Art. 152 project 4P1CQ1 and MP2CQ2 each into a R4 and the polar of C into o; then is C' the centre of the projection of the conic (which is itself a conic, by Art. 151), since its polar is at o ; also P1'C', Q1'C' and P2'C Q2'C' are two pairs of conjugate diameters at Rs. Hence the projection is a circle. 185 CO-ORDINATE GEOMETRY. If now we hold the centre of projection fast, and exchange 11 and H', we shall get the correlate theorem: A circle may be projected into any conic, and at the same time the centre of the circle into any point of the conic. The system of raYs from S to the points of the circle form a circular cone; the p)rojection of the circle on any plane is the intersection of that cone and the plane; hence, any conic is the intersection of a plane and a circular cone, and aniy intersection of a plane and a circular cone is a conic. Hence the name conic section or conic. The student will now readily see that the section of a right circular cone is a circle when the cutting plane is I to the axis of the cone; as the plane turns, the section passes over into an E with increasing eccentricity till the plane gets 1t to the edge of the cone, when the section becomes a P; as the plane still turns, the section becomes an H. Again, if a circle stand upright on a plane, and a centre of projection descend toward the plane, the projection of the circle on the plane will be an E till the centre reaches the level of the highest point of the circle, when it becomes a P; as the centre still descends, the projection becomes an H; as the centre passes through the plane, the H shrinks to a pair of RLs. fallen together, and swells again into an E as the centre sinks below the plane. Properties of a eurve not changed by projection are called projectile properties. By Art. 41 properties connected with the cro0s-ratio are such. But this thought cannot be followed up here. 186 HOMOGENEOUS CO-ORDINATES. CHAPTER VIII. THE CONIC AS ENVELOPE. The path here struck into leads quickly into the higher regions of the subject; we can follow it but a few' steps, to find out its general direction. 154. Art. 30 has already introduced a new kind of Cds. In the Eq. rl-N + v2N2 + v3IV, = 0 of a RL. we may treat any two of the ratios of the three Y's to each other as Cds.; or, still better, we may treat the N's themselves as Cds. In this case the homogeneity of the Eq. in Ns shows that the apparent number of Cds., three, may at once be reduced to the real, two, by dividing bv any one. ID fact, if Te, T2, r3 be the length of the sides of the A of reference, it has been shown that the triangular Cds. of any point are connected by the relation T'X[ + r2N2 +T3YJ 2 A. Hence, any two of such Cds. being known, the third is known. Again, any two determine a point. For all points distant N, from N. = 0 lie on a RL. 11 to it; so all points distant N2 from N, = 0 lie on a RL. 1I to it; and these two RLs. meet in one, and only one, point. It is equally clear that any point determines two Cds., and therewith the third Cd. Thus it is seen that points and triangular Cds. determine each other exactly as points and Cartesian Cds. If anv two ratios of the N's be taken as Cds., then are these Cds. of 0th degree in the N's, and any combination of them will still be of 0th degree; hence any Eq. between them will be homogeneous of 0th degree in N's, and on multiplication by the least common multiple of the denominators will remain homoge- 187 8CO-ORDINATE GEOMETRY. neous of some degree. Hence Eqs. in triangular Cds. are homogeneous. 155. Let us note more closely the Eq. of a RL. in triangu- lar Cds.: VI'I + V2iN2 + vFN3=O. (L) It is seen that the v's enter the Eq. exactly as the N's do. The significance of this fact is now to be developed. Holding the v's fixed and letting the N's vary, we get various points on the same RL.; the RL. is fixed by fixing the v's; any point on it is then fixed by fi-xing the N's. If, now, we hold the NT's fixed and let the v's vary, we shall clearly get various RLs. through the same point; the point is \ A'P P fixed by fixing the AT's; any RL. through it is then fixed bv fixing the v's. Thus it V\ as/ 4seems that the v's determine a RL. precisely as the N's determine a point. The N's determine a point as being its distances from three fixed RLs. forming a A; how do the v's determine a RL. This question is easily answered thus: Take as the fixed point the vertex of the referee A counter to the side X,= 0, i.e., the point (1Vx=0 O V=0); then Eq. (L) reduces to ll = 0. Now for this point N1l is not = 0, hence v, = 0; i.e., when all the RLs. go through the point (NV = 0, N3Y 0), then for all such RLs. r, = 0. Also, we know from Art. 29, that if V1 = 0, then all the RLs. given by Eq. (L) go through the point (NV = 0, X.,= 0). Hence 1, must be a factor of. or proportional to, the distance of the RLs. from the point (N2 = 0, N3 = 0), since when v, vanishes, and only then, the 188 RIGHT LINES OF AN ENVELOPE. RLs. go through that point. Hence we may say v1 = 0 is the Eq. of that point, meaning all RLs. through it are distant from it 0, just as we say N,=0 is the Eq. of a RL., mean- ing all points on it are distant from it 0. Likewise v2 = 0, V3= 0 are the Eqs. of the other vertices of the A, counter to the sides whose Eqs. are N2 = 0, N3 = 0. As we say the point (N2, N3), meaning the junction-point of the RLs. N2 = 0, N3 = 0, so we say the RL. ('2, V3), mean- ing the junction-line of the points v2 = 0, v3 = 0. Had we not assumed the Eq. of the RLs. in the normal form, but taken the general form XL, + X2L2 + A3L3 = 0, the reasoning would have remained unchanged. Hence, Triangular Cds. of a point are (fixed multiples of) its dis- tances from the three sides of a A. Triangular Cds. of ac RL. are (fixed multiples of) its distances from the three vertices of a A. The above interpretation of the v's is indeed clear from Art. 30. Accordingly, Eq. (L) may be interpreted either as the Eq. of a RL., the v's being arbitrary and the N's variable, or as the Eq. of a point, the N's being arbitrary and the v's variable. The RL. fixed by the v's chosen at will holds on it every point whose Cds. (N11 N3, N.) satisfy the Eq. (L). The point fixed by the N's chosen at will holds through it every RL. whose Cds. (v,, v2, v3) satisfy the Eq. (L). The RL. is the locus of the point (N,, N2, N13) ; the point is the envelope of the RL. (V1, V2, 7'3). Thus far have been used two sets of symbols N's and v's as Cds. resp. of a point and of a RL. ; but that was plainly unnecessary, since the two sets enter Eq. (L) in the same way. We might as well say, regarding either set not as Cds., but as arbitraries, the Eq. (L) is the Eq. of a RL. or of a point, according as the other set be taken as the triangular Cds. of a point or of a RL. 156. The picture of an Eq. of higher degree between point- Cds. is a locus, the Cds. of each of whose points satisfies the Eq. 189 CO-ORDINATE GEOMETRY. The picture of an Eq. of higher degree between line-Cds. is called an Envelope, the Cds. of each of whose RLs. satisfies the Eq. What is meant by a point of a curve is well-known; what is meant bv a Rb. of a curve is to be found out. By heaping together ever thicker and thicker points of a curve, the curve itself is not made but shadowed forth; so by heaping together RLs. of an envelope, the envelope is'shad- owed forth. Let 1, 2, 3 be any three such RLs. whose Cds. fulfIl the Eq. of the envelope. 23 Hold 1 fixed, and let 2 turn toward 1, its Cds. all the while fulfilling the Eq. ; its section- point with 1 will move along 1, and will be definite for every position of 2 ; as 2 falls upon 1, becomes coincident with 1, it is named con secutive with 1, and its section-point with 1 is named a point of the envJelope. The student will see at once that this reasoning is quite parallel to that in Art. 64. If now we let 3 turn toward 1, its Cds. all the while fulfilling the Eq., its section with 2, as it falls on 2, will be another point of the envelope; these two section-points (1, 2) and (2, 3) are plainly consecit- tive points of the envelope; hence -2, which ooes through both, is a tangent to the envelope. Hence a RL. of an enrelope is a tangent to it; and on this account the Cds. of the RL. are com- monlv called the tangential Cds. of the envelope or curve. 157. It is now easy to see the meaning of the degree of a tan- gential Eq. Thepoint-Eq. (i.e., the Eq. in point-Cds.) of a RL. being of first degree, for a point-Eq. of a curve to be of nth degree meant there were n points common to the curve and a RL.; so, the tangential Eq. of a point being of first degree, for a tangential Eq. of an envelope to be of nth degree means there 190 TANGENTIAL EQUATION OF THE CONIC. are n RLs. common to the envelope and a point; i.e., through a point may be drawn n tangents to the envelope. Such an envelope or curve is said to be of nth class. Again, the condition that a RL. should touch a curve was that on combining the two point-Eqs., two sets of point-Cds. should fall out equal, two common points be consecutive; so, the condition that a point shall be on an envelope is that, on com- bining the Eqs., two sets of line-Cds. shall fall out equal, two common RLs. be consecutive. This furnishes a general method of interchanging point-Eqs. and tangential Eqs. Combine the point-Eq. (L) of a RL. with the point-Eq. of any curve; express the condition that two roots of the resulting Eq. be equal; the Eq. which states this condi- tion states that the RL. whose Eq. is (L) touches the curve; hence it is the tangential Eq. sought. In it the parameters (or v's) in (L) are tangential Cds. Likewise, combine the tan- gential Eq. (L) of a point with the tangential Eq. of a curve; express the condition that two roots of the resulting Eq. be equal; the Eq. which states this condition states that the point whose Eq. is (L) is on the curve; hence it is the point-Eq. sought. In it the parameters (or N's) in (L) are point-Cds. Note that the analytic work in both cases is the same. 158. For the curve of second degree a more elegant method is this: Common Cartesian Eqs. may be made homogeneous by replacing x and y by x: z and y: z, which plainly we may do, and then multiplying by z. The Eq. of a RL. becomes Ix+nmy+nz= 0; the Eq. of a conic and the tangent to it at (x1, Yi, z1) become kx + 2 hxy +jy2 + 2gxz + 2fyz + CZ2 = -0 (kxe + hy1 + gzl)x + (hx, +jy +fz1)y + (gx1 +fY! + cz1)z =0. 191 192 CO-ORDINATE GEOMETRY. If the RL. he a tangent, then the first and third of these Eqs. can differ onlv bv a constant factor IL; hence kx, + hyh + gZ - l =0, x +jy1 +fzl- pm =0, gxl +fy1 + cz, - Fn =0 . Also, as the point (xl, y1, z,) is on the tangent, the Eq. holds: 1x1 + my, + nz, = 0. The condition that these four Eqs. between xl, y,, z1 consist, is k h g 1 =0, h j f m g f c n I m n 0 or X1 + 2RM + Jn2 + 2Gln + 2Fmn + Cn2 = 0. Such, then, is the tangential Eq. of the curve of second degree. The tangential Cds. are 1, mn, n, which are to be interpreted like the v's, while the capitals are, as always, the co-factors of the like small letters in the discriminant A. The relation of the above determinant to A is to be carefully noted. From this tangential Eq. the original point-Eq. is now to be got exactly as this tangential Eq. was got from the original point-Eq.: 1. m, n will change back into x, y, z, while the capi- tals will change into their own co-factors in the discriminant KJC I of this Eq. The result can differ from the original Eq. by a constant factor only, by which we may divide. This factor itself is readily found thus: Call the co-factors of IC, H, etc., k', h', etc.; then IKJCI=IkjcjV, and Ik'j'c'I=IKJCOV . 'jc'cj =!kjc&4. But Ak, Aj, Ac =IkjC I,4; k'= Ak, etc., whence BRIANCHON'S THEOREM. as the student may readily verify. Hence the constant factor is the discrinzinant A. When, and only when, A 1 will V = k, or the deduced be the same as the original Eq.; A can always be made = 1 by dividing the original Eq. by Hi. 159. Since k, h, etc., are at will, their co-factors K, H, etc., are at will; hence the tangential Eq. of Art. 158 is the most gen- eral tangential Eq. of second degree; since it represents a conic, we conclude that the general tangential Eq. of second degree repre- sents a conic. Again, since KI, H, etc., are at will, so are k', h', etc.; hence the point-Eq. deduced from the tangential is the general point-Eq. of second degree; hence the general tangen- tial Eq. represents every conic; i.e., all curves of second degree are all curves of second class. In general, degree and class of a curve are not of the same number. If n points be taken on a conic and numbered consecutively from 1 to n, and each pair of consecutives be joined by a RL., the nth being joined to the first, the points will form the vertices, and the junction-lines the sides, of an inscribed n-side. If n tangents be taken on a conic and numbered consecutively from 1 to n, and each pair of consecutives be joined by a point, the nth being joined to the first, the tangents will form the sides, and the junction-points the vertices, of a circumscribed n-side. In this way we might now proceed to double by re-interpreta- tion the whole body of doctrine gone before, by making proper changes in words throughout. But such detailed treatment would not be in place here. One special case of great import- ance may serve to illustrate. 160. If A = 0, 2 = 0, H3 = 0, H4 = 0, H =O, H16= 0 are the Eqs. in homogeneous point-Cds. of the sides of a 6-side inscribed in a conic, they are also the Eqs. in homoge- neous line-Cds. of the vertices of a 6-side circumscribed about a conic: the Eqs. of the two conics will be the same in form, but one will be in point-Cds., the other in line-Cds. By Pascal's Theorem the junction-points of HI = 0 and H4 = 0, H2 = 0 193 CO-ORDINATE GEOMETRY. and HI = 0, H3=0 and H1,=0 lie on a RL.; the Eq. which says this, interpreted in point-Cds., says, when inter- preted in line-Cds., that the junction-lines of HI, =0 and H4 = 0, 2 == 0 and H = 0, 113=0 and 16=0, meet in a point. Thus is found Brianchon's Theorem: Bhe three diagonals through the counter-vertices of a 6-side circumscribed about a conic meet in a point. This correlate to Pascal's Theorem was first proved, though not as above, by Brianchon, a pupil of the Polytechnic School at Paris (1827). As there are sixty Pascal lines (Pascals), so there are sixty Brianchon points (Brianchons). By Pascal's Theorem we can find a sixth point of a conic, 2 knowingi 5; by Brian- chon's we can find a sixth tangent, know- / ing5. For be 1 2, 3, 4 the junction-points of the pairs of consec- utive tangents, taken 4 in order; take on the fifth tangent any point, as vertex 5; draw 1 4 and 2 5; through their section and 3 draw a RL.; it will cut the first side at vertex 6; then is 5 6 the sixth tangent. By Pascal's Theorem we could find the tangent at any vertex; 2 oM by Brianchon's, we can find the tangent-pohit on any tan- gent. For, suppose the tan- gent or RL. 5 6 to fall together with 6 1 upon 5 1; then 5 1 touches at 6. Draw 1 4 and 2 5; through their section and 3 draw a RL., the third diagonal; it will cut 5 1 at 6. 194 RECIPROCAL CURVES. 195 161. Line-Cds. are of special use in dealing with loci of poles and envelopes of polars. If the pole (as to any referee) move on any curve L, its polar will turn around some curve E. The junction-line of two points of L is the polar of the junction- point of the polars of those points; if these points be consecu- tive, their junction-line is tangent to L; then their polars are consecutive, and the junction-point of these polars is the point at which they, fallen together, touch E. Hence the polars of all points of E are tangent to L; i.e., as the pole traces L the polar envelops E, and as the pole traces E the polar envelops L. Hence the relation of L and E holds when the terms are exchanged; i.e., it is a mutual relation. L and E are called reciprocal curves as to the referee; this may be any conic, most simply a circle. We may now prove Brianchon's Theorem from Pascal's thus: The poles of the six sides of the hexagon inscribed in L are the vertices of a hexagon circumscribed about E. The junction- points of pairs of counter-sides of the inscribed hexagon are poles of the diagonals (junction-lines) of counter vertices of the circumscribed; since the poles lie on a RL., the polars go through a point. Since L is any conic, so is E. Of course, it is just as easy to prove Pascal's Theorem from Brianchon's; it is done by exchanging clauses in the sentence, "since the poles, etc." Neither theorem is logically first. .These methods of double interpretation and of exchanging the notions of pole and polar have received the names of Prin- ciples of Duality resp. Reciprocity. They are in last analysis one, and their possibility is given in the fact that the plane, whether viewed as full of points or full of RLs., is doubly extended: there are as many points as RLs. in the plane; a point for every RL., a RL. for every point. The referee sets these points and RLs. in relation to each other. In conclusion, it is to note in regard to reciprocal curves that, if a RL. cuts L in n points, throngh its pole go n RLs., polars of those points, all tangent to E; hence the degree (resp. class) of either of two reciprocal curves is of the same 196 CO-ORDINATE GEOMETRY. number as the class (resp. degree) of the other. Hence the point- (resp. tangential) Eq. of either will be of the same itegree as the tangential (resp. point-) Eq. of the other. Hence the reciprocals of conies are coniCs: if the pole traces a conic, the polar envelops a conic; anld conversely. Note on Points and Right Lines at Infinity. In view of the extensive and important use made of the notions of point and RL. at x, it may be well to ground these notions more thor- oughly than could be done in the body of the book without breaking quite the thread of thought. All reasoning is in first intention not about things, but about notions or concepts. It is a familiar fact of every-day life that the same thing may be conceived variously, and that the conclusions that hold about it may vary accordingly. Important illustrations have already met us. Two coincident points are in themselves one and the same point; it is only the mind that thinks the point now as on this curve, now as on that. So con- secutive points are in themselves one and the same; it is only in thought that they are held apart. The obverse of this fact is the less familiar one that things in them- selves different may be, indeed, must be, thought as the same if there lie no mark to distinguish them in thought. The conclusions that hold about them will be the same. Such things are all points on a RL. whose dis- tances from any given point of that RL. are unassignably great. How- ever apart they may be in themselves, they cannot be held apart in thou qht. [fence all such points are, for thouyht, one point, and we speak with strict- est accuracv of the one point at co, the one point, not of the RL. out of thought, but of the RL. in thought. Such, again, are all points at ao on parallel RLs. If y = 0 and y = b be two such RLs., then indeed the o points of the two would seem to have this mark of distinction, that the y of the one is 0 while the y of the other is b. But this difference, while it distinguishes them in their outer being, does not yet distinguish them in thought. For it imparts no property to the one that does not belong to the other. By the side of the oc value of x, the finite value of y loses all distinguishing power. This is clearlv seen on drawing a RL., say through the origin, towards the 0o point of the RL. y = 6; the RL. is clearly none other than the X-axis, y = 0, for any other RL. will not extend toward the so point of y = 6, but toward some definite finite point of it. Hence we say all parallel IlLs. meet at as, meaning that the marks of distinction in our notions of the EXAMPLES. 197 points of the RLs. vanish utterly as the points are thought retiring on the RlLs. without limit, leaving the notions of all the points undistinguished, one and the same. That parallel RLs. meet at so is, then, no merely con- venient form of speech, but states a fact, not of things but of thoughts. Like reasoning applies to the RL. at so. It might indeed be said that there are many RLs. at A, for any RL. might be thought pushed to X while kept in a given direction; but all such RLs. lose all distinction of direction in thought, yea, since each must go through the o point of each axis, they fall together in thought completely. Hence the phrase the RL. at so correctly expresses our thought of them. Let the student beware of confounding the notion or concept of a thing, which is given in its definition and is the subject-matter of thought about it, with its mental image. Of points and RLs. at oo there are no such images at all. EXAMPLES. Centre and Diameters. 1. Find the centre and the pencil of diameters of 5x2+ 12xy-3y2+8x-10y-7=0. A=15 6 4 =5(-4)-6(-22)+4(-18)=40, 6 -3 -5 4 - 5 - 7 C=-51, G =-22, F =-18; /22 18 hence the centre is (22, 518); the pencil of diameters is 5x + 6y + 4 + A(6x - 3y -5) = 0; the central Eq. is 5 x2 + 12 Xy 3y2 _40 =; 51 the curve is an H. 2. Discuss in like manner these Eqs.: 3x2 -8xy + 7y2 -4 x+ Gy= 13; 4X2_(xy+9y2+5x+3y=10; 7 X2 _ 10 Xy + 32- 8 x- 12y= 2; It has been thought best not to interrupt the development of the sub- ject, but to put all the exercises at the close. The teacher may introduce them as he deems fit; they are arranged in the order of the foregoing text. For very many the author has to thank Hockheim's admirable collection. 198 CO-ORDINATE GEOMETRY. 2x2-5xy-3y+ 9x- 13y + 10= 0; 3y2_2 xy+ y2+ 2x+2y +5=0; 13x2+ 14xy+5y2+ 14x+ 10y+5=0. 3. Turn the axes so as to make the term in xy vanish. Be kx2 + 2 hxy +jy2 + C = 0 the central Eq.; put X=x'cosa-y'sina, y=x/sina+y'cosa; on substitution -2k -sin acos a+ 2jsinacosa + 2h (COSa2-_sina) must -0; hence tan 2 a= 2h and the Eq. takes the form [k+j- (k-j)2 +4hz]x2+ [k+j+ V(k_ j)2+ 4h]y"2 +2c=0. Illustrations: 10 X2 _ 6 xy + 7 y2 = 30; 9x2 + 16 xy-20 y2 = 60. 4. Turn the axes till the X-axis is 11 to the axis of the P kx2 + 2 hxy +jy2 + 2gx + 2fy + c = 0. Here C=kj -h=0, or h= v'kj; the axis of Pis 11 to kx+ 'Wy 0 ; hence tan a= - -\ = _k _h the reduced Eq. is j h j k2 +j Y12 +2 ( - k9 2 f kg ),+ c =0. k + kyt2( v/k(2+j -2 Illustration: 9 X2 - 6 xy + y2 + 4 x + 3y + 10 = 0. 5. A diameter of + = 1 is y = sx; what is the conjugate a2 62 6. Find the chord of I +2 = 1 through (1, 2) and halved by 11 13 2y = 3x. 7. A diameter of x-2 - = 1 is ysX; find the conjugate. 8. Find the chord of L + 1 halved by (4, 2). 36 9 The diameter through (4, 2) is 2 y x; the conjugate is 2 y + x 0; hencethechordis 2(y-2)+(x-4)=O, or 2y+x=8. 9. Find the chord of -- - = 1 halved by (5, 3). 4 49 10 n -2 find the 4 between 31 x and its conjugate. EXAMPLES. 199 11. In 25 x2 - 16 y2 = 400 find the conjugate diameters sloped 450 to each other. 12. Find where the diameter through (2, 3) and its conjugate cut 4x2+ 12y2=48. 13. Find lengths of the diameter 4y=6x and its conjugate in 49.y2+ 9 X2=441. 14. Find the slope of the diameter through (x1, YJ) to its conjugate. The direction-coefficients of the two diameters are A and zb2 -x1-- XI a' y, a2 .Yi X1 b2 hence tang b - ' Z=tb, since a2y,2 b2z2= + a2b2. a2 15. In -x + -/2= 1 find the conjugate diameters sloped 1200 to each other. 16. Find the length of the diameter conjugate to 2 x = 5y in -4x2+ 25y2 - 100. 17. Given a' = 7, b- = 10, Up = 110; find a and b. Remember the relations a12 + b'2 = a2 + b2, a'bl sin 4 = ab. 18. Find the slopes to the axis major of the equi-conjugate diameters of 64 y2 + 25x2 =1600. 19. Prove that the diagonals of the parallelogram of tangents at the ends of conjugate diameters are themselves conjugate diameters. 20. Form the axial Eq. of the conic: when a = 13, ae = 12; when a + b = 27, ae = 9; when the conic goes through (1, 4), (-6, 1); when it goes through (X1, 1Y), (X22 Y2); when b2 = - 144, ae = 13; when e = 3; when a = 4 and the conic goes through (10, 25). 21. Find the points of a centric conic for which x and y are equal. When and how can these points be constructed 22. Find the axes of the conic whose vertical Eq. is y2 = 55 x _ 2 x 2. 23. Find the vertical Eq. when b = 6 and the parameter = 5. 24. An H goes through (xl, yr) and the parameter is q; find the vertical Eq. and the length of the real axis. 25. Express the vertical Eq. of H through e and b as known. 26. Show from the central polar Eq. of a conic that the sum of the squared reciprocals of two I diameters is constant. 200 CO-ORDINATE GEOMETRY. 27. Find where x- Z=1 iscutby y =sx+c. a2 b2 28. Interpret geometrically a2s2 + b2 - c 0. Since s = tan 0, we may write a2 a2e2 cos O2 + c2 cos W; hence the RL. cuts the E in real, coincident, or imaginary points, according as the foot of the focal I on it lies within, upon, or without the major circle; vice versa for H. 29. When does kx2 + 2 hXy +jy2 + 2gx + 2fy + c = O touch either or both axes 30. Find the Eq. of the tangents from the origin to kx2 + 2hxy +jy + 2gx + 2fg + c=0. 31. Find the Eq. of the RL. halving each of the positive half-axes of a centric conic; where does it cut the conic 32. Two vertices of an equilateral hexagon inscribed in the E 25 y2 + OX2 = 225 are at the ends of the axis minor; where are the others 33. Inscribe in a centric conic a rectangle of area 2 ab. Tangents and Normals. 34. Tangent and normal form with the X-axis an isosceles A; find its vertex, the point of tangence. 35. At corresponding points of an E and its major circle are drawn tangents; prove that the subtangents are equal. Hence frame a rule for drawing a tangent at any point of an E. 36. Find the Eqs. of tangents to 5y2+ 3x2= 15 11 to 3y-4x-1=O0. 37. Find Eqs. of tangents to 36 y2 + 25 X2 = 900 sloped 30 to the X-axis. 38. Find the Eq. of the tangent to 9y2 + X2 = 9 when the X-tangent is 5. 39. Find the point of tangence whose abscissa equals the subtangent. 40. Form the Eq. of a circle whose diameter is the tangent-intereept between the vertical tangents. Where does it cut the X-axis 41. Form the Eq. of a circle whose centre is on the axis minor, and which has the Y-tangent as a chord. Where does it cut the other axis 42. Find the 4 between the tangents at (x,, yl) and (x2, y2) and the Eq. of the diameter through their intersection. EXAMPLES. 201 43. Find the ratio of the rectangles of the ordinates resp. abscissas of the points of touch of two I tangents. 44. Find the common tangents to a centric conic and its mid-circle (x2 + y2 = ab), also the 4 under which the curves cut. 45. Find the 4 between the tangents from (xl, Yi) to a centric conic. Foci and Directrices. 46. From the foot of a directrix (on the axis) is drawn a tangent to a conic; from any point of the tangent is dropped a I on the axis; from where the I cuts the curve is drawn a ray to the focus; find the ratio of the I to the focal ray. 47. Find the Cds. of the pole of lx + my + n = 0 as to a conic. 48. Draw a tangent to a conic at a given point, a focus and its direc- trix being known. 49. The diameter through P (xi, yl) cuts a directrix at D; find the 4 between the polar of P and the focal ray of D. 50. Given an 4, the counter-side, and the sum or difference of the other sides of a A; find the sides and angles. 51. Find the sum of two focal chords 11 to two conjugate diameters. 52. Find the rectangle of the segments of a focal chcrd. 53. Find the harmonic mean of the segments of a focal chord. 54. Find the harmonic mean of two I focal chords. Asymptotes. 55. Find the 4 between the asymptotes of 4 X2 _ 5 y2 = 100. 56. How long is the focal I on an asymptote How long is the asymptotic intercept between the two Is 57. Find the Eq. of the tangent at (xl, yl) to 4 xy = a2 + b2 58. When does y=sx+c touch 4xy =a2+b2 59. Find the Eq. of the RL. through (xl, yi), (x2, l2) on 4 xy = a2 + 1b2. 60. The asymptotic intercept between two chords joining two fixed points of an H to a variable point of the H is constant. 61. Find the asymptotic Cds. of the pole of the chord through (x,, yl), (x2, Y2)- 202 CO-ORDINATE GEOMETRY. 62. Find the asymptotic Eqs. of the directrices. 63. The asymptotic distance of a point of an H from a directrix equals the focal distance of the point. 64. The focal ray of a point of a conic, the I through the focus on the ray, and the polar of the point go through a point. The Parabola. 65. The vertex of a P is (a, b), the par-amneter is q, the axis is 11 to the X-axis; what is the Eq. of the P 66. The axis of a P is y = 6, the X of the vertex is 2, and the curve goes through (7, - 8); find the Eq. 67. For what point of y2 = 4 qx is i n-tinmes x 68. What is the Eq. of y2= 0l X when w=4', the a3is being a diameter and the tangent through its end 69. Find the rectangle of the ordinates of the ends of a focal chord. 70. Where does y = 5 x + c cut y2 = 4 qx 2 Interpret q sC. 71. Find the side and height of an equilateral A inscribed in a P. 72. Find the Eq. of the chord through (xl, yj) that is halved by (x, y,). 73. Find the Eq. of the P whose axis is 11 to the X-axis, wvhose param- eter is 3, which cuts the X-axis at (12, 0), and touches the Y-axis. 74. Finsl the Eqs. of the tangents from the origin to (y - b)2 = 4 q (x-a). 75. The vertex of a P is at (a, 0), its axis falls on the X-axis, and it is touched by lx + my + n = 0; find its Eq. 76. Find the tangent-lengths and normal-lengths in P. 77. Find the Cds. of the point of touch of a tangent sloped p to the axis. 78. When is the normal-length equal to the difference of subtangent and subnormal When is the rectangle of the tangent- and normal- lengths equal to the square of the ordinates 79. Find the ratio of the tracts drawn fromn any point of a tangent to the focus, and the foot of the I from the point of touch on the directrix. 80. Find the Eq. of a tangent to y2 = 4 qx sloped p to y _ s.r + c. EXAMPLES. 203 81. Find the rectangle of the subtangents of two I tangents. 82. Show that tangents at the ends of a focal chord are 1. 83. Find the Eq. of the focal ray of the intersection of tangents at P1, P2. 84. An isosceles A is circumscribed about a P; show that the vertex, the point of touch of the base, and the focus lie on a RL. 85. An equilateral A is circumscribed about a P; the transversals through the vertices and the points of touch of the counter-sides go through the focus. 86. Under what 4s do X2+ y2 = q2 and xq +y2=p2 Cut y2= 4qx 87. Find a P whose axis falls on the + X-axis, and which touches y2 + (x -a)2 = r2 enclosing it, e.g., y2 + X _ 132 = 25. 88. Find the common tangents to X2 + y2 = r2 and y2 =4 qx. 89. Find the common tangents to the co-axial P's y2 = 4 qlx, y2=4q2 (x-a). 90. Find the 4 between the tangents through (xl, yl) to y2 = 4 qx. 91. What is the pole of lx + my + n = 0 as to y2 = 4 qx 92. P is on the directrix of P; F is focus; the polar of P meets the curve at I, I'; PF is geometric mean of IF and FI'. 93. The diameter of a P through P cuts the directrix at D; show that FD is I to the polar of P. 94. The axial intercepts of the polars of two points as to P equals the axial intercept of Is from them on the axis. 95. The Eq. of a pencil of ki chords of a P is y = sx + b; what is the Elq. of the conjugate diameter' 96. If a chord cuts off equal segments from two diameters of a P, the diameters cut off two equal segments from the chord. General Focal Properties. 97. The focal I on a tangent meets the central ray to the point of touch on the directrix. 98. The focal ray to a point of a conic equals the ordinate of the point, prolonged to the tangent at an end of the focal ordinate. CO-ORDINATE GEOMETRY. 99. The focal polar Eq. of a chord, the slopes of the rays to whose ends are ,1 +2' 61-e2, is 2q:p=-ecosO+see2cos((-61). 100. Hence, show that the tangent at (pi, 0,) is 2 q: p =- e cos 6 + cos (6- 0j). The Eq. of a RL. through (p', 9') and (p", 9") is ppI sin 9-6' + p'p" sin -'- 9" + pfp sin " -e 0= . Put 0, + 02 resp. 61 - 02 for of rcsp. e", and for p' resp. pt' put 2q resp. 1-ecos' -eCos 0t since (pt, 9'), (plf, 9') are on the curve, clear of fractions; so we get, on collecting, p sin (6- 1 + 0) (1-e Cos 01-0,) + 2q sin 2 62 +psin (6-62-6) (I-ecos01+6 ) = 0, 2psin(-02) cos (9- e) + 2q sin 292 -e p sin (60- 1 + 02) Cos O1-62-sin ( -1-62) Cos 1 + O2 = . Applying the addition theorem of the sine to the bracket, we get 2p sin (- 02) cos 6 - 61 + 2 q sin 2 02 + e p sin 2 02 0, whence, on transposition and division by sin 2 02 = 2 sin 62 . cos 02, there results the Eq. of Ex. 99. In Ex. 100, 02 = 0. 101. If 0 be fixed, PF' any chord through it, then tan JPFO. tan PFO is constant. By 99, the Eq. of PP' is 2q: p = -e cos 6 + see , + ,9 cos (6- a + + A). 204 EXAMPLES. 205 Hence, 2q:p1=-ecosa&+23 +sec3+ 13'cos (13-61') forthe point o (pi, a+ 26i). Or, cos 1- : cos 1 + 13' is constant; or, cos ,.-,S1-cos3+1Al: cos6 -13+ cos13+13'I is constant; or, - nsin h = tan 3. tan 13 is constant. Cos's Cos 102. Normals at the ends of a focal chord meet on the It to the axis major through the mid-point of the chord. For they meet in the centre of the circle through the ends of the chord and the other focus; the proof is now readily completed. Hence, find the locus of the intersection of such normals. 103. Find the locus of the intersection of I normals to a P. The magic Eq. of the normal is y - sx + 2 qs - qs3 = 0. The I normal is q + 2 qs2-Xs2-ys3 =0. These Eqs. consist when, and only when, o 0 y x-2q 0 -q =0. o q x-2q 0 -q 0 y x-2q 0 -q 0 0 O- 0 q 0 2q-x y o q 0 2q-x y 0 q 0 2q-x y 0 0 Multiply the first row by y, and the third by q; add to the first q times the fourth; take from the third y times the sixth; there results 0 q2 + y2 xy-2qy 2q2-qx -=0O. y x-2q 0 -q qx-2q2 xy-2qy -(q2+y2) 0 q 0 2q-x y Multiply the last row by q, add to it y times the second, then add to first row 2 q - x times the second; there results y(2q-x) q2 + y2- (2q-x)2 y (x-2q) O 0. q (x-2q) y(x-2q) -(q2+y2) y 2+ q 2 y (x-2q) q(2q-x) Take the second row from the third, set out the factor y2 - qx + 3 q2; this factor equated to 0 satisfies the Eq. of condition, making the determi- nant 0, and is the locus sought: a P one-fourth as large as the original, its vertex where normals at the ends of the focal chord meet. This result may be got more simply otherwise, but the above illustrates the general method and the use of determinants. CO-ORDINATE GEOMETRY. Eccentric Angle. 104. The lengths of two conjugate half-diameters, at and b', when e is the eccentric 4 of a', are a/2=a2cos 2+ b2 sin e-2, 2= a2 sin e +62cos F. 105. rind the length of a chord of an E in terms of eccentric 4s. 1 2 a2 (cos F _cos F2)2 + 62 (sin - sin C2)2, or 12i2=4 1 2i4 a2sin i+ + b2 cos "I +2 The bracket is the squared half-diameter 11 to the chord; call it D3; 12 2 D3 sin 2 . Or, much more neatly, thus: The corresponding chord of the major circle is 1/ 2'= 2 Df sin -fl2. 2 In projection 11 chords are changed in the same ratio; the 11 diameter 2af or 2,J changes to 2 D3, hence 2D' sin' F" 2 to 2D3sin 1l f2- 2 2 106. Find the area A of the A whose vertices are (eF), (F2), (F2). The sides of the corresponding A' are 2 a multiplied by sin "I - sin 2 L3 resp. sin i3 Ii; 2 2 2 the double area of any triangle is the product of the sides divided by the diameter of the circumscribed circle; 2 At= 4 aa sin L2 sin "2 sin 32 - 2 2 2 ' hence A = 2 ab sin "I 2.sin "I sin f-3 c 2 2 2 Clearly, also, 2 Al= aa sin Fl- F2 + sinF2 - f3 + sinF3 - n hence sin F1-F2 + sin F2-F3+ sin F3-El 4 sin "I 2 sin 2-' .sin !f-!I. 2 2 2 If r be the radius of the circle about the A (F1, F2, E3), then 206 EXAMPLES. 207 2A=12-23-31 :2r, or r=12.23.31:4A, r=D3. D. ab A focal chord c, is a third proportional to the axis major and the 11 diameter; i.e., 2a: 2 Di = 2 D1: c1; hence results r2=Clc2-c3, if 2q= b2 16q a 107. Find the area A of the A of tangents touching at (E,), (62), (bE). The area A' of the corresponding L' is plainly Al = aa tan i.' + tan2 3 ' + tan '3 ; .-.A =ab ... . By applying the determinant formula for the area of a A fixed by three RLs. we find A=abtan "I-2 tan f2 2 -. tan g; 2 2 2 hence the sum and the product of the three tangents are equal, a relation hold- ing only when the sum of the Ws is a multiple of jr. 108. Show that the area of the A formed by three normals is as _2 tan tan, tan El 4al) )2 2 2 sin e1 + E3 + sin e, + 63 + sin 63 + e(. Areas. 109. The side of a rhomb inscribed in an E is s, the linear eccentricity is the geometric mean of the half-axes; find the area of the E. 110. In the E 26 y2 + 9 x2 = 225 find the area of a sector whose centric + reckoned from the axis major is 600. 111. Find the ratio of the segments into which the parameter cuts an E. 112. Find the ratio of the parts into which a concentric circle through the foci cuts an E. 113. Find the ratio of the parts into which a confocal P, with vertex at the centre, cuts an E. 208 CO-ORDINATE GEOMETRY. 114. Find the segment cut off from lA2X2 - a2y2 = a2b2 by x = d. 115. Find the area bounded by b2X2 a2y2 = a21b2 and the RLs. Y=c, Y=-c. 116. Find the area bounded by an H, its major circle, and q a, y - a. 117. An equilateral A of side s has for its altitude the axis, for its vertex the vertex of a P; find the segments cut off hy its sides. 118. The segment cut off by a focal chord of a P is one-third of the trapezoid of chord, directrix, and diameters. 119. A focal ray is prolonged ivy twice itself; through the point thus reached is drawn a diameter to the P; compare the triangular areas bounded by the ray, the curve, the diameter, and the axis. If the ray meet the curve at (x, y), the ratio is found to be 28x+ l2q which is 10 when the ray is I to the axis. x + 3 q 120. Find the ratio of the parts into which y2 = 4 qx cuts y2= 16qx-X2. 121. Find the area of the figure bounded by y2 = 4 qx and x2 = 4 cy 122. Find the area between y2 = 4 qx and y2 = 8 q (x - q). 123. Find the area of a focal sector of a P. Determination of the Conic. 124. Determine elements of 3 X2 -4 xy + 5 y2 - 30 x-16 y -20 = 0. A=-2017, C=11, the centre (G:C, F:C) is (91, 54), the curve L (it\ 1 111 is an E, the central Eq. is 3X2-4 y+ 5y2 20l7=0, the Eq. of the 11 pair of RLs. through the centre and the intersections of the E and a con- centric circle is 3x2-47y + 5y2 2017 72 =0; these RLs. fall 11 r2 together in an axis when, and only when, 2017f) 5 (5 201'\ = 4 20 11 7 V 11 r7 or 20172_8 +2017+l =4 --15+l6=5; 1 F1 r, - 11 r2 EXAMPLES. 209 whence 2017 = 4 A; whence the axes are (v'5+ 1)X2+4zxy ( 5-1)y2=0 and (N- )X and (4-lz 4 xy + (51) y2 = , or _2 and Y _,+v/-l0 /5-1 x +ijI 2 while the half-axes are respectively _4 ______ and V4V5.2917 11 11 125. Determine the elements of X- 5 xy + y2 + 8x -201y1+ 15 =0, 36 X2 + 24 xy + 29.y2 - 72 + 126y + 81 = 0, 2x2-5xy - 3y2+ 9x-13y + 10=0. 126. Determine the elements of 9X2 - 6 xy + y2 + 4 x + 3 y + 10 = 0. A =-169: 4, C = 0, the curve is a P; the Eq. may be written (3x -y + A)2 = 2 (3 A-2) x- (3 + 2 A) y + A2- 10, or L2 = LI; the RLs. L =20 and LI= 0 are a diameter and a tangent at its end; they are I, and .-. are the axis and the vertical tangent when 3 2 (3A-2) + 1=0, 20A- 9=0, 2 2xk+ 3 20 The parameter is 4= 4 (3- 2-2)2 +4(1. 9 _ -3)21 4 20 20 2 i IV R31) 432+12 100 127. Determine the elements of 25 X2 - 120 xy + 144y2 -2 x - 291y = 1, 9 x2 -12 xy + 4y2- 24 x + 16y-9=0, 4x2+ 9y2-8x+ 54y+ 85=0. 128. The linear eccentricity of kX2 + 2 hxy + jy2 + 2 gx + 2fy + c = 0 iS Vv(k+j Yj)2+4z2 .A: C. 210 CO-ORDINATE GEOMETRY. 129. Find the conic through (0, 0), (2, 2), (18,6), (32, 8), (72,12) by in Spection. 130. Find the conic through (3, 7), (-2,-8), (11, 31), (9,-2), (17,1). 131. Find the conic through (1,2), (0,1), (1: 4, (v+ 5) :4), (!,,2), (7:8,(15-VY):8). 132. Find the conic through (- 8,0), (3, VI :12), (4, -2: (I, - 1: '), (6, X/7 :3). 133. Find the P touching the axes at (4, 0) and (0, 3). Wehave K=0, J=0, C=0, 16k+ 8g+c=0, 9j+6f+c=0, whence (3 x + 4 y)2 - 24 (3 x + 4 y -6) = 0; to the lower in the double sign corresponds a P, to the upper a double RL. through the points of touch. 134. Find the conic touching both axes, the Y- one at (0, 4) and going through (16: 3, -20: 3), (-3, 10 + 515): . Constructions. 135. Given the conic drawn, to determine its elements. Draw a pair of pairs of 11 chords, and through the mid-points a pair of diameters; these meet in the centre. If the conic be a P, draw two chords I to the axial direction; their halver is the axis; from the foot of any ordinate lay off a subtangent double the abscissa, through its end and the end of the ordinate (on the curve) draw a RL.; it is a tangent to the curve. Through the mid-point of the tangent-tract draw a I to it; it meets the axis at the focus; the directrix is I to the axis and is counter to the focus as to the vertex. If the conic be centric, draw on any diameter of it a half-circle, and from where this meets the conic draw to the ends of the diameter a pair of chords; they are supplemental and I; hence, the diameters 11 to them are conjugate and I; i.e., are the axes. In the E, from an end of the axis minor draw a circle with the half-axis major as radius; it cuts this axis in the foci. But, in the H, the ends of the (minor or) conjugate axis being imaginary, draw the vertical tangents (I to the real axis); draw the diam- eter of any two 11 chords, and at its end a tangent ( 11 to the chords); on the intercept of this tangent between the l! tangents as diameter, draw a circle; it cuts the real axis in the fici. A circle about the centre and through the foci meets the vertical tangents on the as, mptotes. Combining the Eqs. of an asymptote and the major circle, we see they meet on the directrices; but this construction of the latter is available only in case the EXAMPLES. 211 asymptotes are real, i.e., in H. In the E, draw a tangent at the end of a focal ordinate; it cuts the axis major on the directrix corresponding to that focus. 136. To draw a tangent at a point to a conic. In P, lay off from the foot of the ordinate the subtangent double the abscissa; thus is reached a second point of the tangent. In E, draw a tan- gent at the corresponding point to the major circle; it cuts the major axis in a second point of the tangent to E. In H, halve the inner W of the focal rays to the point. The like construction of course holds for P and E. 137. To draw a tangent from a point to a conic. On a focal tract to the point, as diameter, draw a circle; it meets the major circle in two second points of the two tangents. In P the vertical tangent is the major circle. 138. Given the foci and one point (or tangent) ; construct the conic. To give the focus at ao is the same as to give the direction of the axis of P. 139. Given a focus, an axial direction, a tangent and its point of touch. Use the counter-circle of the other focus. 140. Given 2 a, a focus, a tangent and its point of touch (or an asymp- tote). 141. Given a focus, two tangents (and their points of touch in E and H). 142. Given a focus, and one diameter in length and position. Find other focus and the axes, and use the major circle. 143. Given three tangents and a focus. Use major and counter circles. 144. Given a focus, two tangents, and the axial directions. 145. Given the centre, a focus, and a tangent; find where tangents from a point P touch the conic. On the focal tract FP as diameter draw a circle; through its intersec- tions with the major circle draw RLs. from P; they are the tangents; to find points of touch, use the counter-circle. 146. Given the centre, axial directions, a tangent and its point of touch P. Through P and the intersection of tangent and axis minor draw a circle with centre on the axis minor; it passes through the foci. 147. Given the centre, a tangent, and 2 a. Use the major circle. 212 CO-ORDINATE GEOMETRY. 148. Given the centre 0, a point P of an E, and 2 a. Find the corresponding point P'; draw through P a H to axis major cutting 01" at B; OB is axis minor in length. 149. Given the centre, axial directions, and two points of the conic, ', 2. Express a2 and b2 through the Cds. of PI and P2. 150. Given two tangents and their points of touch, and the direction of the axis major. Construct a diameter and apply 149. 151. Given a point P of an E and the axis minor in length and position. Draw 11 to axis major, through P, cutting minor circle at PIc; the cen- tral ray through f11 cuts the ordinate of /' on the major circle. 152. Given a tangent and the axis major in length and position. 153. Given two conjugate diameters, AA', BB', in length and position. Draw a tangent at B; lay off a' on the normal from B; through the end of a' and the centre draw a circle with centre on the tangent; it cuts the tangent on the axes. 154. Given any pole P and its polar L, a directrix, and the position of the axis major. Let L meet the directrix at D; on PD as diameter, draw a circle it cuts the axis major at a focus. 155. Given the asymptotes and the foci, or the transverse axis. 156. Given the transverse axis and a point of the H, X2 = a2 157. Given the asymptotes and a point P of the H. Draw through P tracts ending in the asymptotes; from P lay off on the longer segment the difference of the two segments; so are got any number of points of the H. Draw the tangent at P 1 to the fourth har- monic to the asymptotes and the diameter through P; the asymptotic Cds. of the vertex are each the geometric mean of the asymptotic Cds. of P. 158. Given the asymptotes and a tangent. Halve the tangent intercept. 159. Given the asymptotes and the difference of a and b. 160. Given an asymptote, a tangent, its point of touch, and a secor:! point. EXAMPLES. 213 161. Given the centre, an asymptote, a tangent, and the ratio a: b. 162. Given the centre, an asymptote, and two points. 163. Given an asymptote and three points. 164. Given the vertical tangents, point of touch of one, and a third tangent. 165. Given two tangents and the focus of a P; find the point whose focal ray is the half-sum of the focal rays to the points of touch. 166. Given focus and (1) two points, or (2) one point and a tangent of P. In (1) draw about each point a circle through the focus; either outer common tangent is directrix; in (2) either tangent to the one focal circle from the counter-point of the focus as to the given tangent is directrix. 167. Given the directrix and two points (or one point and axis) of a P. 168. Given the directrix, a tangent, and a point of a P. 169. Given the vertex, the axis, and a point of a P. 170. Given the axis, a tangent, and its point of touch (or the vertex). 171. Given the vertical tangent, another tangent, and its point of touch. 172. Given two tangents and their points of touch, in a P. 173. Given the vertical tangent and two others, in a P. 174. Given three tangents (or two points) and the axis of a P. 175. Given four tangents to a P. Use focal circles. 176. Sides, altitude, base of an isosceles A are tangents, axis, chord of a P. 177. A P touches one side of a A at its mid-point, and the others pro- longed. 178. Given the axis, the parameter, and a point of a P. Draw normal first. 179. Given the directrix (or focus, or axis), a pole, and its polar as to a P. Loci. 180. Find the locus of a point of a tract whose ends move on fixed RLs. 181. Fixed are the base of a A and the point of touch of escribed circle ; find locus of vertex. CO-ORDINATE GEOMETRY. 182. The product of Is from P on two RLs. is k2; where is P 183. Given a side and counter 4 of a A; find locus of its mass-centre. 184. Find the locus of the fifth noteworthy point in a A, given a pair of counterparts, or a pair of adjacent parts. 185. Find the locus of mid-points of chords of an E, through a point. 186. From two points on a tangent to a circle, d apart, are drawn two other tangents; where do they meet 187. An E turns about its centre; where it cuts a fixed RL. tangents are drawn to it; where do they meet 188. A circle intercepts given lengths on two given RLs.; where is its centre 3 189. Find the locus of mass-centre of a A of constant area, two of whose sides are fixed. 190. A A is formed by a fixed RL. and two sides of a given 4 turning about a fixed point; find the locus of centre of circumscribed circle. 191. Tangents from P to P form with the polar of P a A of constant area; find the locus of P. 192. A vertex of a A is fixed, the constant counter-side is pushed along a RL.; find the locus of the centre of the circumscribed circle. 193. The base of a A is given; the vertex glides on y + nx2= mx, whose directrix is 11 to the base; find the locus of centre of mass of the A. 194. Find the locus of pole of tangent to y' = 4 qx as to X2 + y2 = r2. Homogeneous Co-ordinates. 195. The Eq. of x cos a+ y sin a-p =O in homogeneous Cds. (N1, N2, N3) is V1N + v2N2 + v3N3 = 0; find v, Y2, Y3. We have AVl = x cos a, + y sin al-p, = 0, and so for N2, N3,; hence, P, Cos a, + v2 Cos a2 + v3 Cos a3 = Cos a, rl sin a, + v2 sin a2 + P, sin a, = sin a, 'l Pi + v2P2 + P3P3 = P; P,=Cos a sin a2p3:A, v2 cos a, sin aP3: A, v3=,cosa, sin aspI:A,& A=Jcos a, sin ap31. 214 EXAMPLES. 215 196. Base lines are L, = 5x-2y + 1 =0, L2,=2x-y-3= 0, L3 = x + y + I = 0; find the Eq. of x + 2 y-4 = 0 in homogeneous Cds. (L,, L3, L3). Wehave 5A,+2A2+A3_1, -2A1-x3+X3=2, A,-3A3+A3=-4; whence, on finding Al, A2, A3, substituting in AI L,+ A2L2 + A3L3 =O and multiplying by 23, -20 L1 + 39 L2 + 45 L3 =O. 197. Show that the two Eqs. m. 1N2 +2N N3 + T, N, = 0 and N, sin A, + N2 sin A2 + N3 sin A3, = O represent each the RL. at o. 198. Where does Yv N, + v, N2+ V,7 N3= 0 cut the sides of the referee A Find the intersection with N, = 0 from P,2 N2 + V3 N3 = 0, T2 N2+ -SN3: 2A, whence N2 2 AV3 (.r V3 T3 V2), 13 = 2 A2'2: (r3 V2-T2 V), and so for the others. 199. Where do v1 N v2 N2 + ,N3 = 0 and v,'N, v2-N + v,3tN3 = 0 meet Since 1 N + T N2 + T3 N3 = 2 A ,N = 2 v,' V2Tr and so on. 200. Find the Eq. of the RL. through (N'1, N2', N31) and (N",l/ N2'l, Nt) Assume ,, NV '2 N2 + ,v N3 =0; then vP'N,'+ v2 NJ+v3 N1z = 0, vl iV l''+2 -v21'+ V3 N3"= 0 I .N I - 0 201. Find the RLs. through (N,', NJ/, NV31) and the vertices of the referee A. 202. The ELs. N, = 0, N2 = 0, v2 N2 + V3 N3 = O v/1N, + vN3t' = 0 form a four-side; find the diagonals and where they meet. 203. Show that the diagonals of a four-side are cut harmonically. 204. Homogeneous Eqs. of 11 RLs. differ only by constants. 205. Find Eqs. of RLs. through the vertices of the referee 11 to the counter-sides. 206. Find Eqs. of ELs. through the vertices and the mass-centre of the referee. 207. When are v, N, +v2 N2 v3N 3P = 0 and ''VN + V2tN2 +PtN3' = 0 perpendicular 2 208. Find Eqs. of the altitudes of the referee and the Cds. of the orthocentre. 209. Find Eqs. of junction-lines of the feet of the altitudes of the referee. 216 CO-ORDINATE GEOMETRY. 210. Find Eqs. of the mid-perpendiculars to the sides of the referee. 211. Find Eqs. of IlLs. through the vertices of the referee and the points of touch of the inscribed and escribed circles. 212. Find the distance of (N, 1, Nx2/, N1,) from v1 NL + v2N2 + V3 N3 = 0. When but two Cds. are used to fix a RL., call them I and m, and write tie Eq. of the enwrapt point thus: P= III + 1-0. 213. Find the Cds. of the junction-line of P1 = 0 and P2 = 0. Proceeding exactly as if to find the junction-point of LI= 0, L2 = o- we find I = (VI-r2) : I ul'2 1, in =- (u.-u,) : Iu2) 1. 214. Find the distance of P1 = 0 from the ilL. (7,, mnl). To say the Cds. of the RL. are 11, inl, is to say its Eq. in rectilinear Cds. u, v, is 1lu + mlr + 1 = 0; to say the Eq. of the point is ull + vrn + 1 = 0, is to say its rectilinear Cds. are ul, rl, since they fulfil the Eq. of any iL. through it; hence, the distance sought is 11111 1-s7j1 + 215. Find the Eq. of the point that cuts the tract between Pi = 0, and P2 = 0 in the ratio n, : n2. Think of (ual VI) (up, v2) as the points in rectilinear Cds., then (nIu2 + n2H' nl'2 + n2V1 (' nl 2 - I + . l) is the cutting point; the Eq. of this point, viewed as enwrapped by the varying RL. (/, m), is nl12 + n2U`1 + nIc2 m 1=0 I + , m + 1 = O. nl + n2 ni - n2 For n, and n2 like-signed, the point is an inner one, otherwise an outer one. The Eq. may also be written 'niP2 + 1n2PI 0 . aI1 + n. The inner resp. outer mid-point is PI + P2 = 0 resp. P -P =P 0. 216. The vertices of a L are P1 = 0, P2= O, P3= 0; find the mass-centre. From (215) it is seen to be (P1 + P2 + P3) :3 = 0. 217. Three vertices of a parallelogram are P1 = 0, P2 =0, P3= 0; find the fourth. 218. Find Eq. of any point on junction-line of P 0=O and P2 0= It is P1 -kP2 = 0; for this is Eq. of a point, being of first degree in EXAMPLES. 217 1, m, and it lies on a RL. through Pi = 0 and P2= 0, since it is ful- filled where they are. From (215) it is seen that k is the ratio of the distances of the point from P1 = 0 and P2 = 0; varying k gives all points on the RL. 219. When do P= 0, P2= 0, P3 = O lie on a RL. Clearly when I zq,1 l= 0, or when kP, + k2P2 + k2P3 = O. 220. Show that the points at ao on the sides of a A lie on a RL. 221. The product of the ratios in which the sides of a A are cut is 1; then, and only then, the cutting points are on a RL. The cutting points are P1 - 3P2 =, P2-k1 P3=0, P2-kP2r1=o; the determinant of these three Eqs. vanishes only when k1 .2 . k3 = 1. 222. How do Pl + kP2=0 and P2 + kP1 =0 lie on the RL. (P1 P2) 223. The cross-ratio of Pi - k-P2=0, k = 1, 2, 8, 4, is k -2 -.k3-k4:k2- k3. k4-1 All the problems of the text as to rays may be repeated as to points. All the problems in homogeneous point-Cds. may now be paralleled by problems in homogeneous line-Cds. E.g.: 224. Be Pi= 0, P2= 0, P2= 0 three vertices of a A; express P = 0 through them in the form KP, + K2P2 + K3P3 = 0. See (19o). 225. Writing p = P:N + in: , show that in Kl pl + K2P2 + K3P3 = the K's are proportional to the distances of this point from the sides of the referee. 226. Find Cds. of the RL. through KIcPI + K2P2 + Kp3 3 =0 and KIc'PI+ K 2 P2+ K 3 P=O. 227. Find Eq. of junction-point of RLs. (P'p'2, p3) and (p"1, P"2,P"3)- 228. When is KIPI + K2 P2 + K3p3 = O a point at oo The RLs. (Pi'P21 p,) and (p, + d, p2 + d, p3 + d) are 11; since they go through the same point at a, their Cds. satisfy the same Eq.; or, Kl pl1+i2p2+KBp3= and K1(p+d) +c2(P2(+d)+ K3(p3+d)=0; whence K, + K2 + K3 = O 229. Show that the mass-centre of the A Pi = 0, P2 =, p3 = 0 is P1 + P2 + P33=0. 230. The centre of the circle about the referee is PI sin 2 Al + p2 sin 2 A2 + p3 sin 2 A3 = 0. CO-ORDINATE GEOMETRY. 231. The orthocentre of the referee is Pi tan A, + P2 tan A, + p3 tan A3 = 0. 232. Mass-centre, orthocentre, and centre of vertices of a A lie on a RL. In the determinant of the Eqs. of the points multiply first row by 2 sin A, sin A2 sin A3; take from third row; take out 2 cos A1 cos A2 COS A3. 233. The centre of the circle in the referee is p, sin A, + p2 sin A2 + p3 sin A3 = 0. 234. The intersection of the transversals from the vertices to the points of touch of the escribed circles is cotA + P2 cot A, + p cotAs=O. 2 2 2 235. The points of (229), (23.3), (234) lie on a RL. 236. The point KlP1 + K2p2 + K3p3 = 0 is distant from the RL. (P'IP'2,P'3) (KP'1+ K2P2 + K3)'3):(K + K2 + K3)- Envelopes. 237. When does the RL. (1, mn) touch the P whose axis falls on the + X-axis and its focus on the point ql + 1 = 0 The RL. Iu + my + 1 = 0 touches y2 4qu when the roots of v2 I + mv + 1 = 0 are equal, or when mt 4 q 1, which is therefore the 4q Eq. of the P in line-Cds. 238. Through a fixed point P is drawn a RL., to which, where it meets a fixed RL., is drawn a I; find the envelope of the 1. Take the fixed RL. as Y--axis, the I to it through P as X-axis; be u and v the intercepts of the enveloping RL.; then V2 =pa, i.e., the envelope is a P, P is the focus. 239. Find the tangential Eq. of E referred to its axes. The intercepts are u = a2: x, V= b2: yl; whence, on squaring, inverting, and putting x22 a,2 +b2 1 x + f= 1, results -a + -= 1; or, calling the reciprocals of the inter- cepts I and rn, a212 + b2m2 = 1. Otherwise, thus: a= 0 0 ; whence K=-I2' J=7-a2 C= a212 0 1 0 b2 0 0 -1 whence, on substituting and clearing, the same result is got. 218 EXAMPLES. 219 240. About the point (e, o) is drawn a circle with radius 2a, from (-e, o) is drawn a ray to the circle; find the envelope of its mid-perpendic- ular (a e). The Eqs. of the circle and ray are x-e2+y2=4a2, y=s(x+e); that of the mid-perpendicular is ( Va2 (1 + s2)-e2s2-es2\ + / _ s /a2 (1+ s2) -e2s2 + se 0. 1 + S2 + V 1+ S2 The intercepts of this are Va2(1 + S2)-e2s2 = u and Va2(1 + s2)-es2: = whence eliminating s and putting 1, m for 1: u, 1: v, we get a212 + (a2 e2) M2-= the envelope is an E. 241. Through (e, o) and (-e, o) in the circle X2 + y2 = a2 are drawn 1I chords; find the envelope of the RL. joining two ends of the chords on the same side of a diameter. a2t2+(a2 -e2)m2 = 1 242. The 11 s y = a, y =-a, meet the circle x2 + y2 - 2 Ax-e2, and the points of meeting are joined crosswise; find envelope of junction- line when A varies. a2m2+(a2_e2)12 . How does the envelope change as e changes X2+ 243. The pole, as to - Y+ = -, traces the circle x2 + y2-a2 a2 b2 what does the polar envelop Ans. a42+b4m2 = a2 HINT. Eq. of the polar is :1+ + '/' = 1; here xl-= I Y- m- a2 b2 a2 I - 244. Find the envelopes when the pole traces X2 + y2 = b2 and a2 + 62. 245. Find the envelope of the junction-line of the ends of two conju- gate diameters. Ans. The E a212 + b2m2 = 2. HINT. The ends are the points (a cos e, b sin c-) and (- a sin e, b cos s ); the RL. through them is xb(sin e - cos E)-ya(sin - + cos E)+ ab = - ; here (cos e - sin.E)= al, (cos c + sin e)= bmi; hence the above result, on squaring. 246. From the point (e, o) rays are drawn to the circle (x + e)2 + y2 = a2 ; find the envelope of their mid-perpendiculars. 247. The vertex of a right W glides on x2 + y2 = r2, one side en- wraps the point (e, o); what does the other side enwrap Ans. r212 - (e2 - r2)rn' = 1 220 CO-ORDINATE GEOMETRY. 248. What is the tangential Eq. of 7 Xo _ 5y2 + 12 x + 8 y-47 =O 249. Show, in two ways, that the tangential Eq. of H referred to its asymptotes is lm (a2 + h-2) = 1. 250. When does the general tangential Eq. of second degree, KP+2Hlm+Jm2+2G01+2FnP + C=O, represent an E7 when an H when a P. Be kx2 + 2 hxy +jy2 + 2 gx + 2J + c =O the Cartesian Eq. from which the tangential is got by putting for k, Ad, etc., their co-factors K, H, etc., in A; also suppose A 0. Then, as both Eqs. picture the same curves, the criterion is the same for both: the curve is E, P, H, according as C'0, C=0, C0. 251. Putting If, hr, etc., for the co-factors of K, H, etc., in A, show that the tangential Eq. pictures two points when A = 0 and c' K 0, pictures one double point when Ce = 0, h' = 0, t = 0, 252. Discuss 230 and 251 geometrically, remembering that from every point of the RL. at oo may be drawn two tangents to E; only from outer points may they be drawn to H; from every point may be drawn only one tangent to P. since the RL. at o itself touches P; and combine with the given Eq. of second degree the Eq. 1 -Am = 0 of a point at so . 253. From a point on the X-axis are dropped Is on the RLs. x = y and x + 2 y = 10; find the envelope of the junction-line of the feet of the Is. Ans. A P. 254. Two RLs. mutually I turn about a fixed point; find the envelope of the junction-line of their intersections with two fixedl RLs. 255. Through (o, d) is drawn the secant x + y = d of the system of circles X2 + y2 - 2 Ax = d2; find the envelope of the tangents at the points of secancy. 256. Through (o, d) are drawn secants to XI + y2 = r2, and Is drawn to the secants at the points of secancy; find their envelope. 257. Find the envelope of the polars of a point as to a system of con- focal conics. 258. A secant cuts a system of confocals; find the envelope of tan- gents at the points of secancy. 259. The two points at ao in: H are real; P, coincident; E, imaginary. 260. Two As, one formed by tangents to a curve, the other by chords joining the points of tangence, may be called outer resp. inner, and said to correspond. Show that in P the outer is half the inner. EXAMPLES. 221 261. Establish Carnot's Theorem: The product of all the ratios in which a conic cuts the sides of a closed polygon is 1. HINT. Bv Art. 72 the product of the ratios in which the conic F= 0 cuts the side P1P2 is F1: F2 262. If Fl=kIx2+2khxy+jly2+2glx+2.fy +cl, and A be a parameter, then F1 + AF =O is a system of conies: through how many points what pairs of RLs. belong to the systeml what P's when does a circle belong to it what is the locus of the centres how lie the polars of a point as to the members of the system how lie the poles of a given RL. 263. Show that any RL. cuts the system in a system of points in Invo- lution, whosefoci are the points cutting harmonically the chords of the base conics. 264. Show that when the pole traces a RL. the perpolar envelops a P. 265. Find the envelope of normals to an E, an H, a P. 266. Find the envelope of the perpolar when the pole traces a conic. This page in the original text is blank. PART II. OF SPACE. Tnis subject is much more extensive than the Geometry of the Plane, so that any detailed treatment here is out of the question; only the most essential notions can be developed At the same time, the close analogy of the two doctrines permits a much more condensed discussion than was possible in Part I. CHAPTER 1. 1. We say of Space, it is triply extended or has three dimen- sions, meaning that three determinations are needful and enough to fix any element of it. These determinations may be made in many ways, giving rise to as many systems of determining y -x ,'I /1I o -x -z magnitudes, Co-ordinates. Thus, suppose three planes meeting at 0; call their intersections the X-, Y-, Z- axes, the planes CO-ORDINATE GEOMETRY. themselves the rz-, ZX-, XY-planes: when no confusion would arise, omit the words axes an(d planes. A plane 11 to YZ, cut- ting off a tract OA = a on X, has all its points at a dis- tance a from YZ measured 11 to X, and no other points are so distant; hence it is defined completely by the Eq. x=a. So, too, y= b, z=c, are Eqs. of planes 11 to ZX, XY resp. The first pair meet in a RL. 11 to Z, for all points of which, and for no others, the relations hold: x= a, y = b, which are there- fore the Eqs. of a RL. 11 to Z. So,too, y=b, z=c resp. z=c, x=a are Eqs. of RLs. 11 to X resp. Y. As a special case, x=O, y=O, z = O are the Eqs. of YZ, ZX, X Y; y = O, z = O, and z= O, x = O, and x = O, y= O arc the Eqs. of X and Y and Z. The three planes and three RLs. of intersection meet in a point for which, and which alone, hold all three relations: x=a, y=b, z=c, which are therefore the Eqs. of the point. It is most convenient to think XY horizontal, right or east being + X, forward or north + Y. as in Plane Geometry. Either up or down mav l)e taken as + on Z, but up is b)etter, according to the convention, important in Mechanics: T7at side of a plane is + whence positive rotation (as from + X to + Y) appears connter-clockwise. Clearly Space is cut by the three planes into eight re7iorns. The ipper four we name 1, 2, 3, 4, from the quadrants in XY -, +, l+ +, i on which they stand; those below, in the same order, 5, 6, 7, 8. -, -, + -e Then the signs of x, y, z in the eight regions are, as in the diagram, the lower sign of the z referring to the lower region. The X s yz, zx, xy may be denoted by X, qI, w; unless otherwise stated the! will be considered right 4 s. We may call x, y, z tiplavar Cds., and speak of the point (x, y, z). 2. Around any RL. (say Z) as axis, suppose laid a cylinder of radius r=ri. All points of the surface, and no others, are distant r, from the axis, and the surface is defined completely 224 SPHERIC CO-ORDINATES. by its Eq. r=r1. Through the axis pass a half-plane sloped 0 =0, to some base-plane through the axis (say ZX) ; then is this half-plane defined completely by its Eq., 0 = 01. For all points on the RL., the intersection of half-plane and cyl- inder, and for no others, hold the relations: r = r, 0 = 01 which are therefore the Eqs. of that RL. Pass a plane I to Z, hence 11 to XY, and distant z = z1 from this latter. By Art. 1, z = z, is its Eq. For the intersection of this plane and the RL. (rl, O1), and for no other, hold the relations r = rl, 0 = 01, z = zj, which are therefore the Eqs. of the point. We may call r, 0, z cylindric Cds., and speak of the point (r, 0, z) r may always be taken +, 0+ when reckoned counter-clockwise, z + when reckoned up. Connecting the two systems of Cds., the relations hold: x=rcosO, y=rsin, z=z. 3. About any point (say the origin 0) lay a sphere of radius p,; clearly its Eq. is p= p,. Pass a half-plane as in Art. 2; the Eqs. of the half great circle in which the half-plane meets the sphere are clearly p = p1, 0 = O'. About Z lay a cone sloped 8, to Z and do to XY, so that 8,+ += 90; its Eq. is 8 = 8,, or c=q+. For the point where it meets the half-circle, and for no other, hold the z z 81 -r - z , -S relations: p pi, 0 = 01, S51 (or 3= 31), which are therefore its Eqs. '225 CO-ORDINATE GEOMETRY. We may call p, 0, 4 (or 8) polar or spheric Cds., and speak of the point (p, 9, k); p may be taken always +, 0 + when reckoned counter-clockwise, 0 + where reckoned toward + Z (8 always +). Clearly 0 and 0 correspond to geographic Longi- tude and Latitude: 8 mav be called (north) polar distance. On projecting p on Z and on XY, and this last projection on X and Y, these relations become manifest: = p cos s cos 9 = p sin 8 cos 0, y = p cos 4 sin 0 = p sin 8 sin 8, z=p sin4 =p cos8. 4 Hereafter, cosine resp. sine may be denoted by putting a horizontal bar under resp. vertical bar a.fter the argument, thus: wI=slno, (=cos . Call the tract OPfrom the origin 0 to any point P the radius vector of the point, and denote it by p; denote its slopes to P X, Y, Z by a, ,3, y, and call their cosines a, /3, y direction-cosines of p. Then, by definition, the projections of p on the axes are the Cds. of P; i.e., x=ap, y=,3p, z=yp. Squaring, adding, and remembering xI + y2 + Z2 = p2, we get a2p2+J2=1 (A) 226 DIRECTION-COSINES. For one factor in each term put a, Y. Z, multiply by p, and get P P P ax +fjy + yz = p, (B) which simply says the sum of the projections of a train of tracts from 0 to P on OP is OP, as is already known. Now take a plane perpendicular to OP; it will be sloped a, f3, y to YZ, ZX, XY; let it meet X, Y, Z at A, B, C, and call the area of the A ABC A; the projections of this A on YZ, ZX, XY are the A BOC, COA, AOB; their areas are aA, ,8A, yA; squaring, we get a+ B+y2=A 2 since a2+/32.+y2.= 1 i.e., the squared hypotenuse-face of a right-sided tetraeder equals the sum of the squares of the otherfaces. This proposition is the analogue (for space) of the Pythagorean. The distance d between P1 (p,, a,, PI, yl) and P2 (p2, a2, /2, Y2) is plainly the diagonal of a parallelepiped, whose edges are X1-X29 Y -Y2 z1-z2; hence d2 = XI-X2 + YI 32 +Z1- Z2- By the Law of Cosines d2= pI2+ p22- 2PP2 PIP. Hence, PI P2 = (xI2 + y1 Y2 + Z1 Z2) P1 P2 = a, (12 + ,1 P2 + Y1 Y2 (C) This last expression for the cosine of the W between two RLs. in terms of their direction-cosines holds even when the RLs. do not meet, since the W between two non-intersecting RLs. equals the W between two 11 intersecting RLs. COROLLARY. When, and only when, Pi and P2 are 1, al a2 +,8192 + yl y2 = (C1) The 4 between two planes equals the (adjacent) 4 between two Is on the planes, but the slope of a RL. to a plane is the complement of the slope of the RL. to a I on the plane; hence, the slopes to the axes of a plane I to p are A = 90' - a, B=90o-/3, r=9O0-y; hence, AJ2 +B12+Pr2= 1 and 111112=A1A2J +B1B21 +r1r2 (C2) 227 228 CO-ORDINATE GEOMETRY. 4 In case of oblique axes we use the theorem: the projection on any RL. of a tract between two points equals the sum of the projections of any train of tracts between the points; hence p. pi =. xl + y..yl + Z. zi. Take as 1, in turn, the vector p and X, Y, Z; so we get px.xp+31./p+Z.Zp p.xp=x+y./x+Z.zX e p.qp=x.xyf+y+z.zq, p.zp=x.xz+y.yzz+z. On multiplying in turn by p, x, y, z, and adding, there results p2 X2 + y2 + z2 + 2 yz. yz +2 zx. zi +2 2xy.xy. Hence ip, y,' zp, or a, B, y, are found at once by using (D). To express conversely x, y, z through p, a, B, -y, form the determinant A of the four Eqs., remembering yz = X, zx = , wy = We; A =- 1 a B -y, and put 52 =1 w, a, 1 W C o 1 x B1X X1 then, denoting the co-factors in A by like letters accented, X = alp: S2, y = ftp:3 2S, Z = -1p : T. On putting these values in the first Eq. of (D) and clearing, there results X 12a2 +j,112 + WoI 2 X12By-2 it ya-2 w'aS = S2, (A) where X"t, V1, c.f are co-factors of like letters in S2. To find the distance d between two points, P and P1, take PI as a new origin (see Art. 6), then the Cds. of P are x - xi, y - Yi, Z -z1; put these for x, y, z, and d for p in the formula found. To find the cosine of the W POP1, put Pi for 1; then ,-14 1-1 -,A 11 p . pp = X . Xp1 + y . yPi + Z . zp1; on substituting for the cosines on the right, there results PP = xx1 + Yy1 + zz + X(yz + Zy )+ (zx, + xz1) +(xyl + yx1):Pp1. (C,) On comparing (A) with (A) and (C,) with (C,), analogy would sug- gest the following Eq. as (C2): PROJECTIONS OF A TRIANGLE. 229 PP = xI2'a, + '.616 + j2ry1-x"(T,+ aY) - (zya + a.-y)- .'I(al + 0al): SW. (C2) This conjecture is readily confirmed thus: on putting, in (C,), for the Cds. their values found above, the product ppi vanishes, SI becomes divisor, subscribed and unsubscribed letters (a, , y) combine every way in sets of two, and the result, symmetric as to the subscribed and unsub- scribed letters, since p = PIP' is of the form Aaa1 + B1 + Cry1 + AI(B71 + ey8) + B' (7ya1 + trl ) + C(aB1 + S1) = S4'. ppe where A, B, C, Al, B', C' depend only on Xv q1, w. For ppi = 0, this must pass over into (A), for then a = al, etc.; hence A = X 1 2. S2, etc., which on substitution yield C2, as guessed. We have seen that in case of rectang. axes O -C+ COA2+AoB2=ABC2, i.e., the squared area of a A (and hence of any plane figure) equals the sum of the squared areas of its projections on three rectangular planes. To find the general theorem for oblique axes, of which the above is a special case, we note that the corresponding formulae for rectang. and oblique axes are, when W A OB = c, A = o- 2 + 6B2 and A1 B2 =-A2 + 0B2 - 2 OA. OB.cI. On putting OA = a, OB = b, these formulae may be written AB =-0 a b , /2 o a b ba 0 1 b W 1 So, too, putting 0 C = c, we have in case of rectang. axes 4ABC2=- 0 he ca ab; whence 4ABC =-0 be ca ab be 1 0 0 be 1 4, If a 0 1 0 ca ax 1 X ab 0 0 1 ab b 1 X 1 if the analogy holds, a result easily verified. The area of a parallelogram whose sides, a and b, are sloped w is ab sin w, or ab1 w1 230 CO-ORDINATE GEOMETRY. Accordingly we might suspect the volume of a parallelepiped whose edges a, b, c are sloped co, X, to each other to be abc 1 X X1 This conclusion from analogy is readily verified thus: take the paral- lelogram ab. wel as base, project C on X at Al and on XY at C' (the edges being taken as axes); calling the diedral 4 along X A, we have CAI = c , CC' =c. . A 1. By spheric trigonometry A I= -/1- X A- -aw2 + 2xi:41 I; the volume is ab. wi. CC, whence the formula above. The radical of the determinant, which is S of A, is thus seen to be the rolume of a paralielepiped of unit edges, sloped w, X, to each other. Since the factor S turns the product of the edges into the volume of the parallele- piped, just as sin ab turns the prodnet of the sides into the area of the parallelogram, it has been named (by Staudt) sine of the solid angle of the edges, and may be written sin abc or abcj. If ,, t t be the slopes of X to YZ, Y to ZX, Z to XY, then CC-=c. 4; but CC'=c.ij.A[=c.xyzJ: wl; hence xyzl= l.wl, or S=91.XI=7l lg.l To find the area of any A (or other plane figure) in terms of its projections, call these projections I.yz , J.zx , K. xyI (or I.XI, J4, J K.w), and suppose the I p on the plane of the A directed by a, f, -y; then A.a=IxI.4J, or A.a=I.S, since each is the projection of A on a plane I to X. Hence Aa=.SA.,6= J.S, A.-y=KW.S. If h be the altitude and I the edge of a prism standing on A as base, and i, j, k be the projections of I on X, Y, Z. then h = ia + ji + k7-y; hence A.h = volume of prism =(iJ+ jJ+ kK)S. The projections of a tract on the axes, 11 to the Cd. planes, resp. the pro- jections of a plane figure on the Cd. planes, 11 to the axes, are called Cds. of the tract resp. plane. VOLUME OF A PYRAMID. 231 Now take the pyramid whose vertex is 0, whose base is PP1P2; it is I of the prism with base OPP2 and edge OP; the Cds. of the edge and the double Cds. of the base of this prism are x, y, z, and Y121 xI,I Z, xl I 1,I XIY, I' I I 2 Z2 Z2 X2 X2 Y2 Hence by the above formula we have 60PP1P2= x y z S. Xi Y1 Z1 Xs Y2 Z2 To move the origin to (x3, y., z3), it suffices to put x - x3 for x, y-y3 for y, z - z3 for z, etc., and write 6 0PP1P2-6 PP1P2P3 = 0 x - x3 y y3 Z- Z3 S. o X1-X3 Y1Y33 Z1-Z3 o X2-X3 Y23/Y3 Z2 Z3 1 X3 Y3 Z3 On adding the last row to each of the others there results 6 PP1P2P3 = 1 x y z S = 6 T. 1 X Y Z1 1 x2 Y2 Z2 1 X3 Y3 Z3 When, and only when, this six-fold tetraedral volume is 0, does the point P(x, y, z) lie in the plane of P1P2P3; hence T=0 is the Eq. of the plane through the three points P1, P2, P3. 5. By projecting a point 11 to Z (say), its X and Y are not c/tanged; i.e., the x and y of a point are the x and y of its XY-projection, and are the same for all points of a RL. 11 to Z. To find, then, the x and y of a point cutting a tract P1P2 in ratio n1: n2, project the tract on XY; the Cds. of the projection are the Cds. sought: - nX2 + n2XI "=IY2 + n2Y1 and so n l2Z + n2z ni+ n2 n + n2 alnd+o z Transformation of Co-ordinates. 6. For pushing the axes, not changing their directions, clearly X= X+ X0, Y= Y3 +Y, Z=Z'+zo. CO-ORDLNATE GEOMETRY. If the axes (rectangular) be turned from the position X, Y, Z into the position X', Y', Z', so that the 4s XX', XY', XZ', are at, a"t, a"', then the sum of the projections of x', y', z' on X, for any point, is simply the x of that point; or X = X'a+ Y'o.+ Z'af1i; and so y = x'P'+ y'P"+ Z'p"', (E) z = XI-II+ y'y"+ ZI'"'. The nine Xs are clearly not all at will, since, if a' and j3' be chosen, y' is thereby fixed; for all RLs. sloped a' to X lie on a cone about X, and all sloped /3' to Y on a cone about Y, and XI must be a common RL. of these two cones; the axis X' being fixed, so is the plane YZ', and the choice of one more 4 fixes Y' and Z'. So only three of the nine can be chosen at will; hence there must hold six Eqs. of condition among the nine Xs. These are aI2 +,82 +yr2= 1, a IIa+'+ 0, r + yw 2t= o a'12 + fI112 + Ytt2 = 1 a aM'-+ /ft-. flsf + yFI. ./tl= 0, a"1t2 + M1 2 + yFII2 = 1 ; a aF+ A1 AF+.yrf.. y= 0. The first three say X, Y, Z are rectangular, the second three say X', Y', Z' are rectangular, as appears from (A) and (B). The formula for passing from X', Y', Z' to X, Y, Z are plainly x'- Xal+yf3+zy', y' = Xalt+ y/3"+ ZyI, ZI = Xa"'f+ y1ttM+ Zy'tv. Accordingly there must hold these six Eqs. of condition: a'2 + a"x2 + a"M2 =l at. /31+ all. sft+ a1t M1= 0, 312 + /3112 + /31112 = 1 . F'+ /3t. yf1+ f31f ,Y"= Y a-+ = - all+ r- at-t= 0. These six Eqs. must then be eqti'valeit to the first six; this is clear geometrically, and may be proved analytically thus: 232 TRANSFORMATION OF CO-ORDINATES. Form the determinant C = a' all ap" ptr prr pr rl by solving the first three Eqs. directly, we get C. x'= A'x + B'y + r'z, and so on; hence, A'= a'C, B'= fi', and so on; now at At+ a"- A"+ a"'1- A"t= C, and five other like Eqs., while at- B'+ all. B"+ a'". B"'l= 0, and five other like Eqs. whence, on replacing the co-factors, A', etc., result the twelve Eqs. By squaring C according to the Multiplication Theorem of Determinants, it is shown that C2= 1; hence C= 1, according, namely, as X'Y'Z' is congruent or symmetric with XYZ; i.e., according as, when +X' falls on +X and + Y' on + Y, + ZI falls on + Z or -Z. The formulae (E) hold even when X', I', Z' are not rectan- gular, since this rectangularity was not assumed in their deduction. The general formulae for oblique axes are found, precisely as in Plane Geometry, to be X. (x, yz) x'. (', Yz) I+ Y'. (y', yz) I+ Z'. (', Yz), and two got by permuting x, y, z. The nine coefficients are again connected by six Eqs. of condition. Note that the Eqs. of Transformation are linear in Cds. EXERCISE. Show that PIP2 I= al'2 k+ I1721I + 1 1 21 The Plane and the Right Line. 7. A single Eq. in x, y1, z rpresents a stface. For we may assign all real pairs of values to x and y (say), and reckon 2833 CO-ORDINATE GEOMETRY. the corresponding values of z. The pairs (x, y) fix points in XY; over each such point suppose the fixed value of z erected, laid off parallel to Z; the ends of all such z's will lie on and determine the surface, which of course may be real or imag- inary, continuous or discontinuous. Two Eqs. in x, y, z represent a line. For all points whose Cds. satisfy both Eqs. must lie on both surfaces picturing the two Eqs., hence must lie on the intersection. of the surfaces, i.e., on a line. Three Eqs. in x, y, z of mth, nth, resp. pth degree, represent mvp fixed points. For the three Eqs. are fulfilled at once by mnp triplets of values, each pictured by a point. Transformation of Cds. does not change the degree of the Eq. in x, y, z. For the Eqs. of transformation are linear. 8E A line is fixed by the Eqs. of any two surfaces through it; the simplest surfaces are generally two cylinders 11 each to an axis. The Eqs. of these cylinders are the same as the Eqs. of their intersections, each with the plane of the other two axes, since the third Cd. is the same for every point of any given element of one of them. They are clearly the projecting cylin- ders of the line, and their intersections with the planes are the projections of the line on those planes. Hence, as Eqs. of a line may be taken the EqIs. of any two of its projections, on the planes of two axes, 11 to the third axis. If the line be a RL., the projecting cylinders are planes, and the projections are RLs. whose Eqs. may be written y=sx+b', y=tz+b, x=uz+a, any two of which (generally the last) may be taken as Eqs. of the EL. Clearly, a, b', b are the intercepts of projections on X, Y, while s, t, u are direction-coefficients. Symmetric Eqs. of the RL. mar be got thus: Be (x1, y1, z1) and (x, y, z) a fixed and a variable point on the RL., d their distance apart, and let its direction-cosines be a, /3, y; then 234 RIGHT LINE IN SPACE. 235 d _-1 Y-Y1 -'-Z1 _ = ,) = - 7 a P Y or x= x+ad, y=yl+/3d, z=zl+yd. In X-1 IY-yi Z-Z1 a Y instead of a, p3, 'y, may be put any three proportionates, as X, f, v, so that X-X1 Y-Yi Z-Zi x V If X:a= :/8=v:y=f, then f = Vx-A +'2 + v. On comparing the symmetric and the tangential forms, we see s =/3: a, t = /i: y, u = a: y; whence u= t: s. If the RL. goes through (X2, Y2, Z2), then X2-X1 Y2-Y1 Z2-Z1. a i.e., X2 - X1-1 Y2- Y1 Z2- z1, are proportional to a, /, ; hence X-X1 Y-Yi z-Z1 X2-x1 Y2-Y1 Z2-z1 Eq. of a RL. through two points. 9. Two RLs. in space may or may not meet; if they do, the z of the intersection of their projections on YZ must be the same as the z of the intersection of their projections on ZX: i.e., if y=tlz+bb, x=uIz+a1, and y=t2z+ 2, x = u2Z + a be the RLs., then b- - b2t- t2 = a - a2:U - U2. Or the four Eqs. must hold at once; this yields the same result in a determinant. CO-ORDLNATE GEOMETRY. If two RLs., 11 and 12, are directed by al, /1, y7 and a29 127 2 or by their proportionals X1, 7 v1 and X2, p, v2, then 11 12ala2 + 1112 + YI12V=x1+ 2 f+ 1 + t4t9 + Ult2 /1 + t +11 u1 1 +t22+U22 If the RLs. be perpendicular, then X1+2 + IL2+ V .1'2 = 0 = 1 + tt2 + U1U2- If the RL. (t, u) be I to (t', u'), then 1 +tt'+uu'=O (1) if it goes through (x1, y,, z1), then y,=tzl+b, xl=uzl+a; (2) if it meets the RL. (t', u'), then b-b': t-et'= a- a,: u-'t. (3) These four Eqs. fix the values of t, u, a, b. They may be found thus: From (2), a = x1-uz1, b = y1-tzl; put these into (3), whence (u'z, + a'-x1)t-(t'zl + b'-yl)u = (a'-x1)t'-(b'-yl)u'. (4) From (1) and (4) can now be found t and u, thus: Put N= u' (u'z, + a'- x1) + t'(t'zi + b'- y,), M1= t'Z (b'-y) u'- (a'- xj)t' -(u'z, + a'- x1), L = u'I (a'- x1)t'- (b'- y1)u'l - (t'z1+ b'- Yi) then t =if:N, u=L: N; a = -x,- z b b=y - Zi N N1Kz Accordingly, x - x1: L = y- y: M= z - zj: N is the Eq. of the .1 from (x1, yl, z1) on the RL. y = tz + b, x = uz + a. 236 PERPENDICULAR TO TWO RIGHT LINES. 237 The student may now easily show that the Cds. of the inter- section are x1-(L: 1 +t'2+ U'2), 1 _ _,:+t'2+ U'2), Y-7:1 t'2+u'2), ZI -(Nf: 1 + t12 + Ubl2), while the distance from (x1, Y1, z1) to the intersection is VL2+JIM2+ N2: 1 +t'2+ u'2; or, after simplification, (a'- Xzt'- bl- yu')2+ (u'z,+ a'- X1)2+ (t'z1+ b'- yD)2 : V1 +Tt" +u -2 10. If the RL. y = tz + b, x = uz + a, be 1 to the two RLs. y=tlz+bl, x=ulz+a, and y = t2z + b2, X = u2z + a2, then tlt + Ulu + 1 = O, t2t + u2u +1=O whence t = ul- U2: tlu2 - t2u1, U = tl -t2: u1t2 -u2tl. If the RLs. be directed by (a, /3, y), (al, flk yl), (a2, /32, 72), then aia+ 9 + yiy=, a2a+ P213+y2y01 a2 + ,8 + y2 = 1. From the first two Eqs., on solving as to a: -y and /3: y, it results that a: /3: = /3:y2- 2y1: yla2-aY2: abt2-/31a2. If the RL. (t, u) also meets the RLs. (t,, ul), (t2, u2), then (b - b,) (u - u,) = (a - a,) (t - t1) and (b-b2)(u-u2)=(a-a2)(t-t2). CO-ORDINATE GEOMETRY. Form the determinant D= 1 1 1f, U U1 U2 t tI t2 and denote its co-factors, as usual, by like large letters; then bD= b1U1T2-b2U2T1 + (al- a2) U1U2, and so for aD. Thus is determined the common I to two RLs. The length of the intercept on it between the two may now be found by finding the intersections; but that is tedious. f1. The symmetric Eqs. of a RL. x X1 Y -= h _ = Y contain six parameters or arbitraries, A, 1, v, x1, Yi, Z1, which might be called the Cds. of the RL. in Space. But they are not independent, since four arbitraries (Cds.), (t, u, a, b), fix the RL. First, A, p, v are proportional to a, EB y, and these are connected by the relation a2 + 32 + -2= 1; secondly, denot- ing Lz, -vyl, vXT- Xz, Xy11-Ix, by A, M, N, we see the relation holds: AA +jUM+,N==O. These six symbols, A, ft, v, A, M, N, thus connected by two Eqs. of condition, we may call Cds. of the RL. The last three are interpreted geometrically later. (Art. 25.) 12. The Eq. of a plane. is offirst degree in x, y, z. For the Eq. of a plane 11 to a Cd. plane, as XY, is z=z1; this plane may be referred to any other system of axes by a linear trans- formation of Cd-s., and such a transformation cannot change the degree. Accordingly the Eq. of a plane is of the form Ix+my+ nz+d =O. (1) If a, b, c be the intercepts on the axes, then clearly they 238 THE PLANE IN SPACE. 239 equal -d: I, -di: m, -d: n; hence they vary inversely as 1, m, n; the Eq. of the plane may also be written -+6+_ =1. (2) a b c Conversely, an Eq. of first degree in x, y, z, represents a plane. For, if d be not 0, it may be brought into form (2), which is known to represent a plane making intercepts a, b, c on X, Y, Z; if d be 0, push the origin out any distance, say d', on any axis, say X, by putting x + d' for x in the Eq.; then the Eq. may be brought into the form (2), and the previous reasoning applies. 13. Drop a I p from 0 on the plane, directed by a, /3, Y; then p = aa = b,/ = cy, and on substitution results xa + y3+zy-p = 0, (3) the Normal Eq. of the plane, which we may write 2VN= 0. If F be the factor that turns the general into the normal form, then FE a, Fm- =,/, Fn=y; . F2(11+ M2+ n2) = 1, whence F= 1 :V2 + M2 + 2, and o, = l: Vl2+ m 2 + n2; and so for / and 3. In case of oblique axes we have abc. S p Vab. 2 + &e .X 2 + ca-. 12 -2(ab.wl.bc.x Y+ bc.xt.ca. l.Z+ca.4.ab. wiX), (4) since each is the six-fold volume of the tetraeder 0-A JIC, where OA = a, OB = b, O C = c, and X, Y, Z are cosines of the diedral :P along X, Y, Z. Call the Ls BOC, COA, AOB the planar intercepts of the 240 CO-ORDINATE GEOMETRY. plane whose axial intercepts are a, b, c, and denote them doubled by A, B, (; then abc.S= p VA2+ B12+ C2-2(AB.Z+ BC.X+ CA. Y). (5) In (4) put for a, b, c, p their values - d, --, -d, -Fd; hence I m n on v F- S: / . XI + mn. -4I + a.l1 -2(l.xim.iPI.Z+m.r4n.wIX+n.wll.xl'). (6) he analogy of this to the value of F in plane geometry becomes plain writing each in determinant form: F2=j 1w 1: 0 I m F2 = 1 c 4 : 0 1 M (7) m 1 x m X In X 1 By reasoning like that in plane geometry it is now shown that the dis- tance of (x, y, z) from the plane xc + yB + zy -p = 0 is xa + yO + z-y-p, or (x, y, z) is distant N(x, y, z) from N(x, y, z) = 0. The whole body of reasoning as to normal Eqs. of RLs. may now be repeated as to normal Eqs. of planes; and as there the Abridged flotation issued in a system of homogeneous triangular (or trilinear) Cds., so here it issues in a system of homogeneous tetraedral (or quadriplanar) Cds.; and just as the first could also be interpreted as line-Cds., so the second can also be interpreted as plane-Cds., a thought that cannot be developed here. 14. We have found the Eq. of a plane through three points and the six-fold volume of a tetraeder, given by its vertices, to be respectively Ix y z 1 1=0, Xl X, Z2 1 :X3 Y3 ZS 1 and 6 T= x 2X, 1 Xw, These two Eqs. are really one, the first merely saying that the volume is 0 when the fourth point (x, y, z) is in the plane of the other three. The same six-fold volume can he expressed as the product of the double base P1P2P3 by the I 1) from P on the plane of the base. This p is found by bringing the Eq. of the plane into the normal form,. by multiplying the determinant by F, where (8) y z 1 S. Yi zi 1 /2 Z2 1 Y3 Z3 1 TRIANGLES IN SPACE. F= S vSltx-3TTI 2+ T.l 2 -2(x'. XI y'- Z+y'- q1Iz'- WI X+z'. OAX'. xl Y) [ where x', y', z' are co-factors of x, y, z in the determinant. Hence, the radical C\/ is the double area of the A PIP2P1. We note that x' XI y'- pl, z'. wI are the projections of I-PP, on the Cd. planes 11 to X, Y, Z; i.e., they are the Uds. of the A; also, the negativ e terms are their (so-called) inner products ili sets of two; hence, The squared area of a A is the sum of its squared Cds. plus twice the sum of their inner products, in sets of two; or The squared area of a A is the inner squared sum of its Cds. The Cds. of the tract P1P2 being the differences of the Cds. of its ends, by observing signs, the square of the tract may be expressed as above. 15. The direction-4s a, 8, y of a I on a plane are called the Position-4s of the plane; clearly they are also the diedral 2(s of the plane with the Cd. planes, since the 4 between two planes equals the 4 between Is on them; they are the comple- ments of the 4s betweeu the plane and the Cd. axes. Tile position-cosines of a plane are : a12 + rn2 + 42, and two like ones. 15. In case of oblique axes, the position-cosines are still F7, Fin, 1,ss, but the position-4s no longer equal the diedral 4s. But any diedral 4, as between the plane and XY, equals the 4 between the I p on the plane and the I pz on XY. The direction-cosines of p are a, R, y, or Fl, Fnm, Fn; those of p_ are 0, 0, S: w (see Art. 4); put these values in (C,): i/, Fm, Fn for a, 6, -y, and 0, 0, S: X for a,, ap, y1, and get PPI = PPz = WI' 1 Fn -X"Fm -"Fl: S. w I =Fn (1-c2) -m (X-A)-I(4-wx):S. w. The cosines of the other diedral 4s are got by permuting symbols. 16. The 4 between two planes cre, vr2, whose Eqs. are 1x + mn/ +nlz+d1=O , 2X +m2I+n72Z+d2=O, equals the 4 between the Is on the planes; hence 241 CO-ORDINATE GEOMETRY. 7T7r2= (1l12+mm2+ nn2) :I,2+m2+nl2. 122+i22+n22. The slope of a RL. to a plane is the complement of the slope. of the RL. to the 1 on the plane; hence, if the Eq. of the plane be Ix+my+nz+d =O, and those of the RL. be x-Xi Y- YA Z-Z1 A Mu v rLI = (IA + m/l + nv) :V2+ m+ 72.i Hence, the two planes are -1 when 1112 + mM9 + n2n2 0 . They are 11 when 1112 + mm2 +- nln2 = a/112+ m112+ .112 _ \/122 + M 2 + n 2 i.e., when (11m2 - 12m,)2 + (mjn2 - m2n,)2 + (n,12 -n211)2 = 0. This sum of squares is 0 only when each is 0; i.e., only when 4 :12 = MI:M2 = nj 2. This condition is, indeed, geometrically evident, since it declares only that the position-cosines of the two planes are the same. The plane and the RL. are 11 when IA+ mu+nv= 0. They are I when IA + m/ + nv = a/12 +m2+ n2 -A+2+v2; i.e., when 1:A = m: , = n: v. Hence Ix+ my + nz + d = 0 and x - YY = -Z arel. I m n 242 INTERSECTIONS OF PLANES. 17. To find where a line meets a surface, replace two of the Cds. in the Eq. of the surface by their values in terms of the third, taken from the Egs. of the line; thus is got one Eq. in one Cd., whose roots are Cds. of the points of meeting. The number of these roots cannot be greater than the number of common points, though it may be less, since to any value of z (say) may correspond several values of x and y. If the Eq. in the third Cd. reduces to 0 = 0, i.e., is satisfied for every value of that Cd., then, and only then, the line has all of its points on the surface; i.e., the line is on the surface. In the special case of the plane lx + my + nz + d=0 and the RL. y = tz +b, x =uz+a we get (it& + t+ n)z+(la 4+mb +d)=0, and this reduces to 0 = 0, is satisfied for every z, only when lu+mt+n=O and la+mb+d=O, which Eqs. say the RL. lies in the plane. 18. The common point of three planes, 11x + mly + iZ + d1 = 0, 12,x -f vi,/ 4- nz -f d, = 0, ,.X + my1 + n3z + d3 = 0, is found by solving the three Eqs. as simultaneous; the results are x= -d1, m2 n31:111 m2 n3[, y=- 14 i2 n3!: 1l n2 sn3 z = -jl1 m2 d3: ill m2 13l-I Hence, if I 1m2n3I = 0, the common point is at oc, the in- tersections are 11 RLs. ; if, besides, a numerator, as 1,li, rn, ; be 0, the common point becomes indefinite, the intersections fall together, the three planes pass through the same RL. 19. If four planes meet in a point, their four Eqs. are satis- fied by the same triplet of values x, y. z; this can be ehen, and only when, Ill m2 n3 d4l=O. 243 244 CO-ORDINATE GEOMETRY. Or we may reason otherwise, thus: If 77r=0 7r==0, s= 0 be three planes, then Akl7' + A7-2 + A3 == 0 is a plane through their common point; for this Eq. represents a plane, being of first degree in x, y, z; and it is satisfied by that triplet of values sx, y, z, that satisfies the three at once. Four planes 7w,= 0, 72=, 7-,=O, 7-4= 0 meet in a point when four multipliers Al, A2, A3, A4, can be found such that AL7-1 + A27r2 + A3,w3 + A4`74 = 0 identically; for the triplet of values xi Y, z, which reduces any three of the 7r's to 0, must reduce the fourth 7r to 0 also. 20. If N1 = 0, N2 = 0, be normal Eqs. of two planes, then N, - AN2 = 0 is a plane through their common RL.; for any triplet x, y, z, satisfying the first two Eqs., satisfies the third. Also, A = N2: 7N2; i.e., X is the ratio of the distances of any point of the third plane from the two base-planes, or X is the ratio of the sines of the slopes of the third plane to the base- planes. Hence, N2 - N2 = 0 resp. V, +- IV= 0 is the inner resp. outer halver of the 4s between the base-planes. 21. To find the direction-cosines of a RL. haltving the 4 between two RLs directed by a, /, y and a', a', ye, take two i1s to the Rks., through the origin, and on each take a point dis- tant 2 from the origin; the Cds. of the points will be 2a, 2/3, 2yand 2a', 2,/', 2y'; the mid-point of the two will be (a+a', /3 + ,3', X + 2/ ), and will be on the halver sought; hence x Jo z a+ '13+ fl' y+Y' is the Eq. of the halver; hence a + a', /3 +3', y +y', each divided by (,, + at) 2+ (/ + f3)2 + (y + y_)2 are the direction- cosines sought. The radical V reduces to V2+ =2 =k2 -2 rhr itt 2 where 0 is the : betwreen the RLs. SYSTEMSIS OF PLANES. To find the direction-cosines of the outer halver, it suffices to change the signs of a', /3', y'; the radical iV then becomes 2-_2b = 2 . (+. 22. A system of planes through a point may be called a pen- cil of planes; for a system of planes through a RL. no fitting name has yet been used in English; perhaps cluster would answer best to the German Bueschel, suggested by the phrase in Architecture, clustered column. The common RL. of a cluster may be called its axis. Any two planes may be taken as base-planes of a cluster. To find what plane of a cluster 11X + mIy + 1iz + al-X(12X + m2y + n2z + d2) = 0 is I to lx + my + nz + d-O, we have at once, by Art. 16, 1(11 - \12) + m(mI - Xm2) + n(n1 - Xn2) = 0, whence A is to be found, and, on substitution, the Eq. of the plane is found to be (112 + mim2 + nn2) (11x + m,y + nz + d2) - (ll + mm, + nn,) (12x + m2y + n2z + d2) =0. The position-cosines of this plane are proportional to m (lim2 - 1mn,.) - n (n112 - n211), and two like expressions. But 1, mn, n, are proportional to the position-cosines of the given plane, say Xa + y/3 + Zy -p =0, and the paren- theses are proportional to the direction-cosines of the axis, say XF -I ... XX , -y,; = hence the above Eq. of the sought a p plane may also be written (x - x') (/3y'-sty) + (y-y') (- - yya) + (z - z') (a/3'- a-'f) = 0. 245 CO-O:DINATE GEOMETRY. 23. To find the plane through either of two RLs. 11 to the other, regard the RLs. as the axes of the clusters IIH-,X12=O, 113 - H4 = 0. The two 11 planes of these clusters are the planes sought. By Art. 16 l1 - Al12 m, -Am., -l - An, m=r (say). 13 -Kl4 Mi3- Xm4 fn3- Knb4 Clear each of the three Eqs. and solve for A; so we get X =1 l Ml n4 ! f12n13 1n4 , whence HII 11 ml n4 12 11 m3n41 =O andl advancing the subscripts by 2, U3114 ml xn21 -f41 13 ml n2 = 0, are the planes sought. Like results are reached by this reflection: The common I to the two RLs., say XXr Y _ to Z _Z' X - X t y _ Itt z _ Zi = = ...y-yKz- and = = at 1F3' y a" /3" y" is clearly 1 to the plane having the direction of both, 11 to both or through either 11 to the other; the direction-cosines of this I have already been found proportional to ,/3y"- fi"y', etc.; hence the planes are (X -o' (A,-it) + (Y -y') War' - Y lar + (Z- z') (af/3"-a"/'8) = O. and (x-x") (/LI /Y I) + The distance between these planes from any point, as (x', y', z'), of the first to the second, is plainly the (shortest) distance between the Rls.; the same is got by reducing the Eqs. to the normal form, dividing by the second root of the sum of the squared coefficients of x, y/, z, and then putting x', .y', z' for x, y, z in the second Eq. ; now that sum is the squared sine of the X between the RLs. (Exercise, Art. 6); hence, calling the distance d and the X 0, we have 246 MOMENT OF TWO LIGHT LINES. d 01 = (X'-x") (68I"'3 gy') + (y'-y") (y'a"-ylu') + (z'-Z") (a'/3"-a"/B). This expression, called by Cayley the moment of the two RLs., may be written as the difference of two determinants, thus: d .44 = Xl a' a" - XI a' a" r, zr 'YII I I, PI ye I which clearly - - --la'(/3"z"-yfly"l) +/ 3'fXI_aFzr) +y (a'tlyl- f +al(,8'z'--yy,) +,8B(f Y f-alzr) +ytf(a'yI-;rxI) - Now the Greek letters are the direction-cosines of the two RLs., and are but special values of A', ju', Y' and A"', )tM", V"I; while the parentheses are what, consistently with Art. 11, must be denoted by A", M', N" and A', M', N'; hence, disregarding sign, d 01= AFI+ljFM"+vN"+A"A+autAM+v"N', which expresses the moment of two RLs. through their Cds. 24. To find the volume of a tetraeder fixed byfour planes, III = 0, etc., it suffices to repeat, step by step, the reasoning in Plane Geometry as to the area of a A fixed by three RLs. The result is quite of like form: 6 T= [llm2n3d4 1:12n3n4liman4Hlim2n43l m2n3. 25. It is easy now to interpret A = ,/z'- yy', M and N. Let _ - - Z be the RL., then x I = 8 a /3 Y a'/3 y' is the Eq. of the RL. through the origin and (x', y', zI') hence x'= a'r, y'= ,/'8, z'= /'8, and A = 8(fly'- M = 8(ya'-y'a), N = 8(a/8'- a'/3). The multipliers of 8 are the position-cosines of the p)1ine through the two RLs. ; i.e., of the plane through the given RL. and the origin; or, they are the 24, 248 CO-ORDINATE GEOMETRY. direction-cosines of the RL. through the origin 1 to this plane, and hence A, M, N are the Cds. of a point on this RL. distant 8 from the origin; i.e., the C(ds. of a point on this RL. as far from the origin as (x', y', z') is. EXERCISES. 1. The tract PIP, is cut at P' in ratio n: n2, PXP3 is cut at P' in ratio n, + no: n3, PIP4 at P't in ratio n, + n, + n3: n4,; find the Cds. of PI", the centre of proportional distances of Pl, P2, P3, P,. 2. Find distance of (x, y, z) from a RL. through 0 directed by a, fi, Y. 3. If ri, = O. n,= O. n3= O. ,0= do not meet in a point, the Eq. of any plane is of the form ArI, + A2r12 + A3r13 + A4r14 = 0. Show that the n's, regarded as Cds. of a point, are proportional to fixed multiples of its distances from the planes; the A's, regarded as Cds. of a plane, to fixed multiples of its distances from the points Al = 0, etc. 4. Find Eq. of: a plane through two 11 RLs., a RL. through two points. 5. Show that the Eq. of a sphere of radius r, centre (XI, y", Za) is X_2 2 -2 6. What, then, are the Eqs. of a circle in space l 7. Find the centres of the inscribed and circumscribed circles of a A whose vertices are on the rectang. axes. 8. Show that the three median planes of a trieder (through the edges halving the counter-sides orface-.s) meet in a RL. 9. Three planes through the three edges of a trieder meet in a RL.; show that the compound ratio of the sines of the segments into which they cut the counter-sides is 1; and conversely. 10. Any plane through the vertex of a trieder cuts the sides into seg- ments the compound ratio of whose sines is -1; and conversely. 11. The 6 planes of intersection of 4 spheres meet in a point. 12. Three positive rectang. axes pierce a sphere about 0 at X, Y, Z; XI is the pole of the circle through X Y, Z; the point Xis carried up to X' along the great circle arc XXf; find how Y and Z move, and the formule of transformation from axes OX, O Y, OZ to OXt, 0 Y', OZ'. SURFACES OF SECOND DEGREE. CHAPTER II. SURFACES OF SECOND DEGREE. (Quadrics or Conicoids.) 26. The general Eq. of second decree has the form kx2+2hllXy+jy2+2yzx + 2fyz+iz-+21x+ 2 my+2nz+dd 0. It shall be referred to as F(x, y, z; x, y, z) = 0, or Q= 0. Before discussing it, certain general notions shall be premised. A surface may be thought as the locus of a point; it may also be thought as the locus of a line, or traced by a moving line. Suppose two surfaces F(x, C, z; p) = 0, (x, y, z; p) = 0, contain the same parameter p; for any special value of p they fix a single line as their intersection, while various values of p yield various such lines; by eliminating p between the Eqs. we get a relation holding between x, y, z for all points on all such lines; i.e., we get the Eq. of the locus of the line, the surface traced by it noving. If F and 0 contain two parameters p and p', they must be bound together by some Eq., as f(p,p') = ; the number of such Eqs. of condition is, in general, one less than the number of parameters. The moving line is called the generatrix (in any one position it is an element) of the surface. The motion of the generatrix is commonly defined as gliding along fixed lines called direc- trices. Since the Eqs. of generatrix and a directrix hold for the same triplet x, y, z, by eliminating these from the four EEqs. is got one Eq. of condition between the parameters for each directrix; hence, when the Eq. of generatrix contains n param- eters, it must glide on n -I directrices. When the genercatrix is a RL., the surface is called ruled. The Eq. of the RL. contains four parameters, hence three 249 CO-ORDINATE GEOMETRY. directrices define its motion. Ruled surfaces that can be un- wrapped upon a plane are called developable; the generatrix of such a surface always touches a fixed curve called cuspidal edge. Other ruled surfaces are called warped or twisted. 27. A cyliudric surface is the path of a RL. pushed. Be y = tz + b, x = uz + a the RL. ; then, since the direc- tion of the RL. changes not, t and u are constant; let 0 (a, b) = 0 be the Eq. between the parameters a and b, yielded by the Eq. of the directrix; on replacing a and b by their values results O(X - uz, y - tz) = 0 as Eq. of the cylinder. Or, be Ix+my+nz+d=p, lx+ my y+nz+d=p' the Eqs. of the RL.; letting only p and p' vary, we keep each plane 1I to itself, and hence all the intersections 11; if 4(p,p') = 0 be the Eq. between the parameters, the Eq. of the cylinder is O(lx+my+nz+d, I'x+m'y+n'z+d') =0. Hence any Eq. of a cylinder is an Eq. between two functions offirst degree in x, y, z; and the converse is clear. 28. A conic surface is the path of a RL. turned (about a point). Be _X_ . =Y-q the RL., (x1, y1, z,) being the fixed point about which it turns; then, just as above, (X-X1' Yy-Y =0O :Z-Z Z-:)= is the Eq. of the conic surface (or cone). We note the Eq. is homogeneous in Cd. differences; the converse is clear, that every Eq. homogeneous in Cd. differences pictures a cone. If the fixed point, or vertex of the cone, be the origin, the Eq. is homogeneous in x, y, z; and conversely. Both cylinder and cone are developable; the cuspidal edge of the cone is reduced to a point, the vertex, while the cylinder is but a cone with its vertex at x. 250 DISCRlMINANT OF THE QUADRIC. 29. A surface of revolution is the path of a line revolved about a fixed RL., the axis, to which it is supposed rigidly attached. Or, it is the path of a circle whose (varying) diam- eter is always halved by a fixed RL., the axis, at right angles. The generating circle in any position is called a parallel, the revolving line in any position, a meridian, of the surface. If - - Y-Yi = -z1 be the axis, then lx+my+nz=p I m n is a plane I to it, and x-x12+yy2 +zz12=r2 is a sphere about x1, y,, z1; any parallel is the intersection of two of these surfaces; hence, exactly as before, 2 1 2 +(x -xi + y - Y1 z -zi, lx+ My+ nz) = 0 is the general Eq. of a surface of revolution. 30. Returning to the Eq. Q =0, we note it may be written (kx+hy+gz+l)x+ (hx+jy+fz+m)y + (gx+fy+iz+n)z+ (Ix+my+nz+d) = 0. (1) The parentheses may be called, in order, Q., Q., Q,, Q1. Again, we note there are ten coefficients; but, by dividing by any one, the number is reduced to nine; these may be deter- mined by nine independent Eqs. ; hence, nine sinple conditions are needed and enough to determine a quadric. To pass to 11 axes through a new origin, x', y', z', put x + x', y + y', z + z' for x, y, z; then, by reasoning quite like that in Plane Geometry, the result is seen to be kx2 + 2 hxy + jy2 + 2gzx + 2fyz + iz2 +2Q.'.x +2QQ,'y+2Q2z + Q'=0, (2) where Q,'= kx'+ hy'+ gz'+ 1, and so for the others. We note that the coefficients of terms of second degree change not. lWhen, and only when, all the Q's vanish, does the Eq. become homogeneous in x, y, z; but then it represents a cone through 261 CO-ORDINATE GEOMETRY. the new origin x', y', z'. Now Q' = Qyx- x' + Qy y'+ Q'_2.z'+ Qf,; hence, for Q'., Q'F, Q'., Q' each to vanish is the same as for Q'z, Q'y, Q'., Q'i each to vanish; and these four vanish for the same triplet x', y', z' when, and only when, A= k h g I =0. h j f m g f i n I m n d Hence, Q =0 represents a cone when, and only when, A =0. This A may be called the discriminant of Q = 0. If A= 0, then kx2 + 2 hxy +jy2 + 2gzx + 2fyz + iz2 = 0, and this breaks up into two linear factors in x, y, z; i.e., the cone breaks up into two planes, when, and only when, D= k- h g =0, h j f g f i as was proved in Plane Geometry. Hence Q = 0 represents two planes when, and only when, A = 0, D = 0. 31. If Q' =0, Q',=0, Q' =0, but Q'(or Q',) 0, then if any triplet (x, y, z) satisfies the new Eq., so does the counter-triplet (-x, -y, -z), since the Cds. appear only in pairs; i.e, if any point be on the surface, so is its counter-point as to the new origin; i.e., the new origin halves every chord through it, and is the centre. The Cds. of this centre, or xf, y', z', are found from the three Eqs., Q'. = 0, Q'y = 0, Q'2 = 0 to be L: D, M: D, N: D. Hence, if DO, the centre is infinity, and the sur- face is centric; if D= 0, the centre is at co, the surface is called non-centric. One or more of the numerators L, Mr, N may vanish alone with D; the centre is then indeterminate. In case Q' alone = 0, the origin (x', y', z') is on the sur- face. Call the sum of the six terms of second degree S; then the Eq. becomes 2, 5 2 TANGENTS. S + 2 (xQ'. + yQ', + zQ'.) = (3) Draw through the origin any RL. x:a=y:8 /= z: -y=p. To find where it meets the surface, replace x, y, z in (2) ; hence 1p2 + 2(Q'. + fQrY + yQ'.)p = O. (4) One root p, of this Eq. is always 0; the other, p2, is also 0 when, and only when, aQ'. + lQ'Y + -yQ.= =0, or when X .Q' +y- Q'3+z- Q' =0; (5) i.e., the RL. meets the surface only at the origin, and then in two consecutive points, when, and only when, it lies in the plane whose Eq. is (5). A RL. meeting a surface in two consecutive points is tangent to the surface; the plane containing all tangents to a surface at a point is tangent to the surface at that point. Hence (5) is the plane tangent at the origin. To find the Eq. of this plane tangent at (x', y', z') in terms of the old Cds., replace x, y, z by x-x', y-y', Z-Z'; hence, (x- X') Q'x + (y- Y') Q'5 + (z-z') Q'. = 0. (6) Add x'Q'. + y'Q', + z'Q'. + Q'1 = 0, since (xt, yt. z') is on the surface; hence, XQ'r + YQ'Y + ZQ'. + Q'1 =0 = X'Q. + 7;'Q, + Z'Q. + Q1; (7) or, F(x, y, z; x', y', z') = 0 =F(x',y',z; x,y,z), (7) is the Eq. of the plane tangent to F(x, y, z; x, y, z) = 0 at 32. The meaning of the fact that this Eq. is like-formed as to x, y, z and x', y', z' is quite like the meaning of the like fact in Plane Geometry, and is developed in like way. In fact, if (xr, y', z') be not on the surface, the Eq. still represents a plane, i is what S becomes on putting a, 8, y for x, y, z. 253 CO-ORDINATE GEOMETRY. being of first degree; also, if a plane through (x', y', zt) touches Q = 0 at (xi, yi, z1), its Eq. is F(x1, y1, z1; x, y, z) = 0, and hence F(xl, Yi, z, ; x', y', z') = 0; but this Eq. also says that (x,, Yi, z,) is on F(x, y, z; x', y', z') = 0: hence this last Eq. is that of a plane through al points of tangency of planes through (x', y', z')- Such a plane is called the polar (plane) of (x', y',z z) as to Q = 0. 33. Hence it appears that all tangent-planes, and hence all tangent-lines, through a point, touch a quadric (surface of sec- ond degree) along a plane-section of that surface. But a plane- section is clearly a conic; for the section made by the XY-plane is found, by putting x = 0, to be the conic kxl + 2hxy +jy2+ 21 x+ 2 my+ d = 0O and any plane may be taken as XY-plane without changing the degree of the Eq. or its general form. Hence all tangents through a point, or the tangent-cone through a point, touch the quadric along a conic. We may note in passing that 11 plane-sectio is of a quadric are shnilar conics. For they are got by giving different con- stant values to z, as c, c', etc; but these do not affect the first three terms of the conic, on whose coefficients alone, k, h, j, the shape of the conic depends. 34. The whole theory of poles and polars, since it depends solely on the symmetry of the Eq. of the tangent-RL. resp.- plane as to the current Cds. and Cds. of the pole, may now be repeated from Plane Geometry. Poles lving each on the polar of the other are conjugate. Planes each through the pole of the other are conjugate. Tangent-planes along a conic on a quadric go through a point. Poles of planes through a point lie on a plane. As a pole moves about on a plane, its polar-plane turns about a point; as a plane turns about a point, its pole moves 254 CONJUGATES. about in a plane. A tract from a pole to its polar-plane is cut harmonically by the quadric (referee). 35. If (L: D, M: D, N: D) be taken as pole, then Q',, Q'Y, Q'. vanish, and the polar is Ox + Oy + Oz + Q'1 = 0; i.e., the polar is the plane at oc. Hence, the polars of all points at ac pass through this point, the centre. All RLs. through the same point at o are 11; this point being the outer mid-point of all intercepts (chords) of the quadric on these RLs., their inner mid-points lie on the polar of the point at o ; this central plane accordingly halves all 11 chords through its pole (at oc); i.e., halves all its conjugate chords. Hence it is called a diametral plane. Among all these chords is one central one, which is therefore a diameter conjugate to the diametral plane. The section of the diametral plane, being a conic, itself has an QC of pairs of conjugate diameters; any one of these forms with the common conjugate diameter a triplet of conjugate diameters; the three planes fixed by the triplet of conjugate diameters form a triplet of conjugate diametral Planes. Each plane halves all chords 11 to the intersection of the other two. The poles of a system of 11 planes lie on the diameter conju- gate to the planes. The central distances of a pole and its polar, measured on the diameter through the pole (conjugate to the polar), have for their geometric mnean the half of that conjugate diameter. Tangent planes at the ends of a diameter are 11 to its conjugate diametral plane. 36. The notion of diametral plane may be got otherwise, thus: B - xi S-My' Z-Z' Be = y-y z-z a 2'Y or X=X'+ap, Y=y'+/3p, Z= +2'p a RL.; combining with the Eq. of the quadric, we get Jp2+2Tp+YOY ; (1) when E has its former meaning, 255 CO-ORDINATE GEOMETRY. T = Q'z a + and, lastly, Y= QF x' + Q'Yy' + Q..z z + 2 Q'1 +d. Eq. (1) has two roots, pl, p2; i.e., every RL. meets a quadric in two, and only two, points. These roots are counter when, and only when, T = O; i.e., for a, /3, y held constant (for 11 RLs.), the intercept or chord of the quadric is halved by (x', y', z') so long as the point (x', y', z') lies in the plane, T= 0; this latter is the Eq. of a plane, being linear in x', y', z'. From the Eq. of this diametral plane, T= Q'. - +Qfy + Q'- =, it is seen that all diametral planes form a pencil through the intersection of Q'. = 0, Qt1 = 0, Q'_ = 0; i.e., the centre. 37. To find the mutual slope of conjugate plane and chords, put a-, T, V, j (read koppa) for ka + hB + gy, ha +4j/3 +fy, ga+f,3+iiy, la+m /+ny; then is T = cx' + y' + vz' + 0 = 0, and the position-cosines a', /3', y' of this diametral plane are a'= : R, ,B'=T: R, r= v: R, where R2= o-2+72+ v2. Hence, if 0 be the slope in question, -ta'a+13/3f'+yy= (oa+T+vy): R= Y: R. The important question arises: Are conjugates ever perpen- dicular If so, 4= 90, +1= 1, Y = R, a = ,(3=,B', y = y'; whence, a=or:R, /3=r:R, y=v:R; (1) or, (aa + r1 + vy) a = a, (+ oa + +Vy)8 = T, (0ra + T + Vy)y = V 256 PERPENDICULAR CONJUGATES. 257 Here then are the three Eqs. to determine the a, /, y of a chord I to its conjugate diametral plane; there is a fourth Eq. connecting them, a2+B12+y2= 1, but this imposes no new fourth condition, since it can be got from the three by mul- tiplying by a /3, y in turn, adding and cancelling. Actually to find a, /3 y from these four Eqs. would be very tedious; it is better to determine R or Y, and thence a, /3, y. On replacing a-, T, v in (1) by their values there result (k - R)a + h/3 +gy = 0, ha+ (j-1R)/3+fy =0, ga+f13+(i-R)y =0. Divide in turn by hg, fhi, gf; add in turn a :f, /3:g, y: 4; put A, B, C for also put Ufor a:f+/3:g+y: h; hence result U= a(R -A) hg, U= /3(R-B) fh, U=7(R-C) :gf; or, a=hgU:(R-A), f3=fhU: (R-B), y =gqfU: (R- C) . Squaring, adding, and re-membering a2 + /32 + y2 = I, we get 1: U=V/h g:RA!+f2h2: R - le+g2f-: R --C-2 (2p) and on dividing in turn by f, g, h, and adding, we get UJlg U flt U +f U f R-A+ g R-B h vR-C' 258 CO-ORDINATE GEOMETRY. whence 1kg 1 I fit 1 gf 1 or U= 0. (3) On multiplying by (R-A)(1-B)(R-C), and by K = 1: hgf, results (R-A) (R-B) (R-C) i f2(R-A) g2(R-B) h2(R 1C) I This Eq. is of third degree in R, hence has at least one real root. To decide about the other roots, suppose K + and A B C; then, calling the left side of the Eq. E, we see that for R =- a;, E is-; for R =A, E is -; for R = B, E is +; for R= C, E is-; for R= + oo, E is +. In case K is -, a change of sign in E takes place between R = - oc and R = A, instead of between R = C and R= m. In any case E changes sign thrice as R passes through real values from - X to +X ; i.e., in any case the Eq. E= 0 has three real roots: R1, R2, R1. These real values of R give three real triplets of values of a, /3, -y; hence, there are, in general, three, and only three, diametral planes . to their conjugate chords. They are called chief (or principal) planes of the quadric. 38. Each of these roots R1, R1, R3 must satisfy Eq. (3) hence result three Eqs.; take second from first, and multiply bv hqf; hence, f h292 + fh R2 s (R.-A) (R2-A) +(R, B) (K, B) +- (X-) (_ka C) PRINCIPAL PLANES. 259 Since 2 - R1 cannot = 0, the parenthesis must = 0; and two like Eqs. are got by permuting the indices. Now a1 = hg U1: (R1-A), /1 =fkU1: (R1-B), TY = gfUl (R1-C); and so for the indices 2 and 3. Calling the three directions of the diametral planes, or what is tantamount, of their conjugate chords, d1, d2, d3, and dividing by the U's, since they can none of them = 0, we get dc1d2 = 0, d1d8 = 0, ddd = 0; i.e., the three chief planes of a quadric are I to each other. 39. In the special case U= 0, follow also R=A=B= C; hence, k = A + (r4Aa+hgU, R=A+hgfU2, whence U-fU2a = O, U-gU2,f=0, U-hU =0O. Hence a=l:fU, 8=1:gU, y=l:hU, or U=O. Squaring and adding the first three Eqs., we get U=V /h2g2 + fh2 +g2f2: hgf. This is the same value of U as is given by Eq. (2) when A = B = C; but besides the triplet a, f3, y thus got, the prob- lem is solved by any other triplet al, /', 8y' that makes U= 0 or makes -+0+ h O. This Eq. is satisfied in an co of ways and whenever U(aa'+ 3/3' + yy') = 0, since 1 :f= Ua, 1:g=U, 1:h= Uy. Now U is here not = 0; hence aa'+ P3f+ yy'= 0; i.e., every direction 1 to the direction (a, /3, y) is a chief direction. 260 CO-ORDINATE GEOMETRY. It is easy to prove analytically, but it is also clear geometri- cally, that this special case is the case of surfaces of revolution; the direction (a, fB, y) is that of the axis. If k =j = i and h = g =f, the surface is a sphere; every triplet (a, /3, y) fulfils the conditions, every direction is a chief one. 40. Returning to the Eq. of a diametral plane T= 0, on putting for the Q's their values, it becomes clear that a trilpht of conjugate diametrals are 11 to the Cd. planes when, and only when, h = 0, g= 0, f= 0, the reasoning being quite like the corresponding in Plane Geometry. Hence, by choosing as Cd. planes three l)lanes 11 to a set of conjugates, we make the terms in xy, yz, zx vanish. This can always be done. Again, by choosing the centre as origin, we make the terms in x, y, z vanish. This can be done only when the centre is in finity. But when the centre is at x, the origin can be taken on the surface, making the absolute vanish, and also the term in z2, a diameter being taken as Z-axis. Hence the forms reduce to x2+jy2_ iZ2=d and kx2+jy2=2nz. When d and n are not O., the varieties of these are X2+ 1i+ Z2 i X2 y2 Z., and 2 '-= 2z. a b The first is an ellipsoid: real resp. imaginary. The second is an hiyperboloid: single resp. double. rhe third is a paraboloid: elliptic resp. hyperbolic. When d = 0 or n =0, the Eq. becomes homogeneous, and so represents a cone, and. in case another coefficient van- ishes, still more specially a cylinder. These limiting cases the student himself can readily trace out. The rectangular conjugate diametrals recommend themselves DISTANCE-PRODUCTS. as Cd. planes; to indicate that oblique conjugates are chosen, it suffices, as in Plane Geometry, to accent the constants a, b, c. 41. Returning to the Eq. 1p2 + 2Tp +Y = 0, which fixes the distances p1, P2 from (X1, y', Z') in the direction (a, /, y) to the quadric Q = O, we note that . depends only on the direc- tion (a, /3, y), and Y only on the point (x', y', z') ; hence, taking two points PI and P", and one direction, we find the quotient of the distance-products p1'P2': pI fp21' is independent of the direction; or, taking one point and two directions, we find the like quotient is independent of the point. Hence: The rectangles of the segments of two intersecting chords are proportional to the squares of the 11 diameters. Tangents from any point to a quadric vary as the 11 diameters. The areas of sections conjugate to a diameter vary as the rectangles of the segments into which they cut the diameter. Proof is quite as in Plane Geometry. 42. We have seen that the six terms of second degree are unchanged by a mere change of origin; to find whatfunctions of the coefficients are unchanyed by a change of axes, -proceed as in Plane Geometry, thus: Let the coefficients k, j, ...i change into i', j', ..., i', and the sum S of the six terms change into S'; then S = SI; also x2 +2 woxy + y2 + 2 tzx+2Xyz+z2 - Xf2+ 2 W'fxy' + y2+ 2 ql'z'x'+ 2 Xtytz+ z'2, since each is the squared distance D2 from the common origin o to the same point P(x, y, z) or P(x', y', z'). Hence S+1.D2 is not changed by change of axes; hence the values of u which make S + LD2= 0 are the same for all axes, and specially the values of ju that make S + pD2 resoluble into two factors of first degree in (x, y, z) are the same for all axes. This resolu- tion is possible when, and only when, 261 CO-ORDINATE GEOMETRY. 7+JL h+ po g+ 14= 0, h+ uo j+ f+i+x as was proved in Plane Geometry; for the form of S+/,D2=0 is the same as that of F(x, y; x, y) = 0, as is seen on putting z = 1. The roots of this cubic in It, called discriwninatibg cubic when to = = X = 900, are the same for all axes; hence, on making the coefficient of al' 1, the other coefficients must be constantfor all axes. They are UAj-h2+ki- _2 +ji-f2 -2[(7i-gf)w + (gi-fh) f + (fk-gh) X]: S2 k.XV 12+ij jq, o - 2[h W-X) + g (q-FOX) +f(X- -A ) :SX k h g :S2, or D:S2. h j f g f i It is to note that the binomials are all co-factors of elements of either S2 or D. Geometric Interpretation. 43. 1. For rectangular axes, X Q, x vanish, Ad, qt I, xl become each = 1, and S2 becomes 1; hence k +j + i = constant. Sup- pose S 1 the central Eq. of a quadric, then k, j, i are the squared reciprocals of the half-diameters; therefore, the sum of the squared reciprocals of three rectangular diameters of a quadric is constant. 2. If conjugate diameters be taken as Cd. axes, h, g, f vanish, and the constants are (kj + ji + ik) 2, (kX12+ji,12+i._12) :S2, kji:S2. 262 THE SPECIAL QUADUICS. On dividing the first by the third, we get - + - + = constant; i j k i.e., the sum of three squared conjugate (half-) diameters of a quadric is constant. Extracting the second root of the third, and inverting, we get 8 : ji = constant; i.e., the volume of the parallelepiped fixed by three conjugate (half-) diameters is constant; hence, its eight- fold, the volume of the parallelepiped of three conjugate diane- ters, or bounded by six tangent planes at the ends of conjugate diameters, is constant. On dividing the second by the third, there results xF + 1. + (F= constant; 0j ki jk i.e., the sum of the squared parallelograms fixed by three conju- gate (half-) diameters, taken two by two, is constant. The axes of the quadric being 2a, 2b, 2c, the values of the above four constants are 1 1 1 2 22 2 + + -, a +b +c2, abc, a+ Cc +ca. The third constant is abc = a'b'c'S. Hence 8 7ra'b'c'S is constant; or, since S= w , (47ra'bwl) (2cr t) is constant. Here the first factor 47ra'b'wJo is the area of the central section of the plane XY; and the second, 2c'.C, is the projection of conjugate diameter on the I to that plane; i.e., the first factor is the base of a cylinder touching the quadric along a central section, the bases themselves touching the quadric at the ends of the conjugate diameter; while the second is the height of that cylinder. Such a cylinder may be said to be 11 to the diameter. Hence, the volume of a circumscribed cylinder 11 to a diameter of a quadric is a constant: 87rabc. The Special Quadrics. 44. The quadric + + . -1 has no real points, sic b2 C2 since the sum of the squares of no three reals can be - 1. 263 264 CO-ORDINATE GEOMETRY. The section of the XY-plane is got by putting z = O; it is the imaginary E = -1; all 11 sections are also iimag- a2 LA inary E's. The like may be said of sections 11 to the other Cd. planes. Hence the surface may be called Imaginary Ellipsoid, with axes 2ai, 2bi, 2ci. The sections of + Y"+_Z= 1 made by the Cd. planes X2 y2 y2 z2 x2 Z2 are the real E's, . + 62 = 1 + c= 'a + =1. All I sections are similar E's. Hence the surface may be called Real Ellipsoid, with axes 2at, 2b, 2c. We may suppose abc; i.e., 2a the greatest, 2c the least, 2b the mean, axis. For x a, or y b, or z c, the sections become imaginary E's; hence the surface lies wholly in the parallele- piled whose edges are = and 11 to the three axes. 45. The plane sections of the ellipsoid are in general ellipses; are they ever circles That they are, is made clear geometri- cally, thus: Pass a plane through the greatest and mean axes; it cuts out the E 2 + Y2 = 1. Turn the plane about the mean a2 b 2 axis, 2b or Y; the section remains an E of which 2b is still the minor axis, but the major axis gets smaller; when the plane is turned through 90, the section is the E / + = 1, of b2 c- which 2 b is the major axis. At some stage 2 b must have ceased to be minor and become major; at that stage the axes of the E wetre =, the E was a circle. To find the slope 0 of this cyclic Ilane to the greatest axis, we have the Eq. b2 =c2: a2_ c2 .23 ca2 whence - b= ca -a : b Vb2_c2 Of course there are two cyclic central planes, one sloped 0, the other -0, to the greatest axis. All planes 11 to these cut CYCLIC PLANES. 265 the ellipsoid in circles, growing smaller as the cyclic gets farther from the centre, and vanishing in so-called umbilics, or, better, cyclic points, as the planes become tangents. Clearly there are four such points. When a = b, the ellipsoid becomes one of revolution, made by turning the E, whose axes are 2 a, 2 c, about the less axis 2 C; the sections to this less axis are all circles clearly, the two series of cyclic sections falling together in them. This ellipsoid is sometimes called an oblate spheroid. The earth-surface is nearly such an ellipsoid. When b = c, the ellipsoid, called prolate spheroid, is formed by turning an E about its greater axis 2 a. The two series of cyclic planes fall together I to the axis 2a. 46. An important way of looking at the ellipsoid is to look at it as a strained sphere. Suppose a sphere of radius a to have all its chords 11 (say) to Y-axis shortened in the ratio b: a, and all II (say) to Z shortened in the ratio c: a; then, if P'(x', y', z') be any point of the sphere, and OP' be directed by a, f y, we shall have x'=aa, y'=/a, z'=ya, and if P (x, y, z) be the corresponding point on the surface got by working on the sphere as stated, we shall have X=aa, y=/3b, z=yc; whence, on squaring and adding, results X2 2= -j+ v+ - the surface is an ellipsoid. We may call a, 8, y the eccentric :s of P or OP. Since, in the shortening prescribed, 11 and = tracts remain II and =, it follows that conjpgate planes and diameters in the sphere remain conjugate in the ellipsoid; but in the sphere con- jugates are I; hence conjugates in the ellipsoid correspond to Is in the sphere. Hence, if 266 CO-ORDINATE GEOMETRY. ,C 2, 63I t) "Ef, Ci , Cfa , ( i Fit lis (1 (2 (3) ( 1X 29 IL)X( 12X 3 be eccentric Ws of three conjugates, etlel I + 2E 112 + E 3 Ft3 = 0, (ffIC III If + C1f21EIf1 + 1E13I = 0, C( + (3(2 + CC 0.3 C Ifl Eft + f, 2E2 + 'El elf3 = - _ _-I _2 3_ 3 The student may now show that the sum of the squared projec- tions of three conjugate diameters on any RL. or plane is con- stant. xx I Yy ZZI 47. The Eq. -x +v + =T 1 of the tangent-plane at (x', y', zo) becomes in the eccentric form x ', + Y E'2 + z E3 = 0. a- b- c- On squaring and adding the Eqs. of three tangent planes at the ends of three conjugate diameters, the locus of the intersection 22 2 of the three is found to be the ellipsoid x2 + + = 3. a2 b2 48. The Normal Eq. of the tangent plane is -x + fly + yz ='P where a, /3, y are position-cosines. Comparing, we see p)X py' ___ -a= b2 = c2 1 x2 yF z2 and 1 = + + Hence aea' + debt + y2C2 =p2 (2 b Hence the Normal Eq. of the tangent planes is ax + jly + Lyz =Va aa2 + FIbN + ayce. On squaring and adding the Eqs. of three such planes mutu- ally I, the locus of the intersection is found to be the director- sphere X2+y2+z2=a2+b2+c2. THE HYPERBOLOIDS. 267 .T2 +2 _Z2 i 49. The quadric -2 + 2 2 = 1 is cut by the XY-plane a b 2 c2 X2 2 in an E -2+ y=i as is seen on putting z=O. All a 2 62 1I sections are similar E's, only larger the farther from the XY- plane. It is cut by YZ in the H V2-_Z2 =1. The 11 see- tions are similar H's, smaller the further from YZ, till x = a, Y2 z2 when the section becomes a pair of RLs. Y- Z-0. Thence b2 C2 the sections are secondary H's, flattening out towards a pair of 11 RLs. as x nears i a:. Like remarks hold for sections 11 to the XZ-plane. Hence this surface is called an Hyperboloid simple or of one sheet. By reasoning like that in case of the ellipsoid, it is shown that this hyperboloid is cut in circles by a central plane through the greatest axis and sloped 0 to the mean axis, where a2 _ C2:( b + C 62), 1a=cVa2_-2b aVb2+c2 All planes 11 to these are themselves cyclic planes, cutting the surface in ever larger circles. Hence, the single hyperboloid has no cyclicpoints. X2 + 2 z2 50. Thequadric 2 - -=-1 is cut byXYin the imaginary E x + Y- --1' the 11 sections remain imaginary a2 b2 till z = c; thence the E's are real and grow ever larger, with z nearing i a. The section of YZ is the secondary H', y2- _ =2-1 and It sections are similar, with ever larger 6 2 c2 axes. Like remarks hold for sections 11 to XZ. Hence, this surface is called an Hyperboloid double or of two sheets. The student can readily convince himself that the cyclic planes 268 CO-ORDIN'ATE GEOMETRY. of the simple hyperboloid are also cyclic planes of the double; the circular sections shrink into four cyclic points as they retire from the centre. These two hyperboloids have clearly a common asymptotic cone 29 + L - Z2 = 0, which has common cyclic planes with a2 b2 c2 them. 5L The clearest notion of these three surfaces is got thus: Turn an equiaxial H, its conjugate H', and their common asymp- totes around the conjugate axis; the H will trace out an equi- axial simple hyperboloid of revolution, the H' an equiaxial double hyperboloid of revolution, the asymptotes the common equiaxial asymptotic cone of revolution. Now change (say shorten) all chords I to ZX in the ratio b : a; all chords I to XY in the ratio c: a; the resulting surfaces will be the surfaces in question. The circle 2 + a = 1 traced by the vertex of the revolving a2 a" H is called circle of the gorge; the corresponding E a + y-2 = 1 is called ellipse of the gorge. 52. Passing now to non-centric quadrics, we see that the first - + yb = 4 z is cut by XY in the point (0, 0, 0), the origin, a b while all 11 sections are Ls: real for z 0, imaginary for z 0. The section of YX is the P y2 = 4 bz, while that of ZX is the P x2 = 4 az. The surface may be thought made by a vari- able E moving always 11 to XCY, with its vertices on these two P's. The surface is called Elliptic Paraboloid; 4 a and 4 b are its parameters; 0 is its vertex. Suppose the P y2 =4az to turn around its axis, the Z-axis; the surface generated will be the paraboloid of revolution 2 + = 4z. Now suppose all y's, or all chords I to XZ, shortened in the ratio b: a; the x2Y2 surface got so is + -2 = 4z. a2 b2 IMAGINARY CYCLICS. As to cyclic planes, we may reason thus: Turn a plane 11 to XZ about the major axis of its elliptic section; the minor axis grows, and the E tends to a P, as the plane turns through 900; at some 4 the minor must have become equal to the major axis, the E must have passed over into a circle. The slope of the cyclic planes to XY is readily seen to be e when 0=Vb-a, ab. The student will readily see there are two cyclic points. 53. The second non-centric - = 4z is likewise tangent a b to XY at 0, as is seen on writing out the Eq. of plane tangent at (0, 0, 0): x - O _ lb =2 (z + 0), or z = 0, which is a b the XY-plane. But XY cuts the surface along the pair of RLs. v2 All II sections are H's: primary for zO, a6 secondary for z 0. The sections of YZ and ZX are the P's y2= - 4bz, x2 = 4az. The surface may be thought made by an H moving, always 11 to XY, with its vertices on one of these two P's. In all positions the asymptotes of the H are 11 to the pair - =Y. As the H nears the XY-plane, it a W passes over into this pair of RLs., then into its conjugate, while its vertices pass over from one P on to the other. To two counter-values, +z, -z, correspond two conjugate H's. The surface is named Hyperbolic Paraboloid; 4 a and 4b are its parameters, 0 is its vertex, Z is its axis; it is saddle-like in shape. (See Figs. at end.) 54. It is easy to see geometrically that the cyclic planes thus far determined are all the real ones. For the diametral plane of the chords of the circles must be I to them; hence it must be one of the three chief planes; hence the diameter of the cen- tral circle must be one of the axes. This can only be the mean one, 2b, in case of the ellipsoid; for any circle of radius a 269 270 CO-ORDINATE GEOMETRY. resp. c lies wholly without resp. within the ellipsoid. Similar reasoning holds for the other surfaces. But there are other imaginary cyclic planes, as may thus be shown analytically: Bex y2 z2 2 2 2 Be 2 + - + ,=1 an ellipsoid, and + + 2=1 a concentric sphere; then ( 61 2 + ( )y2 + ( __)! 0 is a cone through 0, being homogeneous of second degree, and through the intersection of sphere and ellipsoid, being satisfied whenever their Eqs. are. When, acd only when, this cone breaks up into two planes, the intersection of ellipsoid and sphere is a plane curve; i.e., is a circle. This is the case only when the Eq. is resoluble into two linear factors; and this is the case only when the determinant A vanishes, or when one coefficient (one element in the diagonal of A, the others being 0) vanishes; and this is so only when 2 = a2, or b2, or c2. For r2 = a2 or r2 = c2, the factors, i.e., the planes, are imaginary; for r2 = b2 they are real. The student can easily apply the reasoning to the other surfaces. 55. We have seen that two RLs. lie on the hyperbolic para- boloid: the intersection of that surface and the XYdplane. But the general proposition holds: All surfaces of second degree are ruled. On each lies an oc of RLs. This is clear at once on referring to the condition that a RL. lie on a surface (Art. 17) : on combining the Eqs. of RL. and surface, the resultant Eq. in a single Cd. must vanish iden- tically. This Eq., being of second degree, vanishes thus when its three coefficients each reduce to 0; and these three Eqs. of condition can be satisfied by the four parameters of a RL. in an 0o of ways. 56. Let us apply this argument to the simple hyperboloid: X2 a2 b2 c2 RIGHT LINES ON THE QUADRIC 271 Be y=tz+v, x=sz+u the RL. On substitution results (2 t+ 1 + 2(su + tv)Z+U+ ,=1 This vanishes identically, is satisfied for every z, when 82 t2 1 = s tv O, 2 + 2 1. 'W+2 C2= 0,aW b2 a2 62 Hence result readily the real values s = av: be, t=4: bu: ac. The third Eq. of condition says that every such RL. meets the ellipse of the' gorge, as was to be foreseen. The double sign shows that through every point of this ellipse go two RLs.; there lies on the surface a double system of RLs. Two RLs., one of each system, are waz+u, y=_--z+v, be ac and x=_az+u, y=bu z+ v. be ac The condition that these two RLs. meet is (Art. 9) u-u':v-v' av+ av'(bu +lbu\) be be Sac c) a condition always fulfilled, since a2 62 a2 62 a2+b2 = a2 + 2 =1 Hence every RL. of each system meets every RL. of the other. avi bul Changing the signs of c and b, we see that the con- dition is fulfilled only when (u - u')2b2 = a2(v - V)2; i.e., never. Hence no RL. of either system meets any RL. of the same system. Hence through every point of the surface there pass two, and only two, RLs. on it. The plane of these RLs. is CO-ORDINATE GEOMETRY. clearly the plane tangent at their intersection. For it can meet the surface only on these RLs., which form its conie of inter- section; hence all RLs. in it through the intersection of the pair meet the surface only at that point. As the point of tan- gence glides along either of the RLs., the tangent plane turns about the RL., cutting the surface. The RLs. of either system are called elements (or generators) of the surface. Since no two elements meet, the surface is not torse or developable, but skew or a scroll. This is easy to see, thus: Be 1, 2, 3, 4, ... consecutive elements. If 1 and 2 meet, 2 and 3 meet, 3 and 4 meet, etc., then we may turn the strip between 1 and 2, which is plane, infinitesimally, about 2 till it falls into the plane of the strip 2 3 ; then turn the sum of the strips 1 2 and 2 3 about 3 into the plane of the strip 3 4, and so on. Thus, and thus only, could the surface be turned off, unwrapped, into a plane surface. Now, since 1 and 2, 2 and 3, etc., do not meet, this can not be done. A more rigorous proof would not be in place here. 57. To find the RLs. on an ellipsoid, in the values of s and t, put C2 for -c2, or ic for c; the values then fall out imaginary: there are no real RLs. on the ellipsoid. (On putting ai, bi for a, b, the values of s and t again fall out imaginary: no real RLs. lie on the double hyperboloid. X22 58. Proceeding with the hyperbolic paraboloid - _ Y= 4z n a2b2 exactly as with the simple hyperboloid, we find for s and t the real values: s = a: , t = i ab: u. Hence there lies on it a double system of real RLs. Every RL. of either system cuts every RL. of the other. No RL. of either system cuts a RL. of the same system. Through every point of the surface pass a pair of RLs., fixing the tangent plane at the point, which plane cuts the surface. The surface is not developable. Eliminating z, the student will find the XY-projection of an element to be y= ib(x-2u); 272 CONFOCAL QUADRICS. 273 whence projections of all elements, on XY, are 11 to one of the pair of RLs. y = -b . a x, in which X Y cuts the surface; i.e., all elements are 11 to one of the planes fixed by Z and this pair. These planes contain the asymptotes of the generating hyperbola. Hence all elements of a hyperbolic paraboloid are 11 to an asymptotic plane. (See Figs. at end.) 59. The foci of the chief sections of a quadric are called foci of the quadric. In an ellipsoid with half-axes a, b, c, the sec- tion (ab) is an E with two foci F, F' on the axis 2a, distant Va2 - b2 from the centre; the section (be) is an E with two foci G, G' on the axis 2 b, distant V/b -c- from the centre; the section (ac) is an E with two foci H, H' on the axis 2 a, distant Va2 - C2 from the centre. Thus, on the greatest axis lie four foci, on the mean axis lie two, on the least lie none. If a quadric with half-axes a', b', c' be confocal with this base-ellipsoid, the relations hold: a'2 _b F2 =a2 - b2, b12 _ CI2 = b2-_ c2, a,2 - cI2 = 2_ 2 whence a"2 _aa2 = b'2 be = c'2 _ c2 = (say) X; ie. X2 + Z= , and asi+A + '+c c2 + t = are confocal for all values of X. To trace the system: for X=++o the surface is a sphere with radius o; as X sinks toward -c2, the surface (an ellip- soid) shrinks, and for k - c2 flattens to the inner doubly- laid surface of the so-called focal E in the section (ab), whose half-axes are Va2 - c2, V/b2 - c2, its foci F, F', and its vertices HI, H', G, G'; as A sinks from - c2 towards - b2, the outer doubly-laid surface (thought as a simple hyperboloid) spreads out into a simple hyperloloid, which, as A nears -b2, flattens into the so-called xc(dl H in the section (ac) with foci H, HI and vertices F, F'; as A sinks from - b2, the surface becomes a double hyperboloid, which, as X nears - a2, flattens down to the CO-ORDLNATE GEOMETRY. section (bc); as A sinks from - a2 towards - x, the surface becomes and remains an imaginary ellipsoid. 60. For any triplet (x', y', z') the Eq. of the confocal yields three values of A: Al, X2, As; by reasoning quite like that in Plane Geometry (Art. 148), it is shown that these roots lie between + o0 aid - c2, - c2 and - b2, -2 and-a2, resp.; i.e., through every point of space pass three, and only three, confocals: an ellipsoid, a simple hyperboloid, and a double hyperboloid. The three X's are called elliptic Cds. of the point (x', y', z'). Substituting them in the Eq. of the confocal, in turn, and solv- ing the three Eqs. as to x', y', z', we get x' =-/Va2 + Al - a2+ A2 a2+ A3: -,a2/ b2 - a2- a c2 , and two like expressions for y', z' got by permuting a, b, c. If we divide this x' by a2+ A,, we get the coefficient of x in the Eq. of the plane tangent at (x', y', z') to the first confocal; dividing it by a2+X2, we get the corresponding coefficient in the Eq. of the plane tangent to the second confocal; the product of these two coefficients is (a2+ X3): (a2- b 2) (a2 - Ce). The products of the coefficients of y and of z in the two Eqs. are got by simply permuting a, b, c. The sum of these three prod- ucts is 0. This means, by Art 16, that the two planes are I. Like holds, of course, for the second and third confocals, and for the third and first. Hence three confocals through 'a point are mutually -. Cubature of the Quadric. 6L The part of a surface intercepted between two 11 planes is called a zone. The space bounded by the planes and the zone we may call a segment of the surface (meaning a segment of the space fixed by the snrface). Suppose an equiaxial H, its conj. A', and their common asymptotes turned about the conjugate axis. There will be generated by H, a simple hyperboloid of. revolution; by H', i 274 CUBATURE OF THE QUADRIC. double hyperboloid of revolution; by the asymptotes, a cone of revolution: the Eqs. are XI + y2 _ Z2 = a2, X2 + y2 _ Z2 -a2, 2 + y2 _ Z2 = 0. Sections of the surfaces I to Z are circles, and their areas, they being distant z from XY, are 7r(Z2 + a2), X -(Z2_a2), 7rz2 Hence it is clear that the circle of the cone is the arithmetic mean between the circles of the hyperboloids; it differs from each of these by a ring whose area is ray, the area of the circle of the gorge; these differ from each other by double this area, by 2 7ra2. It is to note that the circle of the double hyperboloid is imaginary, and so does not really come into consideration, for z a. Accordingly, to find the volume of any hyperboloidal seg- ment, it suffices to find the volume of the corresponding cone- segment and then add resp. subtract the volume of the corre- sponding ring-segment in case of the single resp. double hyperboloid. The cone-segment is itself the difference of two cones whose altitudes are (say) z1, z2, and bases -Z2 7rZ22; hence the volume is - (z - z 3Z3); the constant area of a section 3 of the ring-space is t-al, and the altitude is Z2 - ZI; hence the volume is 7ra2 z2 - z1). If the H be not equiaxial, but have axes 2 a, 2 c, then to any altitude z will correspond in the cone a circle of radius not z but az, the surfaces then being c tz + y2 a z2 = 2 2 + y2 2 = 2 C2 X2 + y2 - a . =0. C2 Hence it is enough to change z into az :c. From these last surfaces the most general, viz., 275 276 CO-ORDINATE GEOMNETRY. a2+-2- 2- = 1, a2 + be- _ =-I x2 y2 z2 a2 y2 z2 X2 + y _Z2 _o b2 c2 are got by shortening every y in the ratio b: a; hence it suf- liees to multiply the preceding results by this ratio. 62. The ellipsoid (a, b, c) is got from the sphere (a, a, a) by shortening every y in the ratio b: a, and every z in the ratio c: a; hence the whole volume of the ellipsoid is got from that of the sphere by shrinking it in the ratio be :a2; and the same ratio holds between volumes of corresponding parts of ellipsoid and sphere. The volume of the sphere is 41ra", hence that of ellipsoid is 4abc. On three axes, 2 a, 2 b, 2 c, construct an ellipsoid and the two hyperboloids; also construct a cylinder tangent to the simple hyperboloid along the ellipse of the gorqe, its bases tangent to the double hyperboloid at the latter's vertices. Let us compare the volumes E, C, IC, H of the ellipsoid, cylinder, cone-seg- ment, hyperboloid-segment, the bases of the two latter being the bases of the cvlinder. The volume E is 4-3,Tabc; C is 2ec -7rab or 2 7rabc; K is of C, or is 2g7abc; H is K plus 2 c rab, or H is h-rabc; hence K: E: C: H= 1:2:3:4. 63. To find the volume Vof a segment of the elliptic parabo- loid + Y = 4z, first take the vertex for one base and the a b section of the plane z = z for the other; cut this cap-shaped segment into n thin slices by planes 1i to the base; let the alti- tude of each slice be ; it will have two bases, each an E, a greater and a less; the volnme of each slice will be less than the altitude by the greater base and greater than the altitude by the less base; hence the whole volume will be less than the common z altitude - by the sum of the greater bases and greater than that n METHOD OF SLICES. 277 altitude by the sum of the smaller bases; or, common factors set out, V- -4r-Vab 1+2+3+ +n, n V Z24 4rab 0 + 1 + 2 + -+ n -1, or V 47r-/ab - z - zg 1 v4 rVIb z z1- 1. As n nears x, - nears 0, and there results n V= 2 ir Nfaib. z. z. Now 4 7r\/ab z is the base of the segment, z its altitude; hence 4 7r Vab z z is the volume of the circumscribing cylinder; hence the volume of a cap-segment of an elliptic paraboloid is half that of the circumscribed cylinder. The volume of any segment is the difference of two cap- segments. 64. To find the volume v of a segment of an hyperbolic parab- X2 72 oloid 2 - = 4 z, suppose it bounded by the surface, the (t2 b XY-plane, the YZ-plane, and a plane x = x, 11 to YZ. The section of this last plane with the surface is a parabola; the chord (in XY-plane) of the segment of this P is 2xvb:a, the altitude of the segment is x2:4a; hence the area is 3. a2 Cut the solid segment into n thin slices; then, reasoning exactly as before, we get V H/aM . X 13 + 23 + 33 + ...+ n3j 278 CO-ORDINATE GEOMETRY. V3 a2. n4 03 + 13 + 2 2 Now 13+23+ ... +n3= 4 4 as we know from Algebra; hence V7 V,-, (t( At V Vt0b ' _ 4 1)2 12a2 k W) As n nears a, 1 nears 0, and there results n V=Jib -.x4: 12 a2. Now I- x3 :3a2 is the base of the segment, x its altitude; hence \ab. x4:3 a2 is the volume of the circumscribed cylinder; hence the volume of such a segment of an hyperbolic paraboloid is one-fouirth that of the circumscribed cylinder. The student mav confirm the results as to the ellipsoid and the hyperboloids by this method of slices. The segments thus far treated have been right, i.e., I to an axis of the surface; but like reasoning applies to oblique seg- ments, on observing that the intercept between the bases on the conjugate diameter is not the altitude of the segment but a multiple of it. Varieties of Quadrics. 65. If a quadric be given by its Eq. in the general form, it is of course possible to determine what kind of a quadric it is by reducing its Eq. to the simplest form; but this is tedious. It is possible to establish certain simple tests, however, by some such reasoning as this: By Art. 30 the surface is centric or non-centric, according as D0 or D = 0. If D O, the centric Eq. is k2+ 2hy+ jy2 +2 gzx+ 2yz + iz2+ = 0. VARIETIES OF QUADRICS. If now a RL. through the centre, y = tz, x = uz, meet the surface in finity for all values of t and u, the surface is closed or ellipsoidal; otherwise, it is hyperboloidal. The stu- dent can show that the first holds when both D0 and C or j - h2 0; the surface is then a real ellipsoid, an imaginary cone with one real point (the centre), or an imaginary ellipsoid, according as A 0, = 0, or A 0 (lo being taken + aalways). If D and C be not both 0, the surface is hyperboloidal. We now test whether the RLs. on it be real or imaginary by the former method ; the result is, the surface is a simple hyperbo- loid, an elliptic cone, or a double hyperboloid, according as A0, A 0, AO. In case D=O, the surface is non-centric or paraboloidcl. Putting z = 0, we find the section of the XY-plane is an E or an H, i.e., the surface is an elliptic or an hyperbolic paraboloid, according as CO or CO. If C=O0, the section is a P. and this test fails. In that case, test with ik _ g2 in the same way. If both kj - h2 and 'ik - g2 vanish, then must also ji -J2 vanish; all sections are P's, and the surface is a parabolic cylinder. Lastly, in case one or more of the numerators L, M, N of the Cds. of the centre (L: D, Kf: D, N: D), as well as the common denominator D, vanish, the centre becomes indeterminate, the surface has an x of centres. The surface is then a cylinder. In case the three Eqs. of planes Q, = 0, Q, = 0, Qz =0 which determine the ceiitre reduces to two only, their line of in- tersection is the line of centres, every point on it is a centre of the surface. The cylinder is elliptic, hyperbolic, or breaks up into a pair of planes, according as one of its plane sections is an E, an H, or a pair of intersecting RLs. In case the three Eqs. reduce to one, each represents the plane of centres, every point on it is a centre of the surface. The surface itself consists of two 1I planes, midway between which lies the 11 plane of centres. In case the Eq. of the surface is a perfect square, the surface consists of two planes fallen together in the plane of centres. 279 CO-ORDJNATE GEOMETRY. SIMPLE HYPERBOLOID. AB is Ellipse of the Gorge. EOD is the Asymptotic Cone. Fig. 4 PROJECTIONS OF ELLIPSOID ON XZ AND XY. OD gives direction of cyclic sections. C's are cyclic points. C'C' is locus of centres of cyclic sections. RLs. ON THE S. H. MIP, and NQ, are of 1st system, JIP2 and NQ, are of 2d system. NoTrx. For Figures on pp. 280 ald 281 thanks are due Fort and Schloemilch's A-mlytische Geometrke. 280 Fig. 1 C Fig. 3 ELLIPSOID. OA=a, OB=b, OC=c. Fig. 2 l ( MA I DIAGRAMS. Fig. 5 Iz HYPERBOLIC PARABOLOID. OC and OD are Parabolas. OA and OB are Asymptotic directions for the Hyperbolas. DOUBLE HYPERBOLOID. EOD is the Asymptotic Cone. ELLIPTIC PARABOLOID. RLs. ON THE H. P. FA=2a and GB=2b are half-parameters. 281 Frts5 of Xjrfrmick jmijk, 190stan. This page in the original text is blank. This page in the original text is blank. Peirce's Three and Four Place Tables of Loga- rithmic and 7Trigonometric Functions. By JAMES MILLS PEIRCE, University Professor of Mathematics in Harvard University. Quarto. Cloth. Mailing Price, 45 cts.; Introduction, 40 cts. Four-place tables require, in the long run, only half as much time s five-place tables, one-third as much time as six-place tables, and -one-fourth as much as those of seven places. They are sufficient for the ordinary calculations of Surveying, Civil, Mechanical, and Mining Engineering, and Navigation; for the work of the Physical or Chemical Laboratory, and even for many computations of Astron- omy. They are also especially suited to be used in teaching, as they illustrate principles as well as the larger tables, and with far less expenditure of time. The present compilation has been prepared with care, and is handsomely and clearly printed. Elements of the Differential Calculus. With Numerous Examples and Applications. Designed for Use as a College Text-Book. By W. E. BYERLY, Professor of Mathematics, Harvard University. 8vo. 273 pages. Mailing Price, 2.15 ; Intro- duction, 2.00. This book embodies the results of the author's experience in teaching the Calculus at Cornell and Harvard Universities, and is intended for a text-book, and not for an exhaustive treatise. Its peculiarities are the rigorous use of the Doctrine of Limits, as a foundation of the subject, and as preliminary to the adoption of the more direct and practically convenient infinitesimal notation and nomenclature; the early introduction of a few simple formulas and methods for integrating; a rather elaborate treatment of the use of infinitesimals in pure geometry: and the attempt to excite and keep up the interest of the student by bringing in throughout the whoie book, and not merely at the end, numerous applications to practical problems in geometry and mechanics. James Mills Peirce, Prof. of is general without being superficial; A4zalh., Harvard Univ. (From the Har- limited to leading topics, and yet with- vard Regis/er): In mathematics, as in in its limits; thorough, accurate, and other branches of study, the need is practical; adapted to the communica- now very much felt of teaching which tion of some degree of power, as well as knowledge, but free from details which are important only to the spe- cialist. Professor Byerly's Calculus appears to be designed to meet this want.... Such a plan leaves much room for the exercise of individual judgment; and differences of opinion will undoubtedly exist in regard to one and another point of this book. But all teachers will agree that in selection, arrangement, and treatment, it is, on the whole, in a very high degree, wise, able, marked by a true scientific spirit, and calculated to develop the same spirit in the learner.... The book contains, perhaps, all of the integral calculus, as well as of the differential, that is necessary to the ordinary stu- dent. And with so much of this great scientific method, every thorough stu- dent of physics, and every general scholar who feels any interest in the relations of abstract thought, and is capable of grasping a mathematical idea, ought to be familiar. One who aspires to technical learning must sup- plement his mastery of the elements by the study of the comprehensive theoretical treatises.... But he who is thoroughly acquainted with the book before us has made a long stride into a sound and practical knowledge of the subject of the calculus. He has begun to be a real analyst. H. A. Newton, Prof of Math. in Yak Coli., New Haven: I have looked it through with care, and find the sub- ject very clearly and logically devel- oped. I am strongly inclined to use it in my class next year. S. Hart, recent Prof ofAMath. in Trinity Coll., Conn.: The student can hardlv fail, I think, to get from the book an exact, and, at the same time, a satis- factory explanation of the principles on which the Calculus is based; and the introduction of the simpler methods of integration, as they are needed, enables applications of those prineiples to be introduced in such a way as to be both interesting and instructive. Charles Kraus, Techniker, Pard- ubitz, Bohemia, Austria: Indem ich den Empfang Ihres Buches dankend bestaetige muss ich Ihnen, hoch geehr- ter Herr gestehen, dass mich dasselbe sehr erfreut hat, da es sich durch grosse Reichhalfigkeit, besonders klare Schreibweise und vorzuegliche Behand- lung des Stoffes auszeichnet, und er- weist sich dieses Werk als eine bedeut- ende Bereicherung der mathematischen Wissenschaft. De Volson Wood, Prof. of Math., Stevens' Inst., Hoboken, N.77.: To say, as I do, that it is a first-class work, is probably repeating what many have already said for it. I admire the rigid logical character of the work, and am gratified to see that so able a writer has shown explicitly the relation between Derivatives, Infinitesimals, and Differentials. The method of Limits is the true one on which to found the science of the calculus. The work is not only comprehensive, but no vague- ness is allowed in regard to definitions or fundamental principles. Del Kemper, Prof. of Math., Hampden Sidney Coll., Va.: My high estimate of it has been amply vindi- cated by its use in the class-room. R. H. Graves, Prof of Math., Univ. of Nor/h Carolina.: I have al- ready decided to use it with my next class; it suits my purpose better than anv other book on the same subject with which I am acquainted. Edw. Brooks, Author of a Series /fMath.: Its statements are clear and scholarly, and its methods thoroughly analytic and in the spirit of the latest mathematical thought. Syllabus of a Course in Plane Triqonometry. By W. E. I3YERLY. Svo. 8 pages. Mailing Price, to cts. Syllabus of a Course in Plane Analytical Geom- elry. By W. E. BYERLY. 8vo. 12 pages. Mailing Price, lO cts. Syllabus of a Course in Plane Analytic Geom- eh)y (Advan;ced Course.) By W. E. BYERLY, Professor of Mathe- matics, Harvard University. 8vo. 12 pages. Mailing Price, 10 cts. Syllabus of a Course in Analytical Geometry of Three Dimensions. By W. E. BYERLY. 8vo. so pages. Mailing Price, 10 cts. Syllabus of a Course on Modern Methods in Aoztahrlic Geomwetry. BY W. E. BYERLY. 8vo. 8 pages. Mailing - Price, lo cts. Syllabus of a Course in the Theory of Equations. By W. E. BYERLY. 8vo. 8 pages. Mailing Price, 1o cts. Elements of the Integral Calculus. By WN. E. BYERLY, Professor of Mathematics in Harvard University. Svo. 204 pages. Mailing Price, 2.15; Introduction, 2.00. This volume is a sequel to the author's treatise on the Differential Calculus (see page 134), and, like that, is written as a text-book. The last chapter, however, - a Key to the Solution of Differential Equations, - may prove of service to working mathematicians. H. A. Newton, Prof. of Math., Yale Coil.: We shall use it in my optional class next term. Mathematical Visitor: The subject is presented very clearly. It is the first American treatise on the Cal- cullus that we have seen which devotes any space to average and probability. Schoolmaster, London: The merits of this work are as marked as those of the Differential Calculus by the same author. Zion's Herald: A text-book everv way worthy of the venerable University in which the author is an honored teacher. Cambridge in Massachusetts, like Cambridge in England, preserves its reputation for the breadth and strict- ness of its mathematical requisitions, and these form the spinal column of a liberal education. A Short Table of Integrals. To accompany B YERL Y'S INTEGRAL CAL CUL US. By B. 0. PEIRCE, JR., Instructor in Mathematics, Harvard University. 16 pages. Mailing Price, io cts. To be bound with future editions of the Calculus. Elements of Quaternions. By A. S. HARDY, Ph.D., Professor of Mathematics, Dartmouth College. Crown, 8vo. Cloth. 24o pages. Mailing Price, 2.15; Introduction, 2.00. The chief aim has been to meet the wants of beginners in the class-room. The Elemzents and Lectures of Sir W. R. Hamilton are mines of wealth, and may be said to contain the suggestion of all that will be done in the way of Quaternion research and application: for this reason, as also on account of their diffuseness of style, they are not suitable for the purposes of elementary instruc- tion. The same may be said of Taits Quaternions, a work of great originality and comprehensiveness, in style very elegant but very concise, and so beyond the time and needs of the beginner. The Introduction to Quaternions by Kelland contains many exer- cises and examples, of which free use has been made, admirably illustrating the Quaternion spirit and method, but has been found, in the class-room, practically deficient in the explanation of the theory and conceptions which underlie these applications. The object in view has thus been to cover the introduictory ground more thoroughly, especially in symbolic transformations, and at the same time to obtain an arrangement better adapted to the methods of instruction common in this country. PRESS NOTICES. Westminster Review: It is a The Nation: For those who have remarkably clear exposition of the sub- never studied the subject, this treatise ject. seems to its superior both to the work of Prof. Tait and to the joint treatise by The Daily Review, Edinburgh, Profs. Tait and Kelland. Scotland: This is an admirable text- book. Prof. Hardy has ably supplied New York Tribune: The Qua- a felt want. The definitions are models ternion Calculus Is an instrument of of conciseness and perspicuity. mathematical research at once so pow- erful, flexible, and elegant, so sweeping in its range, and so minutely accurate, that its discovery and Development has been rightly estimated as one of the crowning achievements of the century. The time is approaching when all col- leges will insist upon its study as an essential part of the equipment of young men who aspire to be classified among the liberally educated. This book fur- nishes just the elementary instruction on the subject which is needed. New York Times: It is especially designed to meet the needs of begin- ners in the science. . . . It has a wav of putting things which is eminently its own, and which, for clearness and force, is as yet unsurpassed.... If we may not seek for Quaternions made easy, we certainly need search no longer for Quaternions made plain. Van Nostrand Engineering Magazine: To any one who has labored with the very few works ex- tant upon this branch of mathematics, a glance at the opening chapter of Prof. Hardy's work will enforce the conviction that the author is an in- structor of the first order. The book is quite opportune. The subject must soon become a necessary one in all the higher institutions, for already are writers of mathematical essays making free use of Quaternions without any preliminary apology. Canada School Journal, To- ronto: The author of this treatise has shown a thorough mastery of the Qua- temion Calculus. London Schoolmaster: It is in every way suited to a student who wishes to commence the subject ab initio. One will require but a few hours with this book to learn that this Calculus, with its concise notation, is a most powerful instrument for mathe- matical operations. Boston Transcript: A text-book of unquestioned excellence, and one peculiarly fitted for use in American schools and colleges. The Western, St. Louis: This work exhibits the scope and power of the new analysis in a very clear and concise form ... illustrates very finely the important fact that a few simple principles underlie the whole body of mathematical truth. FROM COLLEGE PROFESSORS. James Mills Peirce, Prof of Afath., Harvard Coll.: I am much pleased with it. It seems to me to supply in a very satisfactory manner the need which has long existed of a clear, concise, well-arranged, and logi- cally-developed introduction to this branch of Mathematics. I think Prof. Hardy has shown excellent judgment in his methods of treatment, and also in limiting himself to the exposition and illustration of the fundamental principles of his subject. It is, as it ought to be, simply a preparation for the studv of the writings of Hamilton and Tait. I hope the publication of this attractive treatise will increase the attention paid in our colleges to the profound, powerful, and fascinating cal- culus of which it treats. Charles A. Young, Prof. of Astronomy, Princeton Coil.: I find it by far the most clear and intelligible statement of the matter I have yet seen. Elements of the Differential and Integral Calculus. With Examples and Applications. By J. M. TAYLOR, Professor of Mathematics in Madison University. 8vo. Cloth. 249 pp. Mailing price, i.95; Introduction price, i.80. The aim of this treatise is to present simply and concisely the fundamental problems of the Calculus, their solution, and more common applications. Its axiomatic datum is that the change of a variable, when not uniform, may be conceived as becoming uniform at any value of the variable. It employs the conception of rates, which affords finite differen- tials, and also the simplest and most natural view of the problem of the Differential Calculus. This problem of finding the relative rates of change of related variables is afterwards reduced to that of finding the limit of the ratio of their simultaneous increments; and, in a final chapter, the latter problem is solved by the principles of infinitesimals. Many theorems are proved both by the method of rates and that of limits, and thus each is made to throw light upon the other. The chapter on differentiation is followed by one on direct integra- tion and its more important applications. Throughout the work there are numerous practical problems in Geometry and Mechanics, which serve to exhibit the power and use of the science, and to excite and keep alive the interest of the student. Judging from the author's experience in teaching the subject, it is believed that this elementary treatise so sets forth and illustrates the highly practical nature Of the Calculus, as to awaken a lively interest in many readers to whom a more abstract method of treat- ment would be distasteful. Oren Root, Jr., Prof. of Math., C. M. Charrappin, S.J., St. Hamilton Coll., A. Y.: In reading the Louis U;niv.: I have given the book a manuscript I was impressed by the thorough examination, and I am satis- clearness of definition and demonstra- fied that it is the best work on the sub- tion, the pertinence of illustration, and ject I have seen. I mean the best the happy union of exclusion and con- work for what it was intended,-a text- densation. It seems to me most admir- book. I would like very much to in- ably suited for use in college classes. troduce it in the University. I prove my regard by adopting this as I (Jan. i2, i885.) our text-book on the calculus.