tC: Digitized by the Internet Archive in 2010 with funding from Boston Library Consortium IVIember Libraries http://www.archive.org/details/analyticgeometryOOdowl Hmericau /IDatbematical Series E. J. TOWNSEND GENERAL EDITOR MATHEMATICAL SERIES E. J. TOWNSEND, General Editor. While this series has been planned to meet the needs of the student who Js preparing for engineering work, it is hoped that it will serve equally well the purposes of those schools where mathematics is taken as an element in a liberal education. In order that the applications introduced may be of such character as to interest the general student and to train the prospective engineer in the kind of work which he is most likely to meet, it has been the policy of the editors to select, as joint authors of each text, a mathemati- cian and a trained engineer or physicist. The following texts are ready : L Calculus, By E. J. TowNSEND, Professor of Mathematics, and G. A. GooD- ENOUGH, Professor of Thermodynamics, University of Illinois. ^2.50. IL Essentials of Calcultis. By E. J. TOWNSEND and G. A. Goodenough. ^2.00. IIL CoIIegfe Algebra. By H. L. RiETZ, Assistant Professor of Mathematics, and Dr. A. R. CraTHORNE, Associate in Mathematics in the University of Illinois. $1.40. IV, Plane Trigfonometry, with Trigfonometric and Logfarithmic Tables. By A. G. Hall, Professor of Mathematics in the University of Michigan, and F. H. Frink, Professor of Railway Engineering in the University of Oregon. $1.25. V. Plane and Spherical Trig-onometry. (Without Tables) By A. G. Hall and F. H. Frink. ^i.oo. VI. Trigonometric and Logarithmic Tables. By A. G. Hall and F. H. Frink. 75 cents. VII. Analytic Geometry of Space. By Virgil Snyder, Professor in Cornell University, and C. H. SiSAM, Assistant Professor in the University of Illinois. ^2.50. VIIL Analytic Geometry. By L. W. DowLiNG, Assistant Professor of Mathematics, and F. E. TURNEAURE, Dean of the College of Engineering in the University of Wisconsin. IX. Elementary Geometry. By J. W. Young, Professor of Mathematics in Dartmouth College, and a'. J. Schwartz, William McKinley High School, St. Louis. HENRY HOLT AND COMPANY NEW YORK CHICAGO ANALYTIC GEOMETRY BY L. WAYLAND BOWLING, Ph.D. Associate Professob of Mathematics University of Wisconsin AND F. E. TURNEAURE, C.E. Dean of the College of Engineering University of Wisconsin BOSTON ooLLEOK,„.„ ^4 NEW YORK HENRY HOLT AND COMPANY ^^. 150563 Copyright, 1914 BY HENET HOLT AND COMPANY NorSiiooli }Pr£03 J. S. Gushing Co. — Berwick & Smith Co. Norwood, Mass., U.S.A. PREFACE Ix accordance with the general plan of this series of textbooks, the authors of the present volume have had constantly in mind the needs of the student who takes his mathematics primarily with a view to its applications as well as the needs of the student who pursues mathematics as an element of his education. The processes of analytical geometry find their application, for the most part, in the scientific laboratory where it is often neces- sary to study the properties of a function from certain observed values. The fundamental concept is, therefore, that of functional correspondence and the methods of representing such correspond- ence geometrically. For this reason rather more than usual attention has been given to these subjects (Chapter III; also Chapter IX, Arts. 135 to 140). An intelligent appreciation of functional correspondence re- quires an intimate knowledge of the relation between an equation and the graphical representation of the functional correspondence determined by the equation. Such a knowledge is most easily obtained by a study of linear equations and equations of the second degree together with their corresponding loci. This knowl- edge is not only of importance to the student of applied mathe- matics, but it has a special disciplinary value for the general student. The standard forms of the equations of a number of important loci are developed early (Chapter IV), and the properties of these loci are discussed in detail later (Chapters VI and VII) by means of the equations already at hand. By this arrangement, it is hoped that some unnecessary repetition has been avoided. The equations of tangents to the conic sections have been derived by means of the discriminant of the quadratic equation whose roots are the ^'-coordinates of the points of intersection with a variable secant, rather than by means of the derivative. This course has been adopted, first, because the geometric inter- iv PREFACE pretation of the discriminant is important in itself ; and, second, because the use of the derivative ought, logically, to be preceded by a chapter devoted to its definition and the methods for finding it, at least for algebraic functions. Moreover, the use of the derivative for finding the equations of tangents is only one of its many applications. No student should feel that his mathe- matical education is complete without a knowledge of the calculus, where he will become familiar with the derivative and can appre- ciate its usefulness in many directions. The present volume is designed for a four-hour, or a five-hour, course for one semester, but may be shortened to a three-hour course by omitting certain parts of the text. For example, Art. 105 may be omitted without marring the continuity of the course. Again, Arts. 110, 111, and 112 contain all that is essential in dealing with the general equation of the second degree in two variables, and the remainder of Chapter VIII can therefore be omitted from the longer course. Parts of Chapter IX can also be omitted according to the needs of the student. The chapters on solid analytic geometry have been added for the benefit of those students who have time only for an outline of the subject matter. ISTo apology is therefore offered for the meager treatment. The authors desire to express their appreciation to their col- leagues of the University of Wisconsin and of the University of Illinois for the assistance and the many helpful suggestions given them during the preparation of the book. They are under especial obligations to Professor W. H. Bussey, of the University of Minnesota; Professor S. C. Davisson, of the University of Indiana ; Professor J. L. Markley, of the University of Michigan ; and Professor E. J. Townsend, of the University of Illinois, for their care and assistance in seeing the book through the press. L. W. BOWLING, F. E. TURNEAURE. University of Wisconsin, July, 1914. CONTENTS INTRODUCTION PAGE A. The quadratic equation 1 B. Trigonometric formulas ......... 1 C. Numerical tables 3 PART PLANE ANALYTIC GEOMETRY CHAPTER I SYSTEMS OF COORDINATES ARTICLE 1. The linear scale 7 2. Directed segments, directed angles 8 3. Addition of directed segments, addition of directed angles . . 9 4. Position of a point in a plane 10 5. Cartesian coordinates ......... 10 6. Rectangular coordinates 11 7. Notation 12 8. Polar coordinates .......... 13 9. Relation between rectangular coordinates and polar coordinates . 14 CHAPTER II DIRECTED SEGMENTS AND AREAS OF PLANE FIGURES 10. Projections upon the coordinate axes 18 11. Inclination and slope of a directed segment 18 12. The length of a segment 20 13. Angle which one segment makes with another .... 22 14. Parallel segments 23 15. Perpendicular segments ,23 16. Point bisecting a given segment 24 17. Point dividing a given segment in a given ratio . . . .25 18. Area of a triangle, one vertex at the origin 26 19. Sign of the expression (xi2/2 — A'22/i) 26 20. Area of a triangle, vertices in any position ..... 28 21. Area of any polygon ......... 30 V vi CONTENTS CHAPTEE III FUNCTIONS AND THEIR GRAPHIC REPRESENTATION ARTICLE PAGE 22. Constants and variables 33 23. Functions 33 24. Notation 33 25. Determination of functional correspondence ..... 33 26. Dependent and independent variables 34 27. Graphic representation ......... 34 28. Single-valued and multiple-valued functions ..... 36 29. Symmetry 37 30. Intercepts 38 31. Graph in polar coordinates ........ 39 32. Algebraic functions 41 33. Transcendental functions 41 34. Graphs of transcendental functions ....... 41 35. Geometric construction of the graphs of trigonometric functions . 42 36. The exponential function 44 37. Graph of the exponential function 44 38. Inverse functions c . 45 39. Graph of an inverse function 46 40. Observation 48 41. Machines 48 CHAPTER IV LOCI AND THEIR EQUATIONS 42. Locus of a point, equation of locus ....... 53 ■ 43. A fundamental problem 53 44. General definition 54 45. The circle 55 46. The equation :>i^ -\- if + Ax + By -\- C = 56 47. The straight line .......... 57 48. The determinant form 59 49. The ellipse 60 50. The axes and eccentricity 62 51. The hyperbola .63 52. Axes and eccentricity ......... 65 53. The parabola 66 54. The cassinian ovals . .67 55. Recapitulation 69 56. Polar equation of a circle ......... 69 57. Polar equation of a straight line .... . . 70 CONTENTS Vll ARTICLE 58. Polar equation of the parabola ..... 59. Polar equations of the ellii^se and the hyperbola 60. Parametric equations 61. Geometrical construction of the ellipse and the hyperbola 62. Recapitulation 71 71 73 74 76 CHAPTER V EQUATIONS AND THEIR LOQI 63. Locus of an equation .... 64. A second fundamental problem 65. Discussion of an equation 66. Example I. The equation x- + 4 y^ = 4 . 67. Example II, Tlie equation x'^ — iy- =4: 68. Example III. The equation xy — x — y = 69. Example IV. The equation y — 70. Example V. The equation y^ = 1 +x- x'(6— x) 3x -\- b Example VI. The catenary, y = ^(e" ) Simple harmonic curves, compound 72. Example VII. curves ...... 73. Example VIII. Damped vibrations 74. Polar equations . . . 75. Example IX. The equation r = cos 2 e . 76. Example X. The equation r- = a^ cos 2 6 77 77 77 79 80 81 82 83 harmonic . 86 . 87 89 90 TRANSFORMATION OF COORDINATES 77. Translation of the axes 91 78. Rotation of the axes ......... 92 79. Removal of terms of first degree 94 80. Removal of the term in xy 95 81. Classification of algebraic curves . • 96 CHAPTER VI LOCI OF FIRST ORDER 82. Linear equations 83. Intersection of two lines . 84. The pencil of lines . 85. The pair of hues 98 99 100 102 viii CONTENTS ARTICLE PAGE 86. The normal form 102 87. Reduction of Ax + By + C — to the normal form . . . 103 88. Distance from a line to a point 104 89. The angle which one line makes with another .... 105 CHAPTER VII LOCI OF SECOND ORDER. EQUATIONS IN STANDARD FORM DIRECTRICES 90. Review 107 91. Directrices 108 92. A fundamental theorem 108 93. Construction of an ellipse or an hyijerbola ..... 109 94. Two common properties 110 TANGENTS 95. Equation of a tangent in terms of the slope . . . . .111 96. Coordinates of the point of contact 11-5 97. Equation of a tangent in terms of the coordinates of the point of contact . . . . . . . , . . .116 98. Normals 110 99. Tangent length, normal length, sub-tangent, sub-normal . . 119 100. Reflection properties 120 DIAMETERS 101. Definition of diameter .......... 123 102. Conjugate diameters 123 103. The locus of the middle points of a system of parallel chords . 12-5 POLES AND POLAR LINES 104. Definition of pole and polar line ....... 127 105. Geometric properties of poles and polar lines .... 129 SYSTEMS OF CONICS 106. The asymptotes of the hyperbola ....... 133 107. Conjugate hyperbolas ..•....,. 134 108. The system of concentric hyperbolas ...... 135 109. The system of confocal conies ....... 136 CONTENTS IX CHAPTER VIII LOCI OF THE SECOND ORDER. EQUATIONS NOT IN STANDARD FORM ARTICLE 110. Translation of the coordinate axes 111. Discussion of the equation ax'^ + hy^ + 2 gx -\- 2fy + c 112. The general equation of the second degree 113. Rsmoval of the terms of first degi'ee 114. First case, «?) —/i- ^t .... 115. Second case, ah — li^ = 0, hf — hg ^ . 116. Third case, ah - h'^ = and hf - bg = 117. Recapitulation PAGE 141 142 145 148 149 152 155 156 TANGENTS AND DIAMETERS 118. Tangents 119. Diameters 157 159 SYSTEMS OF CONICS The pencil of conies ..... The system of circles with a common radical axis The parabolas in the pencil U+ k\' = Straight lines in the pencil U + Vk — 16.3 163 164 165 167 CHAPTER IX LOCI OF HIGHER ORDER AND OTHER LOCI 125. Introductory note 169 126. 127. 128. 129. 130. 131. 132. 133. 134 ALGEBRAIC LOCI The Cissoid of Diodes .... The Conchoid of Nicomedes . The Witch of Agnesi .... The Liraagon of Pascal .... TRANSCENDENTAL LOCI The cycloid The hypocycloid Special hypocycloids The epicycloid The cardioid . 169 170 171 172 173 175 176 177 178 X CONTENTS EMPIRICAL EQUATIONS AND THEIR LOCI ARTICLE PAGE 135. Typical equations 182 136. Loci of typical equations 182 137. Selection of type-curve and determination of constants . . 186 138. Test by means of linear equations 187 139. Examples and exercises 189 140. 'Type, y = a + bx + cx^ + dx^ + ■■■ ^kx" 193 PART n SOLID ANALYTIC GEOMETRY CHAPTER X SYSTEMS OF COORDINATES 141. Rectangular and oblique coordinates 142. Spherical coordinates 143. Cylindrical coordinates . 195 196 197 CHAPTER XI DIRECTED SEGMENTS IN SPACE 144. Projections upon the coordinate axes 145. Length of a segment 146. Direction angles and direction cosines of a segment 147. Relation connecting the direction cosines of a segment 148. Projection of a segment upon any line . 149. Projection of a broken line 150. The angle between two segments . 151. Perpendicular segments, parallel segments 152. Point dividing a given segment in a given ratio 199 199 200 200 202 202 203 203 204 CHAPTER XII LOCI AND THEIR EQUATIONS 153. Surfaces and curves 154. Equations of loci . 155. The sphere 156. Surfaces of revolution 157. Cylinders 158. The right circular cone 159. Plane sections of a right circular cone 207 207 208 209 210 211 211 CONTENTS XI CHAPTER XIII THE PLANE AND THE STRAIGHT LINE IN SPACE ABTICLE PAGE 160. The normal form of the equation of a plane 213 161. The intercept form of the equation 215 162. Equation of a plane through three given points .... 215 163. Determinant form of the equation 216 164. Perpendicular distance from a plane to a point .... 217 165. Angle between two planes ........ 218 166. Pencil of planes with a common axis 219 167. Pencil of planes with a common vertex 220 168. The equations of a straight line in space 221 169. The projecting planes of a line 222 170. The intersection of two planes 223 171. Intersection of a line with a plane 225 CHAPTER XIV EQUATIONS AND THEIR LOCI 172. Second fundamental problem 173. Construction of a surface from its equation 174. The quadric surfaces, or conicoids 175. The ellipsoid ... 176. The hyperboloid of one sheet . 177. The hyperboloid of two sheets 178. The elliptic paraboloid . 179. The hyperbolic paraboloid 180. The quadric cone . 181. CyUnders 182. Pairs of planes 183. Ruled surfaces 184. Equation of generator . 185. Tangent lines and planes 186. Circular sections 187. Asymptotic cones . 188. Projecting cylinders of a curve in space 189. Parametric equations of curves in space 190. The circular helix , , . . . Answers . . , . . . . 228 228 229 230 231 233 234 235 236 237 237 238 239 240 242 243 244 245 246 249 GREEK ALPHABET Letteks Letteks Names Names Capitals Lower Case Capitals Lower Case A a Alpha N V Nu B P Beta E $ Xi r 7 Gamma o Omicron A 8 Delta n TT Pi E e Epsilon p P Rho Z ^ Zeta 2 a Sigma H V Eta T T Tau © Theta Y V Upsilon I L Iota $ 4> Phi K K Kappa X X Chi A X Lambda * ^ Psi M H- Mil o 0, the roots are real and unequal, if 6^ — 4 ac = 0, the roots are real and equal, if 6^ — 4 ac < 0, the roots are imaginary. The expression b- — 4 ac is called the discriminant of the equation, and when placed equal to zero expresses the condition which must hold between the coefficients, if the two roots of the equation are equal. B. Trigonometric formulas. If A, B, and C are the angles of a triangle and a, b, c are respectively the lengths of the sides opposite, then : 1 INTRODUCTION (1) Laiv of sines: sin A _ siii^ _ sin C a b c ' (2) Laiv of cosines : a'^= b^ + c^ — 2 be cos A, b^ = a^ + c^ — 2ac cos B, c'^ = a'^ + b'^ — 2 ab cos C. (3) Lnw of tangents : tan \{A-B) «t - 6' Addition formulas. If A and B are any angles, then sin {A±B) = sin ^ cos J5 i sin B cos ^, cos (^ ± jE») = cos ^ cos J5 =F sin A sin B, tan 4 ± tan B tan (^ ± B) : 1 T tan ^ tan B TABLES C. TABLES Common Logarithms N 10 o 0000 D 43 1 0043 D 43 2 0086 D 42 3 D 42 4 0170 D 5 D 41 6 0253 D 41 7 D 40 8 0334 D 40 9 0374 D 40 0128 42 0212 0294 11 0414 39 0453 39 0492 39 0531 38 0569 38 0607 38 0645 37 0682 37 0719 36 0755 37 12 0792 36 0828 36 0864 35 0899 35 0934 35 0969 35 1004 34 1038 34 1072 34 1106 33 13 1139 34 1173 33 1206 33 1239 32 1271 32 1303 32 1335 32 1367 32 1399 31 1430 31 14 15 1461 1761 31 29 1492 1790 31 28 1523 30 29 1553 1847 31 28 1584 1875 30 1614 28 1903 30 28 1644 1931 29 28 1673 1959 30 28 1703 1987 29 27 1732 29 27 1818 2014 16 2041 27 2068 27 2095 27 2122 26 2148 27 2175 26 2201 26 2227 26 2253 26 2279 25 17 2304 26 2330 25 2355 25 2380 25 2405 25 2430 25 2455 25 2480 24 2504 25 2529 24 18 2553 24 2577 24 2601 24 2625 23 2648 24 2672 23 2695 23 2718 24 2742 23 2765 23 19 20 2788 3010 22 2810 3032 23 22 2833 3054 23 21 2856 3075 22 21 2878 3096 22 2900 22 3118 23 21 2923 3139 22 21 2945 3160 21 2967 3181 22 20 2989 21 21 3201 21 3222 21 3243 20 3263 21 3284J20 3304 20 3324 21 3345 20 3365 20 3385 19 3404 20 22 3424 20 3444 20 3464 19 3483119 3502 >o 3522 19 3541 19 3560 19 3579 19 3598 19 23 3617 19 3636 19 3655 19 3674! IS 3692 L9 3711 18 3729 18 3747 19 3766 18 3784 18 24 25 3802 18 18 3820 18 17 3838 4014 18 17 3856 18 3874 4048 8 3892 L7 4065 17 17 3909 4082 18 17 3927 4099 18 17 3945 4116 17 17 3962 4133 17 17 3979 3997 4031 17 26 4150 16 4166 17 4183 17 4200 16 4216 16 4232 17 4249 16 4265 16 4281 17 4198 16 27 4314 16 4330 16 4346 16 4362 16 4378 5 4393 16 4409 16 4425 15 4440 16 4456 16 28 4472 15 4487 15 4502 16 4518 15 4533 L5 4548 16 4564 15 4579 15 4594 15 4609 15 29 30 4624 4771 15 15 4639 4786 15 14 4654 4800 15 14 4669 4814 14 15 4683 5 4698 4 4843 15 14 4713 4857 15 14 4728 4871 14 4742 4886 15 U 4757 14 14 4829 4900 31 4914 14 4928 14 4942 13 4955 14 4969 4 4983 14 4997 14 5011 13 5024 14 5038 13 32 5051 14 5065 14 5079 13 5092 13 5105 4 5119 13 5132 13 5145 14 5159 13 5172 13 33 5185 13 5198 13 5211 13 5224 13 5237 3 5250 13 5263 13 5276 13 5289 13 5302 13 34 35 5315 5441 13 12 5328 5453 12 12 5340 5465 13 13 5353 5478 13 12 5366 5490 2 5378 2 5502 13 12 5391 5514 12 13 5403 5527 13 12 5416 12 12 5428 13 12 5539 5551 36 5563 12 5575 12 5587 12 5599 12 5611 2 5623 12 5635 12 5647 11 5658 12 5670 12 37 5682 12 5694 11 5705 12 5717 12 5729 1 5740 12 5752 11 5763 12 5775 11 5786 12 38 5798 11 5809 12 5821 11 5832 11 5843 2 5855 11 5866 11 5877 11 5888 11 5899 12 39 40 5911 6021 11 10 5922 11 11 5933 6042 11 11 5944 6053 11 11 5955 ] 6064 1 1 5966 1] 10 5977 6085 11 11 5988 6096 11 11 5999 6107 11 10 6010 6117 11 11 6031 16075 41 6128 10 6138 11 6149 11 6160 10 6170] 6180 11 6191 10 6201 11 6212 10 6222 10 42 6232 11 6243 10 6253 10 6263 11 6274 1 6284 10 6294 10 6304 10 6314 11 6325 10 43 6335 10 6345 10 6355 10 6365 10 6375 6385 10 6395 10 6405 10 6415 10 6425 10 44 45 6435 6532 9 10 6444 10 9 6454 6551 10 10 6464 6561 10 10 6474 6571 6484 9 6580 9 10 6493 6590 10 9 6503 6599 10 10 6513 6609 9 9 6522 10 10 6542 6618 46 6628 9 6637 9 6646 10 6656 9 6665 ] 6675 9 6684 9 6693 9 6702 10 6712 9 47 6721 9 6730 9 6739 10 6749 9 6758 9 6767 9 6776 9 6785 9 6794 9 6803 9 48 6812 9 6821 9 6830 9 6839 9 6848 9 6857 9 6866 9 6875 9 6884 9 6893 9 49 50 6902 6990 9 8 6911 6998 9 9 6920 7007 8 9 6928 7016 9 8 6937 7024 9 6946 9 7033 9 9 6955 7042 9 8 6964 7050 8 9 6972 7059 9 8 6981 9 9 7067 51 7076 8 7084 9 7093 8 7101 9 7110 8 7118 8 7126 9 7135 8 7143 9 7152 8 52 7160 8 7168 9 7177 8 7185 8 7193 9 7202 8 7210 8 7218 8 7226 9 7235 8 53 7243 8 7251 8 7259 8 7267 8 7275 9 7284 8 7292 8 7300 8 7308 8 7316 8 54 7324 8 7332 8 7340 8 7348 8 7356 8 7364 8 7372 8 7380 8 7388 8 7396 8 TABLES Common Logarithms — Continued N 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 o 7404 7482 7559 7634 7709 7782 7853 7924 7993 8062 8129 8195 8261 8325 8388 8451 8513 8573 8633 8692 8751 8808 8865 8921 8976 9031 9085 9138 9191 9243 9294 9345 9395 9445 9494 9542 9590 9638 9685 9731 9777 9823 9868 9912 9956 u 8 8 7 8 7 7 7 7 7 7 7 7 6 6 7 6 6 6 6 6 5 6 6 6 6 5 5 5 5 5 5 5 5 5 5 5 5 5 4 4 5 5 1 7412 7490 7566 7642 7716 7789 7860 7931 8000 8069 8136 8202 8267 8331 8395 8457 8519 8579 8639 8698 8756 8814 8871 8927 8982 9036 9090 9143 9196 9248 9299 9350 9400 9450 9499 9547 9595 9643 9689 9736 9782 9827 9872 9917 9961 D 7 7 8 7 7 7 8 7 7 6 6 7 7 7 6 6 6 6 6 6 6 6 5 5 5 6 6 6 5 5 5 5 5 5 5 5 5 4 5 5 4 5 5 4 4 2 7419 7497 7574 7649 7723 7796 7868 7938 8007 8075 8142 8209 8274 8338 8401 8463 8525 8585 8645 8704 8762 8820 8876 8932 8987 9042 9096 9149 9201 9253 9304 9355 9405 9455 9504 9552 9600 9647 9694 9741 9786 9832 9877 9921 9965 D 8 8 8 7 7 7 6 6 6 6 7 6 6 6 6 6 5 6 6 6 5 5 5 5 5 5 6 5 5 6 5 5 5 4 ^ 4 4 5 3 7427 7505 7582 7657 7731 7803 7875 7945 8014 8082 8149 8215 8280 8344 8407 8470 8531 8591 8651 8710 8768 8825 8882 8938 8893 9047 9101 9154 9206 9258 9309 9360 9410 9460 9509 9557 9605 9652 9699 9745 9791 9836 9881 9926 9969 D 8 8 7 7 7 7 7 7 7 7 ' 7 7 7 7 6 6 6 6 6 6 6 5 5 5 6 5 5 6 5 6 5 5 5 4 5 4 5 4 5 4 5 5 4 5 4 7435 7513 7589 7664 7738 7810 7882 7952 8021 8089 8156 8222 8287 8351 8414 8476 8537 8597 8657 8716 8774 8831 8887 8943 8998 9053 9106 9159 9212 9263 9315 9365 9415 9465 9513 9562 9609 9657 9703 9750 9795 9841 9886 9930 9974 D 8 7 8 8 7 8 7 7 6 6 6 6 6 6 6 6 6 6 5 6 6 6 6 5 6 6 5 6 5 5 5 4 5 4 5 4 5 4 5 4 4 4 4 5 7443 7520 7597 7672 7745 7818 7889 7959 8028 8096 8162 8228 8293 8357 8420 8482 8543 8603 8663 8722 8779 8837 8893 8949 9004 9058 9112 9165 9217 9269 9320 9370 9420 9469 9518 9566 9614 9661 9708 9754 9800 9845 9890 9934 9978 D 8 8 7 7 7 7 7 7 7 6 7 7 6 6 6 6 6 6 6 5 6 5 6 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 4 5 5 6 7451 7528 7604 7679 7752 7825 7896 7966 8035 8102 8169 8235 8299 8363 8426 8488 8549 8609 8669 8727 8785 8842 8899 8954 9009 9063 9117 9170 9222 9274 9325 9375 9425 9474 9523 9571 9619 9666 9713 9759 9805 9850 9894 9939 9983 D 8 8 8 7 8 7 7 7 6 7 7 6 7 7 6 6 6 6 6 6 6 6 5 6 6 6 5 5 5 5 5 5 5 5 5 5 5 5 4 4 4 4 5 4 4 7 7459 7536 7612 7686 7760 7832 7903 7973 8041 8109 8176 8241 8306 8370 8432 8494 8555 8615 8675 8733 8791 8848 8904 8960 9015 9069 9122 9175 9227 9279 9330 9380 9430 9479 9528 9576 9624 9671 9717 9763 9809 9854 9899 9943 9987 D 7 7 7 8 7 7 7 7 7 7 6 7 6 6 7 6 6 6 6 6 6 6 6 5 5 5 6 5 5 5 5 5 5 5 5 5 4 4 5 5 5 5 4 5 8 7466 7543 7619 7694 7767 7839 7910 7980 8048 8116 8182 8248 8312 8376 8439 8500 8561 8621 8681 8V39 8797 8854 8910 8965 9020 9074 9128 9180 9232 9284 9335 9385 9435 9484 9533 9581 9628 9675 9722 9768 9814 9859 9903 9948 9991 D 8 8 8 7 7 7 7 7 7 6 7 6 7 6 6 6 6 6 5 6 5 5 5 6 5 5 5 6 6 5 5 5 5 5 5 5 5 5 5 5 4 4 5 4 5 9 7474 7551 7627 7701 7774 7846 7917 7987 8055 8122 8189 8254 8319 8382 8445 8506 8567 8627 868(5 8745 8802 8859 8915 8971 9025 9079 9133 9186 9238 9289 9340 9390 9440 9489 9538 9586 9633 9680 9727 9773 9818 9863 9908 9952 9996 D 8 8 7 8 8 7 7 6 7 7 6 7 6 6 6 7 6 6 6 6 6 6 6 5 6 6 5 5 5 5 5 5 5 5 4 4 5 5 4 4 5 5 4 4 4 TABLES Trigonometric Functions [Characteristics of Logarithms omitted — determine by the usual rule from the value] Radians De- Sine Tangent Cotangent Cosine grees Talue logio Value login Value logio Value logio .0000 0° .0000 — CO .0000 -00 00 00 1.0000 GOOO 90° 1..5708 .0175 1° .0175 2419 .0175 2419 57.290 7581 .9998 9999 89° 1.5533 .0349 2° .0349 5428 .0.349 .5431 28.036 4569 .9994 9997 88^ 1.5359 .0524 3° .0523 7188 .0524 7196 19.081 2806 .9986 9994 87° 1.5184 .0698 4° .0698 8436 .0699 8448 14.301 1554 .9976 9989 86° 1.5010 .0873 5° .0872 9403 .0875 9420 11.4.30 0580 .9962 9983 85° 1.48.35 .1047 6° .1045 0192 .1051 0216 9.5144 9784 .9945 9976 84° 1.4661 .1222 7° .1219 0859 .1228 0891 8.1443 9109 .9925 9968 83° 1.4486 .1396 8° .1392 1436 .1405 1478 7.1154 8522 .9903 9958 82° 1.4312 .1571 9° .1564 1943 .1584 1997 6.3138 8003 .9877 9946 81° 1.4137 .1745 10° .1736 2397 .1763 2463 5.6713 7537 .9848 9934 80° 1.3963 .1920 11° .1908 2806 .1944 2887 5.1446 7113 .9816 9919 79° 1.3788 .2094 12° .2079 3179 .2126 3275 4.7046 6725 .9781 9904 78° 1.3614 .2269 1.3° .2250 3521 .2309 36.34 4.3315 6366 .9744 9887 77° 1.3439 .2443 14° .2419 3837 .2493 3968 4.0108 6032 .9703 98(59 76° 1.3265 .2618 1.5° .2.588 4130 .2679 4281 3.7321 5719 .9659 9849 75° 1 .3090 .2793 16° .2756 4403 .2867 4575 3.4874 5425 .9613 9828 74'- 1.2915 .2967 17° .2924 4659 ..3057 4853 3.2709 5147 .9563 9806 73° 1.2741 .3142 18° .3090 4900 .3249 5118 3.0777 4882 .9511 9782 72° 1.2566 .3316 19° .3256 5126 .3443 5370 2.9042 4030 .9455 9757 71° 1.2392 .3491 20° .3420 5341 .3640 5611 2.7475 4389 .9397 9730 70° 1.2217 .3665 21° .3584 5543 .38.39 2842 2.0051 4158 .9336 9702 69° 1.2043 .3840 22° .3746 5736 .4040 6064 2.4751 .3936 .9272 9672 68° 1.1868 .4014 23° .3907 5919 .4245 6279 2.3559 3721 .9205 9640 67° 1.1694 .4189 24° .4067 6093 .4452 6486 2.2460 3514 .9135 9607 66° 1.1519 .4363 25° .4226 6259 .4663 6687 2.1445 3313 .9063 9573 65° 1.1.345 .4538 26° .4384 6418 .4877 6882 2.0503 3118 .8988 9537 64° 1.1170 .4712 27° .4540 6570 .5095 7072 1.9626 2928 .8910 9499 63° 1.0996 .4887 28° .4695 6716 .5317 7257 1.8807 2743 .8829 9459 62° 1.0821 .6061 29° .4848 6856 .5543 7438 1.8040 2562 .8746 9418 61° 1.0647 .5236 30° .5000 6990 .5774 7614 1.7321 2.386 .8660 9375 60° 1.0472 .5411 31° .5150 7118 .6009 7788 1.6643 2212 .8572 9331 59° 1.0297 .5585 .32° .5299 7242 .6249 79.58 1.6003 2042 .8480 9284 58° 1.0123 .5760 33° .5446 7361 .6494 8125 1.5399 1875 .8387 92.36 57° .9948 .5934 34° .5592 7476 .6745 8290 1.4826 1710 .8290 9186 56° .9774 .6109 35° .5736 7586 .7002 8452 1.4281 1548 .8192 9134 55° .9599 .6283 36° .5878 7692 .7265 8613 1.3764 1387 .8090 9080 54° .9425 .6458 37° .6018 7795 .7536 8771 1.3270 1229 .7986 9023 53° .9250 .66.32 38° .6157 7893 .7813 8928 1.2799 1072 .7880 8965 52° .9076 .6807 39° .6293 7989 .8098 9084 1.2349 0916 .7771 8905 51° .8901 .6981 40° .6428 8081 .8391 9238 1 1918 0762 .7660 8843 50° .8727 .7156 41° .6561 8169 .8693 9392 1.1504 0608 .7547 8778 49° .8552 .7330 42° .6691 8255 .9004 9544 1.1106 0456 .7431 8711 48° .8378 .7505 43° .6820 8338 .9325 9697 1.0724 0303 .7314 8641 47° .8203 .7679 44° .6947 8418 .96.57 9848 1.0355 0152 .7193 8569 46° .8029 .7854 45° .7071 8495 1.0000 0000 1.0000 0000 .7071 8495 45° .7854 Value login Value logio Value logio Value logjo De- Radians Cosine Cotangent Tangent Sine grees TABLES Exponential Functions e^ e-^ e n e-^ X iogeX Value logio Value log-io X logeo; Value logio Value logj^o 0.0 — 00 1.000 0.000 1.000 0.000 2.0 0.693 7.389 0.869 0.135 9.131 0.1 -2.303 1.105 0.043 0.905 9.957 2.1 0.742 8.166 0.912 0.122 9.088 0.2 -1.610 1.221 0.087 0.819 9.913 2.2 0.788 9.025 0.955 0.111 9.045 0.3 -1.204 1.350 0.130 0.741 9.870 2.3 0.833 9.974 0.999 0.100 9.001 0.4 -0.916 j 1.492 0.174 0.670 9.826 2.4 0.875 11.02 1.023 0.091 8.958 0.5 -0.693' 1.649 0.217 0.607 9.783 2.5 0.916 12.18 1.086 0.082 8.914 0.6 -0.511 11.822 0.261 0.549 9.739 2.6 0.966 13.46 1.129 0.074 8.871 0.7 -0.357 2.014 0.304 0.497 9.696 2.7 0.993 14.88 1.173 0.067 8.827 0.8 -0.223 12.226 0.347 0.449 9.653 2.8 1.030 16.44 1.216 0.061 8.784 0.9 -0.105 2.460 0.391 0.407 9.609 2.9 1.065 18.17 1.259 0.055 8.741 1.0 0.000 2.718 0.434 0.368 9.566 3.0 1.099 20.09 1.303 0.050 8.697 1.1 0.095 i 3.004 0.478 0.333 9.522 3.5 1.253 33.12 1.520 0.030 8.480 1.2 0.182 3.320 0.521 0.301 9.479 4.0 1.386 54.60 1.737 0.018 8.263 1.3 0.262 3.669 0.565 0.273 9.435 4.5 1.504 90.02 1.954 0.011 8.046 1.4 0.336 4.055 0-608 0.247 9.392 5.0 1.609 148.4 2.171 0.007 7.829 1,5 0.405 4.482 0.651 0.223 9.349 6.0 1.792 403.4 2.606 0.002 7.394 1.6 0.470 4.953 0.695 0.202 9.305 7.0 1.946 1096.6 3.040 0.001 6.960 1.7 0.531 5.474 0.738 0.183 9.262 8.0 2.079 2981.0 3.474 0.000 6.526 1.8 0.588 6.050 0.782 0.165 9.218 9.0 2.197 8103.1 3.909 0.000 6.091 1.9 0.642 6.686 0.825 0.150 9.175 10.0 2.303 22026. 4.343 0.000 5.657 log.a; = (logiox) -^ ill ; M= .4342944819 PART I PLANE ANALYTIC GEOMETRY CHAPTER I SYSTEMS OF COORDINATES 1. The linear scale. Analytic geometry is based upon a geo- metric representation of numbers. Choose a straight line AB of indefinite length and upon it a fixed point 0. With a convenient unit lay off the equal distances 0P„ P,P„ PoP„ ... to the right, and 0Q„ Q,Q„ Q.Q^, •••to the left. We will now agree that the points Pi, Po, P3, ••• shall represent the positive integers 1, 2, 3, •••, respectively, and the points Qi, Q2, Qs, •■' shall represent the negative integers — 1, — 2, — 3, •••, respectively. -3 -2 -1 1 iH 2 3 1 1 1 ( 1 1 1 1 ^ •i Q, Q, Q, r, i? R P3 B Fig. 1 The intervals along the line AB can be divided into fractional parts of the unit, thus obtaining points representing fractional numbers. For example, the point B bisecting the segment P1P2 represents the positive number 1.5. The subdivision can be carried on indefinitely and we may infer, finally, that the following statement and its converse hold concerning this representation. Each of the points on the line AB represents a number ; namely, that number lohich expresses the distance and direction of the point from in terms of the unit chosen. Conversely, if x is a positive (or negative) number, it is repre- sented by a point x units to the right (or left) of 0. 7 8 SYSTEMS OF COORDINATES [Chap. I. The line AB, together with the points constructed as explained, is called a linear scale. It is the geometric equivalent, or graphic representation, of the system of real numbers. The point is called the origin ; it represents the number zero. The scale on a thermometer is an example of a linear scale. Here the points on the scale represent the numbers expressing degrees of temperature. EXERCISES 1. Construct a linear scale using half an inch for the unit. On this scale, mark the points representing 3, ^, — 2, — 2|. 2. Is the scale on an ordinary carpenter's square a linear scale ? Where is the origin ? 3. If the origin of the scale is moved two points to the left, how will this affect the numbers represented by the scale ? If the origin is moved two points to the right, how will the numbers be affected ? 4. The freezing and boiling points on a Fahrenheit thermometer are at 32° and 212° respectively, while on a centigrade thermometer they are placed at 0° and 100°. Compare the units of these two scales. Five degrees below zero on the centigrade is equivalent to what reading on the Fahrenheit ? Con- struct the two scales in this exercise. 2. Directed segments, directed angles. It is frequently necessary to distinguish be- tween the two directions in which a seg- ment may be laid off on a given straight line. This is done by calling one direction positive and the other negative. Thus, if ^ Fig. 2 4. n we agree to call the direction from A to B (Fig. 2) posi- tive, then we shall call the direction from jB to -4 negative. Expressed in symbols, \ ^x AB^- BA. Segments of a straight line to which a direction has been attached are called directed segments. In a similar way, an angle can be directed. For example, the acute angle Arts. 2,3] DIRECTED SEGMENTS, DIRECTED ANGLES 9 shown in Fig. 3 can be described by a line rotating with the hands of a clock, or clockwise, from OB to OA ; or counterclockwise, from OA to OB. It is customary to consider counterclockwise rotation as positive, and clockwise rotation as negative. Thus, the acute angle AOB is considered as a positive angle and the acute angle BOA, as a negative angle. In symbols, AOB = - BOA. 3. Addition of directed segments, addition of directed angles. If AB, BC, and AC are three directed segments along the same line (Fig. 4), then the equation AB + BC=AC Fig. 4 has the following simple interpretation ; namely, a point moving along the line from A to B and then from J5 to C is in the same final position as it would have been had it moved directly from A to C. With this interpretation, the above equation is readily seen to hold however the points A, B, and C are situated with respect to each other. Again, we have AC- AB = AC+ BA = BA + AC= BC. Similarly, the equation AOB + BOC= AOC (Fig. 5) is interpreted as follows ; rotation through the angle AOB and then through the angle BOC is equivalent to rotation through the angle AOC. Again, AOC - BOC = AOC + COB = AOB. Fig. 5 10 SYSTEMS OF COORDINATES [Chap. I. EXERCISES 1. Construct a linear scale, using half an inch for the unit, and mark the points A, five units to the right of the origin, and B, three units to the left of the origin. State ^the geometric meaning of OA — OB, of OB — OA, of OA + AB. What directed segment is equivalent to each ? What is the numerical value of each ? Is there a directed segment in the figure equiva- lent to 0A+ OB? 2. What is the difference in absolute temperature between — 5° Fahren- heit and 20"^ centigrade ? 3. Represent geometrically the difference in time between 10 a.m. and 3 P.M. as the difference between two directed angles. 4. In surveying, the azimuth of a line is its direction expressed in degrees, measured from the South point around towards the West, or clockwise. Thus the azimuth of a line running due North is 180° ; of a line running due East is 270° ; etc. What is the azimuth of a line running N 25° E ? Of a line running N 10° W? 5. What is the difference in azimuth between two lines, one running S 40° W and the other S 10° E ? 6. What is the difference in azimuth between two lines, one running S 40° E and the other N 25° E ? 4. Position of a point in a plane. When we have once chosen a unit of distance, one number is sufficient to locate a point on a line ; namely, the number expressing its distance and direction from a fixed point, the origin (Art. 1). It requires two numbers to locate a point in a plane and these numbers are called the coordinates of the point. The coordinates of a point may be chosen in many different ways. Any particular way of choosing them gives rise to a system of coordinates. There are two systems of coordinates in common use ; namely, cartesian coordinates, named after Rene Descartes, who first used this system (1637), and polar coordi- nates. These systems of coordinates will be explained in the succeeding articles. 5. Cartesian coordinates. Choose two linear scales OX and OY (Fig. 6) with their origins coinciding at 0. Through any point P in the plane XOY draw parallels to OX and OY, meet- ing OY and OX in E and D, respectively. The numbers repre- Arts. 4-6] RECTANGULAR COORDINATES 11 sented by D and E on their respective scales are the cartesian coordinates of P. By article 1, these coordinates express the dis- tances and directions oi D and E from in terms of the unit chosen ; or the distances and directions of P from Y and OX measured along par- allels to OX and OT, re- spectively. Thus, if X and y denote the numbers repre- sented by the points D and E, respectively, the coordi- nates of P are x=OD = EP and y = OE — DP. Fig. 6 In the same Avay, the coordinates of P^ are a-i = OA = E,P^ and y, = OE, = AA- Hence any point in the plane XO T has an aj-coordinate and a ^/-coordinate represented by points on the scales OX and OY, respectively. Conversely, any two numbers x and y serve to locate a point in the plane. For, let D and E be the points representing x and y upon their respective scales. Through D draw a line parallel to OY, and through E a line parallel to OX. These parallels meet in a single point P ; the point whose coordinates are x and y. The scales OX and OY are called the coordinate axes, or sim- ply the axes. OX is called the X-axis, and OY the F-axis. The segment OD is often called the abscissa of the point P; and the segment DP the ordinate of P. The units of distance used in constructing the two linear scales OX and Y are usually taken to be the same, but it is not neces- sary to take them so. In many cases it is more convenient to use different units. 6. Rectangular coordinates. The coordinate axes may intersect at any angle, but it is generally simpler to take them perpendicu- 12 SYSTEMS OF COORDINATES [Chap. I. lar to each other. In this case, cartesian coordinates are called rectangular coordinates. In rectangular coordinates, the axes divide the plane into four quadrants named first, second, third, and fourth quadrant as indicated in Fisr. 7. II. Quadrant + 3 III. Quadrant I. Quadrant rV. Quadrant Fig. 7 The algebraic signs of the coordinates of any point depend upon the quadrant in which the point lies. Thus, Points in Qfadea.vt a; y I + + II — + III - — IV + — Conversely, the algebraic signs of the coordinates determine the quadrant in which the point lies. For example, if x=—6 and ?/=3, the point lies in the second quadrant as indicated in the figure. 7. Notation. The notation P = (a, &), or P(a, &), indicates that the coordinates of the point P are x = a and y = b. The x'-coordinate is always written first. For example, to indi- cate the position of the point P in Fig. 7, we write P=(— 5, 3), or P(- 5, 3). Akts. 7, POLAR COORDINATES 13 EXERCISES 1. Draw the axes OX, OY and locate the following points: (|, 3), (2, -I), (0,5), (5,0). 2. Where are the points located for which x = ? For which x = 1 ? For which y = 0? For which y =— 1? 3. By means of a geometrical construction, locate accurately the points ( V2, 3), (VI, V2), (\/5, a/G). Can the point (0, w) be located accurately ? 4. The axes OX, OF are perpendicular to each other; locate the points Pi = (l, 2), P2 = (5, 5), and P3=(5, 2). Find the lengths of the sides of the triangle P1P2P3. 5. Let the axes OX, OY make an angle of 60 degrees with each other ; plot the points in the preceding exercise and find the lengths of the sides of the triangle FiPoPs- 6. "With rectangular coordinates, show that the points (2, 3), (2, — 1), (—2, — 1), and (—2, 3) form a rectangle. Find the lengths of the sides, the lengths of the diagonals, and the area of this rectangle. 7. With rectangular coordinates, show that the points (1, 1), (3, 1), and (2, 2) form an isosceles triangle which is half a square. Find the coordi- nates of the fourth vertex, the lengths of the sides, the lengths of the diago- nals, and the area of the square. 8. Polar coordinates. The position of a point P in a plane is also determined by its distance ?*, in terms of a given unit of dis- tance, from a fixed point 0, called tlie origin or pole, and the angle $ which OP makes with the positive direction of a fixed linear scale OX, called the initial line or axis (Fig. 8). OP is called the radius vector, and the angle XOP the vec- torial angle, or simply the angle, r and 6 are the polar coordinates of P. The nota- tion P = (r, 9), or P(r, 6), means that r and 6 are the polar coordi- nates of P. A given number r and a given angle determine uniquely the position of a point in a plane with reference to a fixed origin and initial line. For, imagine the initial line OX (Fig. 9) to be rotated Fig. 8 14 SYSTEMS OF COORDINATES [Chap. I. through the given angle $ about into the position OX'. On OX' mark the point which represents the given number r. There is but one such point. Por example, the point P(-~ 5, — 30°) is obtained by rotating OX through the angle — 30° and marking the point 5 units from on the negative end of the scale OX'. ^\. Fig. 9 On the other hand, a given point has many sets of polar coor- dinates. For example, the point P in Fig. 9 is (—5, —30°), (5, 150°), (-5, 330°), (5, -210°), etc. It is always possible, however, and usually most convenient, to choose the polar coor- dinates of a point so that the radius vector shall be a positive number, and < 0^2 it. 9. Relation between rectangular coordinates and polar coordinates. In Fig. 10, let be the origin and OX the initial line, so that the polar coordinates of any point as P are : r = OF and XOP. Fig. 10 Let OX and OF be the X- and y-axes, respec- tively, so that the rec- tangular coordinates of P are X = OB and y = DP. Now, wherever the point P may be located in the plane, we always have oc = r cos 6 and ?/ = r sin 6. (1) Art. 9] RELATION BETWEEN COORDINATES 15 From equations (1), by squaring and adding, we obtain a;2 + 2/' = r\ (2) Also from equations (1), we have = arc cos - = arc sin ^ • (3) r r From (2) and (3), r = ± Vx2 + ?/2 and 8 = arc cos . = arc sin -^ ± Vic2 + 2/2 ^ y/^2 + yt = arc tan — . (4) Equations (1) serve to change the polar coordinates of a point into rectangular coordinates ; and equations (4) are used to change rectangular coordinates into polar coordinates. For example, the rectangular coordinates of the point P(—5, — 30°) are a; = - 5cos(- 30°)=~^^^y = -5sin (- 30°) = |- (See Fig. 9.) Again, the polar coordinates of the point P(— 3, — 4) are r = V9 -I- 16 = 5, ^ = arc cos (- f)= arc sin (- 4) = 233° 8'. In solving this problem, r was taken to be the positive square root of 25. With r = — 5, 6 = arc cos (|) = arc sin (i ) = 53° 8'. EXERCISES 1. Plot the points (.3, -30°), ^-4, ^V (3, 2 radians). Find the rectangular coordinates of these points. 2. Find the polar coordinates of the points whose rectangular coordi- nates are (3, — 7), (4, 2), (- 3, — 5). Plot the points. 3. Where do the points lie for which the radius vector is constant? For which the vectorial angle is constant ? 4. If a; = 4 and r = 5, find y and 0. Is there more than one point satis- fying the given conditions ? 16 SYSTEMS OF COORDINATES [Chap. I. 5. With a centigrade scale on the X-axis and a Fahrenheit scale on the F-axis, plot a number of points whose coordinates represent the same abso- lute temperature. For example (0, 32), (5, 41), etc. Try to show that all these points must lie on a straight hne, and to find where this straight line meets the X-axis. 6. Plot a number of points for which the radius vector is twice the abscissa. Join the points plotted. Do they lie on a straight line ? 7. Elevations of points on the ground above a fixed datum plane are sometimes expressed in meters, while the distances of these points from a given place of beginning may be expressed in feet. Plot the points whose elevations and distances are as follows : Distance Elevation 100 feet 3.2 meters 200 feet 6.0 meters 250 feet 8.0 meters 300 feet 7.0 meters 400 feet 5.0 meters 500 feet 3.0 meters 600 feet -2.0 meters A broken line drawn through the points thus determined is called a profile. 8. Reduce the elevations of exercise 7 to feet and plot the same profile. 9. From a point 0, the azimuths and distances to three points A, B, and C ai'e as follows : Azimuth Distance A B C 120° 180° 240° 10 rds. 15 rds. 12 rds. Make an accurate map of the triangle ABC. With as origin and OB as F-axis, compute the rectangular coordinates of A, B, and C. With as pole and OB as initial line, compute the polar coordinates of A, B, and C. 10. Construct a scale on the X-axis, the unit of measure representing one foot ; and a scale on the F-axis, the unit of measure representing one meter. Take one meter equivalent to 3.28 feet. Plot a number of points whose coordinates represent the same distance, for example (3.28, 1), (6.56, 2), etc. Show that these points lie on a sti'aight line passing through the origin. Art. 9] RELATION BETWEEN COORDINATES 17 11. Construct a scale on the X-axis representing British money, and a scale on the I'-axis representing American money. Take £ 1 equivalent to §4.87. Plot a number of points whose coordinates represent the same value, for example (1, 4.87), (2, 9.74), etc. Show that these points lie on a straight line passing through the origin. 12. Plot the points whose polar coordinates are (—6, 20^), ( — 5, —315°), [ — 4, - j , ( — 3, — ). Change the coordinates of these points so that r and e shall be positive, and d less than 3G0°. 13. Change the polar coordinates of the points in exercise 12 to rectan- gular coordinates. 14. With a convenient unit, mark the points U and B on the X-axis, representing the numbers 1 and &, respectively. On the T-axis, mark a point A, representing the number a. Join A to Z7, and through B draw a parallel to A U, meeting the F-axis in C. Prove that C represents the number ab. 15. With a convenient unit, mark the point U, on the X-axis, represent- ing the number 1 ; and on the T-axis the points B and A, representing the numbers b and a, respectively. Join B to U, and through ^4 draw a parallel to BU, meeting the X-axis in the point C. Show that G represents the number — . b 16. With a convenient unit, mark the points U and A on the X-axis, representing the numbers — 1 and a, respectively (a being a positive number). On UA as diameter, draw a circle and prove that it meets the F-axis in points representing the numbers ± Va. In this way construct geometrically \/2, V3, V5. CHAPTER II DIRECTED SEGMENTS AND AREAS OF PLANE FIGURES 10. Projections upon the coordinate axes. Let PjP^ be any directed segment. Through Pi(xi, y^) and P-ii^^, y^) draw parallels to the axes as shown in Fig. 11. The segments D^Do and E^E^, thus determined upon the axes, are called the projections of P1P2 upon the X-axis and upon the y-axis, re- spectively. The pro- jections themselves are directed seg- ments, and therefore (Art. 3) Fig. 11 j>iD2 = D^o + oi>2 = - oi>i + on.2 E^E.^ = E^O+ OE^ = - OE^ + OE.2 2/2 - Vl' (1) If we call Pi the initial point and P^ the terminal point, the projection of P1P2 upon the X-axis is found by subtracting the x'-coordinate of its initial point from the x-coordinate of its terminal point. Similarly, the projection upon the T'-axis is found by subtracting the y-coordinate of the initial point from the ^/-coordinate of the terminal point. 11. Inclination and slope of a directed segment. Let the coordi- nate axes be rectangular. Through the initial point of a directed segment draw a line parallel to the X-axis, having its positive direction the same as that axis. The line P^D (Fig. 12) is this parallel. The positive angle through which it is necessary to rotate this parallel to make it coincide with the given directed segment is the inclination of the segment. Thus the angle DP^P^ 18 Art. 11] INCLINATION AND SLOPE 19 (a) ^D ^X (h) is the inclination of each of the directed segments in Fig. -12 The inclination may have any value from 0° to 360° inclusive. The tangent of the inclination is called the slope of J'2 the directed seg- y/^ ment. Through P,, ^ '-P3 | ^ ^x the terminal point of the segment, draw a line paral- lel to the r^axis and let it meet the parallel to the X- axis in P3. The tangent of the angle DP1P2 is then — ^^. But P3P2 is equivalent to the projection of P^P^ upon the F-axis and P1P3 is equivalent to the projection upon the X-axis. Hence, ?/2 - 2/1 >X Fig. 12 slope of F^Po = tan DPiP^ OCa — OC't (2) 2 ~ •* 1 For example, the slope of the segment joining (—4, —2) to (2, 5) is 5 -(-2 ) ^7 2 -(-4) 6- Although reversing the direction of a segment changes its inclination by 180°, it does not change its slope. For, slope of PoPi =^^^^^= slope of P.P^. EXERCISES 1. Determine the projections, tlie inclination, and the slope of each of the following directed segments : («) (-2,4), (3,6); (6) (-5,7), (-4, -2); (c) (3, -2), (5,6); (d) (-3,2), (-2, -3). Draw each segment. 20 DIRECTED SEGMENTS AND AREAS [Chap. II. 2. If the coordinate axes make an angle of 60° with each other, determine the angle which the directed segment (2, 1), (4, 2) makes with each axis. 3. Draw the triangle whose vertices are (1,2), (5,4), (2,6), usino- rectangular coordinates. (ff) Find the lengths of the projections of the sides upon the ^-axis. What is the sum of these projections ? (6) Find the inclination of each side. How can the angles of the triangle be found from these inclinations ? 4. Show that the sum of the projections of the sides of any triangle upon either axis is zero, provided that the sides be taken in order around the triangle. 5. Fig. a represents a railroad cutting in a side- hill. The slope of the natural surface is 1 : 4 and that of the proposed cut- ting is 1:2. At what heights above the bottom of the cut and at what dis- tances out from the center line are the points of inter- section a and b ? 6. Fig. b is the outline of a roof truss of 80-f t. span and 20-ft. rise. The spaces ab, be, etc., are equal and the members bf, eg, and dh are perpendicular to the member ae. Calculate the slopes of ae, bf, fc, and ge with respect to a horizontal axis ak. 7. Calculate the slopes of cf and ch with respect to the line ae taken as the horizontal axis. 12. The length of a segment. The problem to find the dis- tance between two points whose coordinates are given, that is, the length of the segment join- ing them, depends upon the problem of finding the length of one side of a triangle when the other two sides and their in- cluded angle are given. Thus, with cartesian coordinates, let Fig. 13 Aet. 12] THE LENGTH OF A SEGMENT 21 -Pi(^'i) yi)i ^2(^2? 2/2) be the given points, and let the angle XOY be w (Fig. 13). Draw parallels to the axes through Pj and P^ forming the triangle PyP^Pf The sides P^P^ and P ^i- When 62 < 9y, the angle which Q1Q2 makes with P1P2 is given by the formula as the student may easily verify. In either case tan ^9 — tan 6, tan .Y (a) Fig. 16 ->X (h) (Art. 11). For example, the segment joining (1, 2) to (—2, —3) is parallel to the segment joining (2, —1) to (o, 4),. since the slope of each is f . 15. Perpendicular segments. When two segments are perpen- dicular to each other, their inclinations differ by an odd multiple of 90° and therefore, in every case, 1 tan 62 = — cot 61 = tan ^1' 'tn.^ = — in^ (1) Thus, the slope of each segment is the negative reciprocal of the slope of the other. Conversely, if the jjroduct of the slopes of two segments is — 1, the segments are perpendicular to each other. For then the tangent of the inclination of one of them is equal to the negative of the cotangent of the inclination of the other. Hence, their inclina- tions differ by an odd multiple of 90° ; that is, the segments are perpendicular to each other. In Fig. 17, PjPo niakes an angle of 90° with Q1Q2, but Q1Q2 makes an angle of 270° with r f. Q.T^ y 1 s-H k ^«, ^ -v Fig. 17 24 DIRECTED SEGMENTS AND AREAS [Chap. II. EXERCISES 1. Find the angle which the segment (—3, 2), (4, — 1) makes with the segment ( — 3, 2), (8, 5). Draw the figure. 2. Compute the lengths of the sides and the angles of the triangle whose vertices are (—3, 2), (4, — 1), and (8, 5). Draw the figure. 3. Show that the triangle whose vertices are (3, 2), (0, 3.5), and (1, 5.5) is right-angled. 4. Show that the segments (—3, 5), (3, 2) and (—1, 6), (3, 4) are parallel. Draw the figure and compute the perpendicular distance between the segments. 5. Join the extremities of the segments in the preceding exercise and compute the area of the quadrilateral so formed. 6. Draw the diagonals of the quadrilateral in the preceding exercise and find the acute angle which one makes with the other. J- ^2 E ^ r^\ p ^--^ Pi 1 \ i ) 1 \ Fig. 18 and whence oc = OD = OD^ + J>iD = xj + y = OE = OE^ + E^E = y^ + 16. Point bisecting a given segment. Let Pi = (a^i, y^ and Po = {x^, yz) ; it is required to find the coordinates of the point P=(x, y) bisecting the segment PiPo (Fig. 18). The parallels to the axes through P must bisect the projections AA and AA in D and E, respectively. Hence (Art. 10) 2 ' 2/2 - Pi OCi + ijC2 2/i + 2/2 (1) For example, the coordinates of the point bisecting the segment (1,3), (-3, -l)are 2 = -1, y = = 1. Art. 17] POINT DIVIDING A GIVEN SEGMENT 25 17. Point dividing a given segment in a given ratio. The results of the preceding article can be generalized. Thus, suppose the point P (Fig. 19) divides the segment P1P2 so that PP. Then the points D and E divide the projections of PiPo in the same ratio. Hence, P,P^D,D^ {x-x,) ^^, PP^ DD, {x,-x) ' Y P2 ^^ i: p ^^-^ ^: A B, D J). x and PR E,E_ (y~y,) ^^, _ EE, {y, ~ y) Solving these equations for x and y, we have y^ (1) For example, to find the coordinates of the point dividing the segment Pi = (2, 4), P, = (- 3, - 2) in the ratio 2 : 3, we have r = f . Substituting in the above formulas we find « = and y = \. Hence the point (0, f) divides the given segment in the ratio 2 : 3. EXERCISES 1. Find the coordinates of the points which bisect the sides of the tri- angle (2, 5), (-2,2), (4, -5). 2. In the preceding exercise, join the vertices to the mid-points of the sides opposite and show that the points dividing each segment from vertex to opposite side in the ratio 2 : 1 coincide. 3. Generalize the preceding exercise and thus prove that the medians of any triangle meet in a point. 4. Show that the points (2, 3), (4, 1), (8, 2) and (6, 4) form a parallelo- gram. Find the coordinates of the mid-points of the diagonals. 5. Find the coordinates of the points which trisect the segment (6, 4), (-3,1). 26 DIRECTED SEGMENTS AND AREAS [Chap. II. 6. Find the coordinates of the point P dividing the segment Pi = (.3, 4), Fi = (— 2, — 6) in the ratio 3 : 5. Prove the result by calculating the lengths of the segments PiP, PP-^ and showing that their ratio is |. 7. The segment in the preceding ex- ercise crosses both axes. Find the co- ordinates of the points of crossing. 1 8. Area of a triangle, one vertex at the origin. To find the area of the triangle OP^P., (Fig. 20), let ^1= ('^'i, Vi), P2^{^'2, Vi), and A A, the projection of P1P2 upon the a;-axis. Then, if A represents the required area. Fig. 20 A = trapezoid P^P^D^D^ + triangle P^OD^ — triangle OD^P^ Ml 2 ^ (Vi + JhX^i - -^2) I X2y2 2 2 Hence, we have (^1^2 - ^2^1) (1) The expression x^^y^ — %Vi is a determinant and is often written thus; X, y^ In determinant notation, the formula for the area of the triangle is then 2/1 ^2 V'l (2) For example, the area of the triangle formed by joining the extremities of the segment Pj = (3, 1), P2 = (1? 3) to the origin is 3 1 1 3 = 4. 19. Sign of the expression {oc\V^2 — *2?/i)- The sign of the expression {x-^y^ — x^yi) is not the same for all positions of the segment P1P2. Thus, if Pj = (3, 1) and P^ = (1, 3), the expres- sion has the value -\- 8, Avhile for the segment P^ = (1, — 2), Art. 19] SIGN OF THE EXPRESSION ixiy2 - 2/2X1) 27 P2 = ( — 1, 1), which has the same length and the same slope as the "former, the expression (xiy2 — x^yy) has the value — 1. Changing to polar coordinates by means of the relations Xi =■ i\ cos 61, 2/1= r^ sin ^1, x^ = r^ cos 02, y-i = r^, sin B^ (Art. 9), the expression {x-^y^ — ^^\) becomes ri9-2(cos di sin 62 — cos 62 sin ^1) = r{i\ sin (82— Oi). Since r^ and 7*2 may be considered as positive numbers (Art. 8), the sign of (x^y2 — x^yi) will be positive when sin (^2 — ^1) is posi- FiG. 21 tive ; that is, when 62 — &i is an angle in the first, or the second, quadrant. In either case, the segment PiP^ has a position such that, in passing from Pi to P2> the origin lies to the left as at (a). Fig 21. On the other hand, the expression {x{y2 — x^^ will be negative when ^2 — ^1 is ^"^ angle in the third, or the fourth, quadrant ; and then the segment P1P2 has a position such that, in passing from Pi to P2, the origin lies to the right as at (&). Conversely, if the segment P1P2 has a position such that the origin lies to the left (or the right) when the segment is traversed from Pi to P2, the sign of {Xyy2 — X2yi) will be positive (or nega- tive). For then the angle O2 — &i must lie in the first, or the second, quadrant (or in the third, or the fourth, quadrant). Con- 28 DIRECTED SEGMENTS AND AREAS [Chap. II. sequently the area of the triangle OPyP^, which is \ (.x*i?/2 — ^iVx), is positive when the origin lies to the left, as at (a), Fig. 21, and negative when the origin lies to the right, as at (h). EXERCISES 1. P\ = (5, 3) and P2 = (— 1, — 3) ; determine the area of OP1P2, being the origin. Explain the sign of the result. Draw the figure. 2. If is the pole, show that the area of the triangle OP1P2 is 1 rir2 sin (6I2 - di)-, where Pi = (n, Oi) and P2= (r2, dt). 3. If Pi =[5, -"j and P2= (3, - 30°), find the area of OP1P2. 4. Given Pi = (3, - 60°) and P2= (3, 4), find the area of OP1P2. 5. When the segment P1P2 passes through the origin, what is the value of the expression. (;i-i2/2 — X2J/1) ? 6. If Pi = (— 3, 1) and P2 = (l, —2), in which quadrant is the angle ^2 — ^1 ? Draw the figure and find the area of OP1P2. Fig. 22 20. Area of a triangle, vertices in any position. Join the ver- tices of the triangle to the origin 0. Let P^, P^, P^ (Fig. 22) be the vertices, taken in counterclockwise order about the triangle. The area of P^P^P^ is then given by the formula area PiP-zP^ = area OP1P2 + area OPiP^ + area OP3P1. (1) Art. 20] AREA OF A TRIANGLE 29 Thus in (a), each of the component triangles has a positive area, by the preceding article, and their sum is obviously the area of F^P^P^. In (&), the areas of OP^Pz and OP^P^ are positive num- bers, while the area of OP^Pi is a negative number. The alge- bl-aic sum of these numbers is clearly the area of P^^P^Pz- Finally, in (c), the axea of OPxP-i is a positive number, while the areas of the remaining two triangles are expressed by negative numbers. The algebraic sum of these numbers is again the area of P^P^P^. Replacing the area of each component triangle in (1) by its value in terms of the coordinates of the vertices (Art. 18), we have area PiP->Ps=l[(^iU-2-'X-2Ui) + (JC'iUi--^s!/i) + (xsUi-^iUs}]- (2) The area can be expressed in determinant notation. Thus (3) \^l 2/1 1 area jPi PoPs =-\^2 Vi 1 \X'A 2/3 1 since, if the determinant is expanded, the result agrees with formula (2). The following is a convenient rule for computing the area. Let P1P2P3 be the vertices, taken in counterclockwise order about the triangle. Arrange the coordinates in rows, thus Vv 2/2 Vz Vi multij)ly each x by the y standing in the next column to the right and add the products, thus ^iVi + X2I/3 + -Ml ; multiply each y by the x in the next column to the right and add the products, thus y^Xo + ?/o>-3 + ^/3.^•l ; subtract the latter sum from the former and take hcdf the difference, the residt is the area of the triangle PjPo-fs- For example, to find the area of the triangle whose vertices are Pi = (- 1, 3), P2=(3, 2), P3 = (5, 4) (Fig. 23), arrange the coordinates as 80 DIRECTED SEGMENTS AND AREAS [Chap. II. Fig. 23 J_ _ ___^ . ,^r^^ . -i£ ^ 1 . 1 in the rule, being careful to note that the vertices are taken in counterclockwise order ; thus - 1 3 5 - 1 3 2 4 3. The area is then 1 [(_ 2 -f 12 + 15) _(9 + 10-4)]=5. If the vertices are taken in clockwise order about the tri- angle, the result obtained by using formula (2) or formula (3) or the rule just stated will be numerically the same but will be negative in sign, as the student may easily verify. 6, ^ ' 3 6, Draw the figure. EXERCISES 1. Find the area of the triangle whose vertices are (2, — 6), (— 9, 7), (8, 3). 2. Find the area of the triangle whose vertices are (—1, —2), (2, 1), and (3, 2). Explain the result. 3. Find the area of the triangle whose vertices in polar coordinates are '¥)• ^"^ {'' T 4. The vertices of a quadrilateral are (— 1, 6), (8, 10), (10, —2), and (—5, — 8). Compute the area of the quadrilateral by dividing it into two triangles. Draw the figure. 5. When three points are in the same straight line, they are said to be collinear. Show that the points (1, 3), (3, 1), and (4, 0) are coUinear. 6. Where will the line joining the points (2, 5) and (3, 6) meet the axes ? 7. If Pi{xi, 2/i), P2{x2, 2/2), and P3(xs, 2/3) are three collinear points, show that = 0. State and prove the converse. 21. Area of any polygon. Formula (2) of article 20 can be extended to find the area of any polygon when the coordinates of the vertices are given. Thus when the vertices, taken in counter- clockwise order about the polygon, are joined to the origin, a Xl 111 1 X2 2/1 1 X3 2/3 1 Art. 21] AREA OF ANY POLYGON 31 1 ']Z ^ ; - "i^ ^' 2 ~ S ^' ^ S^ ..o,^Z__7 . _ S,. _ . iSiS ^l. 5 2 g " S _ . ^Z ^ j' _ - S^- , / '' S C-. 5 ■' ? ^ ^?: : :=*£ 2 ^f-. ^t. . _ ,.-! 4 Z ^" ^--^ Z 2 ,' ,^i!.. _" " t^ , — ;--: __,^',' , -- _ V ■' -^' ; iziiZ": : i_ ^ ' ~^f-3 ^t- Jia '^^ X - - - s -.4- ^ 1 .i __ _ X _ ' ... ± - _. Fig. 24 number of component triangles are formed (Fig. 24). It is geo- metrically evident that the algebraic sum of the areas of these tri- angles is the area of the polygon. A con- venient rule for com- puting the area of a polygon is, therefore, obtained by extending the rule in Art. 20. Thus, write the x's over the ?/'s and form the cross-products : 1 2 3 **^4 * " • to *^'1 ) Vl 2/2 ^3 2/4 ••• Vn VV The required area is then (1) For example, to find the area of the quadrilateral whose vertices are, in counterclockwise order (8, 10), (-1, 6), (-5, -8), and (10, -2), we have 8 _ 1 - .5 10 8 10 6 -8 -2 10 and the area is, therefore, 1 [(48 + 8 + 10 + 100) - ( - 10 - 30 - 80 - 16)] = 151. EXERCISES 1. The vertices of a hexagon are (6, 1), (.3, —10), (—3, —5), (—12, 0), (— 4, 6), and (9, — 4). Draw the hexagon and compute its area. 2. A surveyor finds that the corners of a four-sided field are situated, with respect to a north and south road and an east and west road, as follows: ^ = (25, 32), £=(48, 65), C=(94, -10), and D = (30, -40). Distances are measured in rods. Make a map of the field and compute the number of acres it contains. (160 square rods = 1 acre). 3. From a point in a quadrangular field, the distances and directions to the corners are as follows : J[= 120 feet, N. 65° E. ; J5 = 216 feet, N. 32° W. C= 320 feet, S. 74° W. ; 2) = 65 feet, S. 23° E. Make a map of the field and compute its area. 32 DIRECTED SEGMENTS AND AREAS [Chap. IL 4. In surveying, points are frequently located by azimuth and distance from a given point, tliat is, by polar coordinates. It frequently becomes necessary to plot the outlines of tracts of land determined in this way and to calculate areas. Dravs^ the polygonal figure whose vertices are determined by the following azimuths and distances : AZLMUTH DiSTAXCE Azimuth Distance 125° 115 feet 342° 175 feet 170° 160 feet 15° 40 feet 250° 200 feet 73° 10 feet (a) Calculate the coordinates of the vertices of this figure referred to N. and S. and E. and W. lines through the given fixed point, as origin. (b) Calculate the directions of the sides of this figure. (c) Calculate the area of the polygon. 5. Fig. 24 A represents a cross section of one side of a railroad cutting. Calculate the area of this section, using coordinates as shown. In railroad field books the data for this problem would generally be recorded as follows : Center \ i2\,.--'9\/'ii\/^\/'o ; ] /\5/\l2/\20/'\8 \ The ordinate, or depth of cutting, is writ- ten above and the distance out from the center line (abscissa) is written below. The coordinates (|) are not actually re- corded as the number 8 is the fixed width of the bottom of the cut. Arranging the coordinates in the above manner, the correct result is obtained by taking positive products along diagonal lines sloping downwards towards the right (shown by full lines) and negative products along the other diagonals (shown by dotted lines) . 6. Compute the area of cross section given by the following cross section notes, left and right of the center line : Center fO\ _4 _6_ a 1_0 J_Q 12 18. 16 (0\ IvS/ 1? iO 5 '0 5 I'S ^5 32 \S/ What side slope of the finished cut has been assumed in this problem ? 7. Drop perpendiculars from the vertices of a polygon upon the X-axis, as in Fig. 24. Show that the area of the polygon is the algebraic sum of the areas of the trapezoids thus formed. Compute the area of the hexagon in exercise 1 by this method. Fig. 24 a CHAPTER III FUNCTIONS AND THEIR GRAPHIC REPRESENTATION 22. Constants and variables. The numbers and magnitudes considered in mathematics are either constants or variables. The coordinates of a fixed point are constants ; the coordinates of a moving point are variables. 23. Functions. If to each given value of a variable x there corre- spond one or more values of a variable y, then y is called a function of X. As examples, the cost of a money order is a function of the amount; the temperature at a given place is a function of the time; the cost of insurance is a function of the age of the in- sured ; the distance a body falls freely in space is a function of the time the body has been falling. 24. Notation. To denote that ?/ is a function of x, the notation y =f{x) (read y equals / of x) is used. When several functions are to be considered in the same prob- lem, different symbols are used. Thus, y=f(x), y=f-,(x), . . . (read y equals f^ of x, y equals fo of x, . . .). Or use is made of Greek letters, as y = <^(.v), y = ^(x), • • • (read y equals phi of X, y equals psi of x, . . .). 25. Determination of functional correspondence. A functional correspondence can be- established, or set up, between two variables in different ways. Thus, the correspondence may be primarily established : I. By an equation connecting the two variables, as y = x^; II. By a table exhibiting corresponding values of the variables, as a table of logarithms ; III. By a curve drawn automatically, thus exhibiting graphi- cally the correspondence between two variables. 33 34 GRAPHIC REPRESENTATION [Chap. III. 26. Dependent and independent variables. If functional corre- spondence is established by au equation, the value (or values) of the function can, in general be readily compiited for any value assigned to the variable x. Thus, for example, if y = 2a^, the value of y is easily computed for any assigned value of x. In general, y (the function) is called the dependent variable, and x, the independent variable. 27. Graphic representation. A table of corresponding values of a function and the independent variable can be derived from the equation by assigning to the independent variable a series of values, arbitrarily chosen, and computing the corresponding values of the function. With these corresponding values of x and y as rectangular coordinates, a series of points can be con- structed. The ordinates of these points, taken together, form a graphic representation of the function. For the functions con- sidered in this book, a curve can be drawn through all the points constructed as above. This curve is called the graph of the function. Thus, from the equation y = 2x^, we obtain the following table of corresponding values : x = -3, -2, -1, 0, 1, 2, 3, 4, y= 18, 8, 2, 0, 2, 8, 18, 32, The process of constructing the points whose coordinates are given in the table and drawing the curve through them, is called plotting. Figure 25 shows the completed graph. In constructing this graph, the unit of the scale on the X-axis was taken four times as great as the unit on the Y'-axis in order to represent more of the curve within a small compass (cf. Art. 5). The curve shows at a glance the change in value of the function for any given change in value of the independent variable x. For exam- ple, as X changes from — 3 to -f 3, the point which represents it moves from D^ to D^. At the same time the function y first decreases from 18 to and then increases from to 18. When a function ceases to decrease and begins to increase, or vice versa, it is said to have a turning point. Thus, the function y = 2 x^ has a turning point at the origin. Art. 27] GRAPHIC REPRESENTATION 35 When a function has no turning points, it is called a monotone function. It is important to know whether a function has turning points or not, and, if it has, to know for what values of the independent Fig. 25 variable they exist. At a turning point, the function has a max- imum or a minimum value. EXERCISES 1. Draw the graph of the function y = 2 x + 3. Find the coordinates of the points where the graph crosses the axes. Does the function have turn- ing points, or is it a monotone function ? 2. Draw the graph of the function expressing the law of falling bodies, s = l gt^. Take g = 32 and corresponding values of s and t as ordinates and abscissas respectively. Make the unit of the scale on the f-axis ten times as great as the unit on the s-axis. Where is the turning point of the function s ? 3. Draw the graph of the equation expressing Boyle's law, pv = k. Take A; = 4 and corresponding values of p and v as ordinates and abscissas respec- tively. Make the units the same on each axis. Is ^j a monotone function of V or not ? 4. Make careful drawings of the graphs of the function ?/ = a:" for n=—1, 0, 1,2, and 3. Use the same axes and preserve the figures. For which of the given values of n is y not a monotone function of a; ? 36 GRAPHIC REPRESENTATION [Chap. III. 5. Draw the graph of the function y = 4: — x^. Determine the value of X for which this function has a turning point. Has the function a maximum or a minimum value at the turning point ? 6. Draw the graph of the function y = x^ — ix + S. Determine the coor- dinates of the turning point. Has the function a maximum or a minimum value at the turning point ? 28. Single-valued and multiple-valued functions. When there corresponds but one value of the function to each given value of the independent variable, the function is called single-valued. If there is more than one value of the function corresponding to any given value of the variable, the function is called multiple-valued. For example, the function y = 2 x^ is single-valued ; but the func- tion y^ = 2x is multiple-valued, since to each value of x there correspond two values of ?/. The following is a table of corresponding values for the func- tion y^ = 2x. x^ -2, -1, 0, 1, 2, 3, 4, .... 2/ = Imag., Imag., 0, ±V2, ±2, ± a/6, ± V8, ..-. _ - - _p TpL " " it II III 1 — — " " -r''"^ ^-- ^ ^^ ^n. ;? ^ ^ 7 z ~ ~ • 1 " " 1 '.T ! ' 1 ■' i ^ <\ ' T 1 ! 1 X ^ -_ j^ - S. _ ± . . ■*,, _ _■=; " _ ^ _ _ - ^,^ ^ S j^ ^.■.... ■* .*J^ """■.- ' Fig. 26 Arts. 28, 29] SYMMETRY 37 The graph is shown in Fig. 26, where the same unit is used for the scale on the I'-axis as for the scale on the X-axis. The curves in Figs. 25 and 26 are called parabolas. 29. Symmetry. A curve, or graph, is symmetrical -with respect to a straight line when the line bisects all the chords of the curve drawn perpendicidar to it. For example, the parabola shown in Fig. 25 is symmetrical with res]3ect to the I^axis and the j)arabola in Fig. 26 is sym- metrical with respect to the X-axis. As another example, consider the single-valued function 7/ = 5 .i* — 6 — or. (1) The following is a table of corresponding values : x=. 0, 1, 2, I, 3, 4, 5, .... y = -Q, -2, 0, i 0, -2, -6, .... The graph is shown in Fig. 27, where the units on the two axes are the same. We now see that the curve is symmetrical with Fig. 27 38 GRAPHIC REPRESENTATION [Chap. III. respect to a line parallel to the F-axis and passing through the point (f, 0). The symmetry is also shown, without plotting, by solving equation (1) for x. Thus, x = ^± Vi-y, (2) and therefore, for a given value of 7j, x has two values represented by points equidistant from, and on opposite sides of, the point cc = f . We thus see that the line parallel to the F-axis through the point X = ^ bisects the chords of the curve drawn perpendicular to it. Notice that the function has a turning point at x = ^, that the value of y is there equal to \, and that this is the maxiraum value of y. A curve is symmetrical tvith respect to a point if the point bisects all the chords of the curve drawn through it. For example, consider the func- tion y = a^. The graph is shown in Fig. 28, where the unit of the scale on the X-axis is taken five times as great as the unit of the scale on the F-axis. We see that the origin bisects all the chords of the curve drawn through it. Hence the curve is symmetrical with re- FiG. 28 spect to the origin. 30. Intercepts. The distances from the origin to the points where a graph crosses the axes are called the intercepts. Thus, in Fig. 27, the curve crosses the X-axis twice, at two and three units to the right of the origin. The X-intercepts are -}- 2 and 4- 3. The curve crosses the Y-axis six units below the origin. The y-intercept is — 6. The X-intercepts are the roots of the equation of the graph when y is pxit equal to zero, and the Y-intercepts are the roots of the equation lohen X is put equal to zero. it ^^l t L V ± t X : 7 ' f 1P5- /- ^ ->-' ' ^3 ._ _.. _ .^ t-Jt- V 2 ' i f--t 4 ^ t t^t ' > 2 -, '/ I Jt ^?- _ _ _ Jj. t 1- Tt ^ J A^tie,* «v''<^' t r. ■Si''''' jyi^'^ y// ^^"^^ ^It / '/ J J f X J if^ / / 4 t ' t i i~ ~ -- - t /- f- t J t - ' ^ -J - J t : i/ 4 .. _ _ w t ' 4 f t-j JZ t 1 / 1 / / y ± Art. 31] GRAPH IN POLAR COORDINATES 39 EXERCISES 1. Draw the graph of the function y = x"^ - 2 x — S. Find the position of the line of symmetry, the intercepts, and the coordinates of the turning point. 2. Draw the graph of the function y, when y^ — 2y = 2x—l. Find the position of the line of symmetry and the intercepts. Is y a. single-valued, or multiple-valued function of a* ? 3. Given yx = 4. Show that the line bisecting the first and third quad- rants is a hne of symmetry. Find the coordinates of the points where this line meets the curve. Is y a monotone function of a; ? A single-valued func- tion oi X? 4. Show that the graph of y =(x— 1)'^ + 2 is symmetrical with respect to the point (1 , 2) . 5. Show that if an equation contains only even powers of y, the graph is symmetrical with respect to the X-axis ; and if it contains only even powers of X, the graph is symmetrical with respect to the F-axis. 6. If 2/ = ax"^ + bx + c, find the coordinates of the turning point. 7. A rectangle is inscribed in a circle of radius 5. Express the area of the rectangle as a function of the length of one side. Draw the graph of the function thus found, and find the coordinates of the turning point. What is the length of the side of the rectangle of maximum area inscribed in the circle ? 8. A box is to be constructed having a square base and containing 108 cubic feet. The box is to have no cover. Express the number of square feet of lumber required as a function of the length of the side of the base. Draw the graph of the function obtained and locate the turning point. What are the coordinates of the turning point ? What is the size of the box requiring the least amount of lumber to construct it ? 31. Graph in polar coordinates. Let r be given as a function of 0, then corresponding values of the independent variable and of the function can be regarded as polar coordinates of points. When r and 6 are connected by an equation, a table of corresponding values can be computed and plotted as in rectangular coordinates. The totality of radii obtained in this way forms a graphical repre- sentation of the function, and a smooth curve drawn through the plotted points is the graph of the function in polar coordinates. For example, let the function be given by the equation 40 GRAPHIC REPRESENTATION [Chap. III. The following is a table of corresponding values, 6 being measured in radians : /J p, TV TT TT 2i IT b TV ^" ' 6' 3' 2' T' "6"' ''' ■■■• r = 0, 1.0472, 2.0944, 3.1416, 4.1888, 5.2360, 6.2832, .... A part of the graph is shown in Fig. 29. The curve is called a spiral of Archimedes, after its discoverer. Fig. 29 EXERCISES 1. Draw the graph of the equation r = - and compare with the graph in the preceding article. 2 2. Draw the graph of the function r — -■ The curve is called the recip- rocal spiral. 3. How does the graph of r = 2 9 + 1 differ from the graph in the preced- ing article ? 4. If the abscissa of every point in Fig. 27, Art. 29, is diminished by 2\ units, how will this affect the graph ? How will it affect the equation ? Write the new equation and draw the graph. Compare the graph with those in Arts. 27 and 28. 5. In the spiral of Archimedes, let the radius vector rotate in the negative direction. Draw the curve and compare with the graph in Art. 31. Arts. 32-34] TRANSCENDENTAL FUNCTIONS 41 6. Draw the graph of ?• = 3 6. How does this curve differ from the spiral of Archimedes in Art. 31 ? 7. In Fig. 29, the curve vflll cross the initial line when 6 is any integral multiple of w. Why ? What is the distance between any two consecutive points of crossing ? 8. Draw the graph of a; = a6, where a is any constant number. For what values of d does the graph cross the initial line ? What is the distance be- tween any two consecutive points of crossing ? 32. Algebraic functions. If the function and the independent variable are connected by an algebraic equation, that is, an equa- tion involving only a finite number of the fundamental operations of addition, subtraction, multiplication, division, involution, and evolution, the function is called an algebraic function. Thus, for example, in each of the equations y=2 x-, y-=2 x, y"-— oy -\- x=0, cc? — 3 xy 4-2/^ = 0, y is an algebraic function of x. To find the value, or values, of an algebraic function for any given value of the independent variable, it is usually necessary to solve an algebraic equation. For example, if ?/^ — 5 ?/ + ^ = 0, it is necessary to solve a quadratic equation to find the values of y for any given value of x. 33. Transcendental functions. In many cases of great practical importance, the function and the independent variable are not connected by an algebraic equation, and then the function is called a transcendental function. The simplest examples of trans- cendental functions are furnished by the trigonometric functions and logarithmic functions. Thus, y = sin x and y = log x are transcendental functions. To find the value of a transcendental function for a given value of the independent variable, use is made of a table. We thus have tables of logarithms and tables of trigonometric functions. 34. Graphs of transcendental functions. Corresponding values of function and independent variable can be taken directly from the table and the function exhibited graphically in rectangular coordinates or in polar coordinates, as in the preceding articles. 42 GRAPHIC REPRESENTATION [Chap. III. EXERCISES 1. Draw the graphs of the following functions. State which are algebraic functions and which are transcendental functions. (a) y = tan x, (h) y^ = x'^, (d) y^ = 4 a;2, (e) y = log x, 2. Draw the polar graphs of the following functions (ffl) r = sin d (&) r = 2 a(l — cos i (c) y = cos X, (/) x^ + y- = 2x. (c) r = (2(1 + cos ( 3. Using the relations between rectangular and polar coordinates (Art. 9), change the equations in exercise 2 to rectangular coordinates and plot ?/ as a function of x. 4. Change the equation (/) of exercise 1 to polar coordinates and plot r as a function of d. ^r ^^•"^ ^"^-s^ y^^ "^ y^ Jr" r 1 \ / A\X \ / ( / \ TT 1 A D IB D' F\ ' \^ _y w Fig. 30 35. Geometric construction of the graphs of trigonometric func- tions. The graphs of trigonometric functions can be constructed geometrically without the use of tables. For example, to con- struct the graph oi y — sin cc, let be the origin (Fig. 30) and F the point representing -k on the scale OX. With any point A on OX. as center and the unit of the scale on the Y'-axis as radius, draw a circle. Let BAP be any angle x measured in degrees. The perpendicular DP is then sin x. Take the distance OJy so that OU : OF :180° then the point U represents the angle x measured in radians on the scale OX. Through D' draw the perpendicular to OX and through P the parallel to OX. Let these lines meet in P', then Art. 35] GEOMETRIC CONSTRUCTION OF FUNCTIONS 43 D'P' = DP= sin X. Hence, as P describes the circle, P' describes the graph oi y = s\n x. For convenience, divide OF into a number of equal parts and erect perpendiculars to OX through the points of division, then divide the quadrant BPC into the same number of equal parts and draw parallels to OX through the points of division. Each perpendicular meets its corresponding parallel in a point of the graph, as indicated in the figure. The graph of y = sin x is called the sinusoid, or wave curve. Trigonometric functions are periodic functions ; that is, the value of the function is repeated again and again for values of the variable which differ by a constant. Thus sin x has the same value when x is increased or decreased by any integral multiple of 2 TT. Many of the phenomena in nature are also periodic. For this reason, trigonometric functions are of great importance in the applications of mathematics. EXERCISES 1. By measuring angles from the line CJ., Fig. 30, instead of from the line BA^ show how to construct geometrically the graph oi y = cos x. 2. In Fig. 30, draw the tangent to the circle at B and let it meet the radius AP produced in K. Then BK is tan BAP {AB = 1); show how to construct geometrically the graph oiy = tan x. 3. Taking CA for the initial line, show how to construct the graph of y = cot X. 4. Devise a method for constructing geometrically the graphs oi y = sec x and y = cosec x. 5. How can the graph of a trigonometric function be used to find the value of the function for any given value of the variable ? 6. A point P describes a circle of radius a with the uniform velocity of k radians per second. Show that the period^ that is, the time of one complete 9 -jT revolution, is T = - — A; 7. Let the center of the circle in the preceding exercise be the origin of rectangular coordinates. Show that, at the end of t seconds, the coordinates of the point P are ^ X = a cos kt = a cos - — . T y = a sm M = a sin 44 GRAPHIC REPRESENTATION [Chap. III. The kind of motion described by eitlier of these equations is called a simple harmonic motion (S. H. M.), a is called the amplitude and T the period of the S. H. M. 36. The exponential function. When the function and the in- dependent variable are connected by the equation y is called an exponential function of x. The constant a is called the base. The exponential function is transcendental, since y and X are not connected by an algebraic equation. 37. Graph of the exponential function. The graph of the ex- ponential function can be constructed as follows : From a point A on the X-axis (Fig. 31) lay off a unit AB and erect the ordinate BBi equal in length to the base a. Draw the line AB^ and also the line AZ making an angle of 45° with the X-axis. Through jBj draw the parallel to the X-axis meeting AZ in C2, and through O2 the perpendicular to the X-axis meeting AB^ in C^ and the X-axis in C. The segment CCi is equal in length to a^; that is, the value of the function when x is 2. For, by similar triangles, AB:BB,::AC:CO„ov 1: a: : a: CCi, since AC= CC^ = BB^^ = a. Hence, CC^ = al Similarly, drawing the parallel to the X-axis through Ci and the perpendicular through C3, we can prove that DD^ is equal in length to al Thus all the positive integral powers of a can be constructed geometrically. The negative integral powers can also be constructed by means of the parallels and perpendiculars. Thus, JOfi = cr'^, KK^ = cr"^, etc. Let be the origin of coordinates and OT the F-axis. Con- struct parallels to the F-axis at intervals of a unit, thus forming a series of rectangles with the parallels to the X-axis. The graph oi y= a" cuts through opposite corners of these rectangles, beginning from the point (0, 1) and running each way. The exponential function has no turning points and is therefore a monotone function (Art. 27). It is important in representing physical phenomena which are not periodic, such, for example, as Arts. 37, 38] INVERSE FUNCTIONS 45 the retarding effect of friction, the pressure of the atmosphere as a function of the altitude, etc. The exponential function is also E ■ / z D. / / C, c, / 2?. / / ^3 a' a' il/i / C\ a'- / ^,^-^ d K,£ ,-^1 ^m -|«-ia-l ^ A K. ^i B C 7 J 1 7 -2 -1 I \ - X Fig. 31 important in computing interest tables, since the amount y, of $ 1 for X years, at rate i compound interest, is given by the formula y = (l + iy. Frequently the base is taken to be e = 2.71828 •••. The num- ber e is the base of the natural, or naperian, system of log- arithms. 38. Inverse functions. If two variables are connected by an equation, or otherwise, either variable may be regarded as a function of the other. For example, in the equation y = 2 x^, we think of y as the function and x as the independent variable, but we may regard x as the function and y as the independent variable. Either of these functions, y or x, is called the inverse of the other. It is convenient to retain the notation " y means function and x means independent variable." Hence, to obtain the equation de- fining the inverse of a given function y, we have but to inter- change X and y in the given equation and then express y in terms 46 GRAPHIC REPRESENTATION [Chap. III. of X. Thus, in the above example, the inverse function is defined by the equation ,- x = 2y\ovy = ± ^| Similarly, the equations y = sin X and x = sin y, or y = arc sin x, define a pair of inverse functions. Again, the equations y = a" and x = a^, or y = log^ x, define a pair of inverse functions. EXERCISES 1. Draw the graph representing the amount of $1 at [>% compound in- terest as a function of the time, interest being compounded annually. 2. Show that the following pairs of equations represent inverse functions : (a) 2/ = 3 x^ and y = 'Y o' (6) y = 5x — 6 — x^ and ?/ = | ± Vi - x, (c) y = a^^ and y = -"'' , (cl) y = tan 2 x and y = ^ arc tan x. logft a 3. Write the inverse of each of the following functions. (a) 2/ = cos 3. T, (6)?/ = -^tan| (c) y =r loge -, (d) 2/ = x2 - 5 iC + 6. 4. Show how to construct the graph of y = ffl"^ from the graph oi y = a^ in Art. 37. How will changing the sign of x affect any graph ? 5. Given the graphs oi y = a^ and y = a~^ on the same coordinate axes, how can one construct geometrically the graph oi y = ^ ~*" ^ — ? 6. With the graphs oi y = sin x and y = cos x on the same coordinate axes, construct geometrically the graph of ?/ = sin x + cos x. 39. Graph of an inverse function. Since the inverse of a given function is obtained by interchanging x and y, the graph of the inverse function can be constructed by interchanging the coordi- nates of every point on the graph of the given function. Thus, if P (Fig. 32) is a point on the graph of the given function, P' is a point on the graph of the inverse function when OD' = DP and D'P' = OD. Art. 39] GRAPH OF AN INVERSE FUNCTION 47 By this construction, P and P' are symmetrically situated with respect to the line OA bisecting the first and the third quadrants. As P describes the graph of the given function, P' de' scribes the graph of the inverse function. Hence, having given the graph of any function, we can ob- tain the graph of the inverse function by plotting points sym- metrically situated to the points of the given graph with re- spect-to the line OA. Or we may consider the entire jDlane ro- tated through 180° about the line OA, carrying the given graph with it. The new position of the graph is the graph of the in- verse function. Por example, let JSLX be the graph of y =a^ (Fig. 32), where a is taken to be 2. Eotating the plane about OA, the curve assumes the position RS, symmetrical to MN with re- spect to the line OA. Therefore RS is the graph of the inverse function y = log^ x. Fig. 32 EXERCISES 1. Given y = 5 x — 6 — x-, draw the graph of the inverse function. 2. Construct the graphs of the following functions : (a) y = arc sin x, (6) y = arc tan x, (c) y = arc cos x, 2 3 (fZ) y = x^ and y = x^. 3. Show that the graph oi y — - and the graph of the inverse function X coincide throughout. What condition must be satisfied in order that the graph of any function shall coincide with the graph of its inverse ? 48 GRAPHIC REPRESENTATION [Chap. Ill, 40. Observation. A functional correspondence between two variables is often established by observation when no relation between the variables is known. Thus the temperature at a given place can be observed throughout the day and the results tabulated. The temperature can then be regarded as a function of the time, and the functional relationship can be graphically exhibited as in the preceding articles. In such cases the functional relation- ship is given in the form of a table of corresponding values. 41. Machines. Machines are devised to draw the graph automatically and thus avoid the necessity of making repeated observations. For example, tlie weather bureau has an instrument to graph the temperature as a func- tion of the time. Coordinate paper is wound upon a clock-driven drum, and a pen is connected with a thermometer in such a way that the rise and fall of temperature is recorded upon the paper at the proper time. The record exhibits the functional relationship in the form of a graph. Corresponding values of the function and the independent variable can be read from the graph as readily as from a table. Other records exhibit functional relationship in the form of a graph upon polar coordinate paper. EXERCISES 1. The following table shows the length of a rubber cord in centimeters when stretched by a weight in kilograms attached to one end. Draw a curve representing approximately the graph of the length as a function of the weight. Weight .... Length .... 10 Weight .... 3.5 Length .... 12.2 2. The number of deaths per hundred thousand lives, according to the American experience table of mortality, is as follows : .5 1.0 1.5 2.0 2.5 3.0 10.1 10.3 10.6 10.9 11.3 11.7 4.0 4.5 5.0 5.5 6.0 12.7 13.3 13.9 14.6 15.3 Age Number of Deaths Age NuMBEE OF Deaths 20 781 60 2669 25 807 65 4013 30 843 70 6199 35 895 75 9437 40 ■ 979 80 14447 45 1116 85 23555 50 1378 90 45455 55 1857 95 100000 Art. 41] MACHINES 49 Draw a curve representing the graph of the number of deaths as a func- tion of the age. 3. The net annual premium for an assurance of § 1000 for life, according to the American experience table of mortality, interest at 3 %, is as follows : Age Premium Age Pkemu M 20 $ 14.41 40 .$24.75 25 ^16.11 45 $29.67 30 $18.28 50 $36.36 35 f 21.08 Draw a curve representing the graph of the premium as a function of the age. 4. The cost of a money order depends upon the amount as follows : Amount Cost Amount Cost $0 to .'$2. 50 3ct. $ 30 to $ 40 15 ct. $2.60 to $5 5ct. $ 40 to $ 50 18 Ct. $ 5 to $ 10 8 ct. $ 50 to $ 60 20 Ct. $ 10 to $ 20 10 ct. $ 60 to $ 75 25 ct. $20 to $30 12 ct. $ 75 to $ 100 30 ct. Draw a curve representing the graph of the cost as a function of the amount. 5. Figure 33 A (p. 50) represents a thermograph for April 12, 13, and 14. The ordinates are made curvilinear to allow for the pivotal motion of the drawing pen. Determine the maxima and minima temperatures between noon of April 12 and noon of April 14. When was the temperature highest ? When lowest ? 6. Figure 33 B (p. 51) is a steam pressure gauge on polar coordinate paper. The radii are made curvilinear to allow for the pivotal motion of the drawing pen. Determine the time of greatest pressure. The time of least pressure. 7. By means of the table of exponential functions (page V), make a careful drawing of the graph of y = e^. From the graph thus made con- struct the graph ot y = e~^. Compare the readings from the graph with the values of e~^ taken from the table. From the graph of y = e^, construct the graph of y = — e''. 50 GRAPHIC REPRESENTATION [Chap. Ill- 8. Having given the grapli of y = f(x)^ show how to obtain the graphs oiy=f{-x), y =-f(x), and y =-f(-x). 9. According to Boyle's law, tlie volume of a gas is inversely propor- tional to the pressure which it sustains. If a volume of 4 cubic feet sustains a pressure of 1 atmosphere, write the equation expressing the volume as a function of the pressure. Draw the graph of this function. 10. The increase in length of a metal bar is proportional to the tempera- ture to which the bar is subjected. If the bar is 1 foot long at 0° temperature Fig. 33 A and 1.0004 feet long at 20° temperature, write the equation expressing the length as a function of the temperature. Draw the graph of this function. 11. The intensity of light is inversely proportional to the square of the distance from the source of the light. Write the equation expressing the intensity as a function of the distance. Draw the graph of this function. If the intensity of light at a point on the earth directly underneath the sun is taken as the unit of intensity, calculate the intensity of light on the planet Venus at a point directly underneath the sun. Take the distance from the earth to the sun as 93,000,000 miles and the distance of Venus from the sun as 67,000,000 miles. Art. 41] MACHINES 51 Fig. 33 B 12. The water rates of a certain city depend upon the amount consumed and are as follows : Consumption per Day Rate pek 1000 Gallons to 499 gallons .35 500 to 999 irallons .32 1000 to 1999 gallons .28 2000 to 2999 gallons .24 3000 to 3999 gallons .20 4000 to 4999 gallons .15 5000 to 5999 gallons .12 6000 and over gallons .10 52 GRAPHIC REPRESENTATION [Chap. III. Calculate the monthly (30 clays) bill of a consumer and draw a curve representing the graph of the amount of the bill as a function of the number of gallons consumed per month. 13. A certain mixture of concrete contains 1.4 barrels of cement per cubic yard of concrete. If the cement costs $ 1.20 per barrel and the sand and crushed stone costs 82.10 per cubic yard, write an equation expressing the cost of the concrete as a function of the number of cubic yards. Draw the graph of this function. 14. Express the area of a circle as a function of the radius and draw the graph of the function. 15. Draw the graphs of y = sin a; and y = cosx on the same coordinate axes. From these graphs construct the graph of the function y = 2smx + cos x. CHAPTER IV LOCI AND THEIR EQUATIONS 42. Locus of a point, equation of locus. When a point P{x, y) moves in the plane, the path it describes is called the locus of the point. The coordinates x, y are then variables (Art. 22). If the point P{x, y) moves according to a given law, this law will lead to an equation connecting x and y called the equation of the locus. The equation of the locus defines ?/ as a function of x, and the locus itself is the graph of this function. As an example, suppose P moves so that it is constantly at a fixed distance from a fixed point A. We know, then, that the point describes a circle. This circle is the locus of the point P. The point P moves according to the given law AP = constant, and we shall see that this law leads to an equation connecting the variable coordinates x and y. 43. A fundamental problem. When the law which governs the motion of a point is given, a fundamental problem presents itself • namely, to find the equation of the locus. For example, suppose a point moves so that it is always equidistant from the points F = (l, 2) and F,={3, 1) (Fig. 34). To find the equation of the locus, let P{x, y) be any point equidistant from F and F-^. Then, by the given law, ^^ ^ ^^^^ for all positions of P. But PF=-V{x - iy-\-{y - 2)2 and PF, = V(x - 3y+ {y - If. Therefore we have V(a.- - 1)2+ (,y - 2f=^ix - 3Y+{y - If, (2) 53 54 LOCI AND THEIR EQUATIONS [Chap. IV. which reduces to 4:X — 2y — 5 = 0. (3) This is the required equation. The locus is the perpendicular bisector of the segment FF^. The following property and its converse are characteristic of this locus and its equation ; namely, the coordinates of every point on the locus satisfy equation (3). For, if the point is on the IJ- ' \ Ml -It - ^ 1 M 1 ' ; 1 1 ' 1 III /^ 1 1 i 1 / 1 ■ ! ! ' / 1 : 1 ] 1 1 ,j JlCxilti 11 1 ' • n 1 II ^■^ 1 7 y . y^ i / < / -^ - - : J' SsJ / 1* C - - ^ 55> X : TT ^r 7 ^^ " "Z . ...__!rt- - U- t A. t t -- t ~ V " ._ : : t .. _ A t i t' t 1 t f ^ Fig. 34 locus, it is equidistant from F and F^. Therefore its coordinates satisfy (2) and consequently (3). Conversely, if the coordinates of any point satisfy equation (3), the point is on the locus. For then the coordinates of the point also satisfy equation (2), and the point is therefore equidistant from F and F^ ; that is, the point is on the locus. 44. General definition. The property just proved for the special locus in Fig. 34 leads to the following definition: The equation of the locus of a point is an equation in the variables x and y which is satisfied by the coordinates of every point on the locus ; Arts. 44, 45] THE CIRCLE 55 and conversely, every point whose coordinates satisfy the equation lies on the locus. The locus of a point moving according to a given law is, in general, a curve, and we shall often speak of the equation of the locus as the equation of the curve.* The problem to find the equation of the locus wlien the law governing the motion of the point is given will be illustrated in the succeeding articles, where the equations of a number of im- portant curves are found and methods given for constructing the curves. 45. The circle. A x>oint moves so that it is alivays r units from ajixecl jioint (a, 6). Find the equation of the locus. The locus is evidently a circle whose radius is r and whose center is the point G = (a, h) (Fig. 35). To find the equation of the locus, assume that P{x, y) is any jDoint r units from C. The point P is then on the locus. The statement of the law gov- erning the motion of P is then CP=r, (1) for all positions of P. But P{^,y) Fig. 35 and therefore GP = ^(-x-af + iy-by, (2) (ic-a)2+ (I/- 6)2 = ^2. (3) If the center of the circle is taken at the origin of coordinates. then a = and b = 0, and equation (3) becomes a52 + ?/2 = /'2. (4) Equations (3) and (4) are standard forms of the equation of a circle. The student should test equation (3) by the definition given in Art. 44. *The word "curve" will henceforth be used to denote any continuous line, straiffht or curved. 56 LOCI AND THEIR EQUATIONS [Chap. IV. EXERCISES 1. Find the equations of tlie following circles : (a) Center (0, 1) and radius 3. (6) Center (— 2, 0) and radius 2. (c) Center (— 4, 3) and radius 3. (cl) Center (1, 2) and radius 6. 2. Find the equation of the circle whose center is (2, 3) and which passes through the origin. 3. What is the equation of the circle which has the line joining the points (3, 2) and (— 7, 4) for a diameter ? 4. Find the equation of the circle which passes through the three points (0, 1), (5, 1), and (2,-3). 5. A point moves so as to be equidistant from the points (3, — 1) and (— 2, 3). Draw the locus and find its equation. 6. Find the equation of the perpendicular bisector of the segment joining (a, 6) to (c, cl). 7. A point moves so that the ratio of its distances from the points (8, 0) and (2, 0) is constantly equal to 2. Find the equation of the locus. 8. A point moves so that the sum of the squares of its distances from (3, 0) and (— 3, 0) is constantly equal to 68. Find the equation of the locus. 9. A circle circumscribes the triangle (6, 2), (7, 1), (8, - 2). Draw the figure and find the equation of the circle. 46. The equation x^ + j/^ + Asc + By + C = When equation (3) of the preceding article is expanded and arranged according to the powers of x and y, it takes the form x^ + 'if + Ax + By + C = 0, (1) where A, B, and C are constants depending upon the radius of the circle and the coordinates of tlie center. A second problem now arises : Is equation (1) the equation of a circle for all possible values of A, B, and C? To answer this question, we shall complete the squares of the terms in x and y separately, and thus put equation (1) in the form In this form, the equation states that the length of the segment joining [ , ) to {x, y) is constantly equal to Arts. 46, 47] THE STRAIGHT LINE 57 for all positions of the point (x, y). Let D stand for the ex- pression under the radical in (3) ; then we can draw the following conclusions : 1. If Z) > 0, (1) is the equation of a circle whose center is the ■ , f A B\ . . point ( — -, — -^ ] and whose radius is VZ). 2. If Z) = 0, (1) is satisfied by the coordinates of a single point ; namely, the point ( — — , )• In this case the locus is called a null circle. 3. If Z) < 0, there is no point in tha plane whose coordinates satisfy (2) and consequently no point whose coordinates satisfy (1). In this case the locus is called an imaginary circle. We shall find it convenient to say that, in any case, equation (1) is the equation of a circle, but that, in particular cases, this circle may be a null circle, or an imaginary circle. EXERCISES 1. Find the coordinates of the center and the radius of the following circles. Construct the figure when possible. {a) x^ + 2/2-6 X - 16 = 0. (5) ^2 + y2 _ 6 x + 4 y - 5 = 0. (c) 3x2 + 32/2-10x-242/ = 0. {d) (x + l)2 + (?/ - 2)^= 0. (e) x2 + 2/2 = 8 X. (/) 7 x2 + 7 ?/2 - 4 x - ?/ = 3. ig) «2 + 2/2 _ 2 X + 2 2/ + 5 = 0. {h) x~ + 2/2 + 16 X + 100 = 0. 2. Eind the coordinates of the center and the radius of the circle which passes through the points (5, — 3) and (0, 6) and has its center on the line 2x-32/-6 = 0. 3. A point moves so that the sum of the scjuares of its distances from two fixed points is constant. Prove that the locus is a circle. 4. A point moves so that the ratio of its distances from two fixed points is constant. Prove that the locus is a circle if the constant ratio is different from unity, and a straight line if the constant ratio is equal to unity. 47. The straight line. A point moves on the straight line joining the fixed points P^ =(x-^, y^ and P, = (^'2; y-i)- Find the equation of the locus. Choose any point P(x, y) on the straight line joining P^ to Pj (Fig. 36). Then, 58 LOCI AND THEIR EQUATIONS [Chap. IV. slope of segment 1\P = slope of segment P1P2. But (Art. 11), slope of FyP = -^^^ , and slope of P.P^ = ^ ~ ^^ Therefore 2/ - 2/1 _ 1/2 -2/1 (1) iC — if 1 iCg — a?i The expression ^- ~ ■^^ is called the slope of the line. Eepresent- 1' m'. ing the slope of the line by m, equation (1) becomes (2) If the point P^ =(0, 6), that is, the point of intersection of the line with the F-axis, equation (2) assumes the form y = mx + 6. (3) If Pi = (0, b) and Po={a, 0), then m= , and equation (2) a reduces to ^2(3:2, y 2) a b (4) ^x Fig. 36 Equations (1), (2), (3), and (4) are all standard forms of the equation of a straight line. Equation (1) is called the two- point form, equation (2) is the slope-point form, equation (3) is the slope form, and equation (4) is the intercept form. From these equations, we conclude that the equation of a straight line is of first degree in the variables x and y. Conversely, it may be shown that any equation of the first de- gree in the variables x and y is the equation of a straight line. Eor, let . -r, . ^ rv /ex ' Ax + By + C=Q (5) be such an equation. Solving for y, we have A C y = X . ^ B B = (1) Arts. 47, 48] THE DETERMINANT FORM 59 Comparison with equation (3) shows that (5) must be the equa- 4 tion of a straight line whose slope is — — and whose intercept on C ^ the F-axis is This reasoning fails when B is zero. In IB that case, however, equation (5) reduces to Ax -\- C =0, which is the equation of a straight line parallel to the F-axis, since x has C the constant value for all values of y. Hence, in every case (5) is the equation of a straight line. 48. The determinant form. The equation of a straight line can be written in the form of a determinant. Thus, the equation X y 1 'Vi yx 1 X2 2/2 1 is the equation of the straight line joining the points Pi = (a,'i, y^ and P2 = (x2, yo). For, equation (1) is of the first degree in x and y and therefore is the equation of some straight line, by the pre- ceding article. Moreover, the equation states that the area of the triangle whose vertices are [x, y), (x^, y{), and (x.^, yo) is zero (Art. 20). Hence, the point P(x, y) is on the line joining P^ and EXERCISES 1. Write the equations of the lines passing through the following pairs of points : («) (0, 1) and (5,6); (6) (1,-2) and (-3,4); (c) (5, -2) and (_ 4, — 1) ; (d) (— 1, 3) and (3, — 4). Draw the figure in each case. 2. Find the intercepts which each of the lines in exercise 1 makes upon the coordinate axes. Write the equations in intercept form. 3. With the intercept on the F-axis and the slope, write the equation of each line in exercise 1 in the slope form . 4. Write the equation of each of the lines in exercise 1 in the determinant form. 5. Find the slope and the intercepts of each of the following lines : (a) 2ij + BX-1 = x + 2.-:^ (b) r — 1 x-3 2 3 (e) y^ = S. (d) aLzil=2x_pi 60 LOCI AND THEIR EQUATIONS [Chap. IV. 6. Write the equation of each of the lines in exercise 5 in the intercept form. In the slope form. 49. The ellipse. A point moves so that the sum of its distances from tivo fixed points F and F^ is constantly equal to 2 a. Construct the locus and find its equation. In the first place, 2 a must be greater than the length of the segment FF^, otherwise no locus is possible. Lay off a line AiB^, ^Y M c Fig. 37 2 a units in length (Fig. 37). Take C, any point on A^B^, and with A^C as radius describe a circle about F. With CB^ as radius describe a circle about F^. The two circles meet in the points 3f and iV. These points are on the locus, since the sum of the radii of the two circles is 2 a. Taking the smaller circle about F^ and the larger about F, two more points, M' and JSf', are found on the locus. By taking C at. different places on AiB^, as many points of the locus can be found as may be desired. Another construction of the locus is made as follows : stick pins in the paper at the points F and F^. Tie the ends of a string together so that the loop is just equal to 2 a plus the Art. 49] THE ELLIPSE 61 distance from F to Fy. Drop the loop over the pins and stretch it taut with a pencil point. Keeping the string stretched, move the pencil around ; it will describe the locus. This locus is called an ellipse. The fixed points F and F^ are called the foci of the ellipse. The distances from any point on the ellipse to the foci are called the focal radii of the point. To find the equation of the ellipse, let the line joining the foci be the X-axis, and the perpendicular bisector of the segment FF^, the T^axis. Let P{x, y) be any point on the ellipse ; then PF=: r and PFi = Vi are the focal radii of P. By definition we have r + r, = 2a (1) for every position of P. Let 2 c denote the length of the segment FF^ ; then the coordi- nates of F and Fj^ are ( — c, 0) and (c, 0), respectively. Then j-2 = (c + xf -{-y^ = c^ + 2cx + x'' + tf, and 7\^ = (c - xf ■i-if = c--2cx + x'^ -f- y\ ^"^ By subtraction, we obtain 7-2 - j\^ = (r — ?'i) (r -f rj) = 4 ex. Hence, since r + r^ = 2 a, r^r, = y^ = ^-^. (3) 2 a a From (1) and (3) we get. r = « + ^, (4) a Substituting the value of r in the first of equations (2), we obtain, after reduction, „ , x" o?- a^ — c^ A further simplification is obtained by putting a'~-c'' = ¥, (6) and then the equation assumes the final form Equation (7) is the standard form of the equation of an ellipse. ^2 ..2 a^ 6^ 62 LOCI AND THEIR EQUATIONS [Chap. IV. 50. The axes and eccentricity. The segment of the line joining the foci and limited by tlie curve is called the major or transverse axis of the ellipse. That part of the perpendicular bisector of the segment joining the foci which is contained within the curve is the minor or conjugate axis of the ellipse. Thus, AB (Fig. 37) is the major axis and CD the minor axis. The axes intersect in the center, and cut the curve in the vertices. When the equation of the ellipse is in the standard form, the axes of the curve coincide with the axes of coordinates (Art. 49). Hence the lengths of the axes of the ellipse can be determined from the intercepts (Art. 30) made by the curve upon the coordi- nate axes. From equation (7) of the preceding article we find that the intercepts on the X-axis are ± a and the intercepts on the F-axis are ± b. Therefore the length of the major axis is 2 a and the length of the minor axis is 2 b. The segments OB and OD (Fig. 37) are called the semimajor axis and the semi- minor axis, respectively. The ratio of the distance between the foci to the length of the major axis is called the eccentricity of the ellipse. Since the distance between the foci is 2 c and the length of the major axis is 2 a, the eccentricity is e = '-' (1) a From equation (6) (Art. 49), c = Va^ — b\ Therefore e = ^^^. (2) a Since a is always greater than c, the eccentricity of the ellipse is necessarily always less than unity. Combining equations (4) Art. 49, with equation (1), we see that the lengths of the focal radii of the point P(x, y) are r = a + ex and ri = a - ex. (3) EXERCISES 1. Find the equation of the ellipse for which the sum of the focal radii is 8 and the distance between the foci is 6, the origin being at the center. What is the eccentricity of this ellipse ? Construct the ellipse. 2. An ellipse passes through the points (- 5, 0) and (0, 3) and is sym- metrical with respect to both axes. Find the coordinates of the foci and draw the curve. Arts. 50, 51] THE HYPERBOLA 63 3. "Write the standai'd form of the equation of the ellipse having given : (a) the length of the transverse axis is 10 and the distance be- tween the foci is 8 ; (6) the sum of the axes is 18 and the difference of the axes is 6 ; (c) transverse axis is 10 and the conjugate axis is i the trans- verse axis ; (fZ) transverse axis is 20 and conjugate axis is equal to the dis- tance between the foci; (e) conjugate axis is 10 and distance between the foci is 10. v'i 9/2 4. The equation of an ellipse is 1--^ = 1. Find the lengths of the ^ 64 15 focal radii of the points whose abscissa is \. 5. Find the lengths of the semiaxes and the eccentricity of each of the ellipses whose equations are : (a) 3 x2 + 2 2/2 = 6 ; (6) | + ^^ 1 ; (f ) X- + 32/2 = 2; {d) 42/^ + 2 z'^ = 2. 6. The latus rectum, or parameter, of an ellipse is the double ordinate, or double abscissa, passing through a focus. Find the length of the latus rectum for each of the ellipses in exercise 5. 51, The h3rperbola. A point moves so that the difference of its distances from ttvo fixed points F and Fj^ is constantly equal to 2 a. Construct the loctis and find its equation. Here 2 a must be less than the length of the segment FFi. For, if P is any point in the plane, it is shown in geometry that the difference be- tween any two sides of the triangle PFF, is less than the third side. Lay off a line AB (Fig. 38) 2 a units in length and take any point C on this line produced. AVith BC and AC as radii and F and F^^ as centers, draw arcs of circles intersecting, in M and N. These points are on the locus, since the difference Fig. B 38 64 LOCI AND THEIR EQUATIONS [Chap. IV. of the radii of the two circles is 2 a. With the same radii, but interchanging centers, two more points, M' and N', are obtained. Taking G at different places on AB produced, as many points on the locus can be constructed as may be desired. The locus is called an hyperbola, the points F and jF\ are its foci, and the distances from any point on the curve to the foci are called the focal radii of the point. The two parts of the curve are the branches. To find the equation of the hyperbola, we proceed as in the case of the ellipse. Let the line joining i^and F^ be the X-axis, and the perpendicular bisector of FF^, the I''-axis. Let F be c units to the left of the origin and F^, c units to the right. Take P{x, y), any point on the curve, and let r and r^ be the lengths of its focal radii (r>?'i). Then, by definition, /• - ri = 2 a. (1) Equations (2) of Art. 49 hold for the hyperbola, and we obtain from them, by subtraction, (r — ri) (r + 7\) = 4 ex. (2) Combining (1) and (2), we have , 2 ex /Q\ a From (1) and (3) we get (4) a Substituting the value of r in the first of equations (2), in Art. 49, we obtain, after reduction, 2 2 ^ ^^— = 1. (5) 2 2 a c — a A further simplification is obtained by putting c2 - tt" = W (6) and the equation assumes the final form Equation (7) is the standard form of the equation of an hyperbola. Art. 52] AXES AND ECCENTRICITY 65 52. Axes and eccentricity. The hyperbola meets the line join- ing the foci iu two points B^ and A^ (Fig. 38) which are equidis- tant from the mid-point 0, as may be seen from the definition of the curve. The segment B^A^ is called the transverse axis. Since the intercepts on the X-axis (when the equation is in the standard form) are ± a, the length of the transverse axis is 2 a. The curve does not meet the perpendicular bisector of FF^, since every point on this bisector is equidistant from i^'and F^, but a segment extending b units above and 6 units below O is called the conjugate axis. is the center of the curve, and the transverse axis meets the curve in the vertices, B^ and A-^. The ratio of the distance between the foci to the length of the transverse axis is called the eccentricity. Since FF^ = 2 c, and B^Ai = 2 a, the eccentricity is ^ = --. (1) From (6), Art. 51, we have c = Va^-f b% and therefore (2) ^^^i^Tb' From (1) or (2) we conclude that the eccentricity of an hyper- bola is always greater than unity. Combining equations (4) of Art. 51 Avith equation (1), we have the lengths of the focal radii in terms of the eccentricity, namely : r = ex + a and r^ = ex — a. (3) EXERCISES 1. Write the standard equation of the hyperbola for which the difference between the focal radii is 6 and the distance between the foci is 8. 2. "Write the standard equation of the hyperbola for which the transverse axis is 12 and the distance between the foci is 16. 3. Find the length of the focal radii of the point whose ordinate is 1 and whose abscissa is positive, the equation of the hyperbola being ^ = 1 . 4. Find the semiaxes and eccentricity of each of the hyperbolas whose equations are : (a) 4a;2-9?/2 = 36; {h) ^-^=1; (c) 16 a-2 - ?/2 = 16 ; (d) ^—-y'i=m. 66 LOCI AND THEIR EQUATIONS [Chap. IV. 5. When the origin of coordinates is taken at the center of an hyper- bola and the foci he upon the T-axis, the standard equation is - — ^ ==— 1. Find the le-ngths of the semiaxes and tlie eccentricity of each of the follow- ing hyperbolas : (a) 3 2/2 _ 2 X" = 12 ; (6) 4 x2 - 16 xfi - -64 ; (c) y"- - my? = n. 6. The length of the double ordinate, or double abscissa, through a focus is called the latus rectum of the hyperbola. Find the length of the latus rectum for each of the hyperbolas in exercises 5 and 6. 53. The parabola. A jyoint moves so as to he equally distant from a fixed point and from a fixed straight line. Construct the locus and find its equation. Let F be the fixed point and AH the fixed straight line (Fig. 39). Draw AF perpendicular to AH and a series of lines parallel to AH, as PD, P,D„ F^D^, etc. With AD as radius and F as center, draw an arc cutting PD in P and Q. These points are on the locus, since PF=AD = FQ. E-epeating the process with ADi, AD^, etc., as radii, a series of points on the locus is ob- tained. The locus is called a parabola (cf. Art. 28). The fixed point F is called the focus of the parabola and the fixed line AH is called the directrix. The point is the vertex. To find the equation of the parabola, let AF be the X-axis and OY, the perpendicular bisector of AF, the Y-axis. Let OF — p, and P{x, y) be any point on the curve. Then by definition. H kY c, when a<^c, and when a = c. 3. Show that a cassinian oval is necessarily symmetrical with respect to both axes. 65. Recapitulation. The results of the preceding articles are so important that they are brought together here in compact form. The standard forms of the equations should be memorized. Locus OK Curve Standard Forms of the Equation in Rectangular Coordinates The straight line. (a) Two-point form ; ^"^1 = 2/2- 2/i _ ^ ^ X — Xi X2 — X\ (b) Slope-point form ; y — yi = m (x — Xi). (c) Slope form ; y = rax + b. (d) Intercept form ; - -|- 1 = 1. The circle. (a) (x-a)2-}-(2/-6)^ = j-2. (&) x2 + y^ = 7-2. The ellipse. X2 ^2_ a2 + ft2 '■ The hyperbola. x^ y2 «2 6-2 - '• The parabola. ?/2 = 4:pX. 56. Polar equation of a circle. Let C = (b, a) be the center of a circle of radius a, and P= {r, 0), any point on the circle (Fig. 41). In the triangle COP, we have OC = b, OP=r, and the angle COP = ± {0 — a), depending upon the position of P. But in either case the law of cosines applies and we have r^ + b^ -2 hr cos(9 - a) = a^ (1) 70 LOCI AND THEIR EQUATIONS [Chap. IV. nr,e) Fig. 41 This equation expresses the relation be- tween r and d for any point on the circle and is, therefore, the -polar equa- tion of the circle. If the initial line passes through the center of the circle, a = and (1) re- duces to r^ -I- fe" — 2 hr cos 9 = a^. (2) If the pole is taken on the circle, !"(>'. 0) Fig. 42 b = a and (2) becomes the im- portant form r = 2a cos 6. (3) This equation is also immedi- ately deduced from Fig. 42, since XPO is a right angle. If the pole is taken at the center, b — and (1) becomes r = a, (4) which is the simplest form of the polar equation of a circle. 57. Polar equation of a straight line. Let AB be any straight line, the pole, and OX the initial line (Fig. 43). Let p be the length of the perpendicular OM let fall from upon AB, and a the angle XOM. r(r 0) Take P (r, 6), any point on the line AB. Then ^. ,< rcos(e-a)=_p (1) is the polar equation of the straight line AB. EXERCISES 1. The center of a circle is at the point whose polar coordinates are i^,-\ and the radius is 4. AYrite the polar equation of the circle and Fig. 43 find the length of the se£?ment of the initial line within the circle. Arts. 57-59] POLAR EQUATION OF THE ELLIPSE 71 2. The perpendicular from the pole upon a line is 5 units long and makes an angle of 60° with the initial line. Write the polar equation of the line. With origin at the pole and X-axis coinciding with the initial line, write the rectangular equation of the same line and find the intercepts on the axes. 3. A circle is tangent to the initial line at the pole, its radius is 4 units long, and its center lies above the initial line. What is the polar equation of the circle ? What is the rectangular equation, the origin being at the pole, and the X-axis coinciding with the initial line ? 4. Change the intercept form of the equation of a straight line to polar coordinates. Show that _ P P fr a = -^ — and b = cos a sin a, p and a having the same meanings as in Art. 57. 5. Discuss the polar equation of a straight line (Art. 57) for a =0^, 90°, 180°. Also for j3 = 0. 6. A circle passes through the origin and has its center on the line bisect- ing the first and third quadrants. Find the polar equation in each of the two possible positions. Also the rectangular equation. 58. Polar equation of the parabola. The polar equation of the parabola assumes the simplest form when the pole is taken at the focus and the initial line is perpendicular to the directrix (Fig. 44). Let P {r, 9) be any point on the parabola. The length of the focal radius PF is (Art. 53) r= X + p. (1) But, from the figure, X = OD = r cos 6 +p. Eliminating x between (1) and (2) and solving the resulting equation for r, we have r = 2p (1- cosO) (2) Fig. 44 (3) 59. Polar equations of the ellipse and the hyperbola. Take the pole at the left-hand focus and the initial line coincident with the transverse axis of the curve (Figs. 45 and 46). Then, for either curve, the length of the focal radius PF is given by the formula (Arts. 50 and 52) r — a -\- ex. (1) 72 LOCI AND THEIR EQUATIONS [Chap. IV. But, from either figure, x= OD = r cos 6 - c. (2) Eliminating x between (1) and (2) and solving the re- sulting equation for r, we have (g - ec) Fig. 45 (1 — e COS 6) Replacing e by its value (3) (3) becomes {a — ccosO) (4) For the ellipse, a^ — c^ = If (Art. 49) and a > c. Hence we con- clude that r is positive for all values of 6. For the hyperbola, a^ — c^ — — If (Art. 51), and a Kc Hence a — c cos 6 will be negative, and therefore r positive, when cos B'> - and then the point c P{r, 6) lies on the -right branch of the curve. For example, when ^ = 0, cos ^ = 1, and T — a + c = FB. The expres- sion a — c cos 6 will be posi- tive, and therefore r negative, when cos ^ < - and then the c point P(>-, ff) lies on the left branch of the curve. For example, when and r = —{c — a) = — FA. 180°, cos ^ = - 1, EXERCISES 1. The sum of the focal radii is 8 and the distance between the foci is 6. Write the polar equation of the ellipse and sketch the curve from this equation. 2. The difference between the focal radii is 4 and the distance between the foci is 6. Write the polar equation of the hyperbola and sketch the curve from this equation. Arts. 59, 60] PARAMETRIC EQUATIONS 73 3. Show from the polar equation that the radius vector for the ellipse is always finite in length. 4. Show from the polar equation that the radius vector for the hyper- bola becomes indelinitely long for two values of d, each less than 180'^. Find these values. 5. The focus of a parabola is 6 units from the directrix. Write the polar equation and sketch the curve from this equation. 6. Show from the polar equation of the parabola that the radius vector never becomes indefinitely long except for ^ = 2 rnt, where n is any integer including zero. 7. Show that the polar equation of the lemniscate is r- = 2 c- cos 2 d, the pole being at the origin and the initial line coinciding with the A'-axis. Sketch the curve from this equation. 8. Change the standard forms of the equations of the ellipse, hyperbola, and parabola to polar equations, making use of equations (1), Art. 9. Why do not the equations thus found agree with the polar equations in Arts. 68 and 59 ? 9. Derive the polar equation of the hyperbola, assuming the right-hand focus as pole and the transverse axis as initial line. 10. Derive the polar equation of the ellipse, making the same assump- tions as in the preceding exercise. 11. Compare the equation r = with equation 4, Art. 59. 5 — 3 cos d Does the given equation represent an ellipse or an hyperbola ? What is the eccentricity and the length of the transverse axis ? 12. If the semiaxes of an hyperbola are equal, the curve is called the rectangular hyperbola. Write the polar equation of the rectangular hyperbola. 60. Parametric equations. It is frequently useful to express the X and y coordinates of a point on a curve in terms of a third variable called the parameter. For example, the x and y coordi- nates of a point on the circle X^ -\- y^z= cO- (1) can be expressed as follows : X = a cos t, y= a sin t, (2) since these values of x and y satisfy (1), whatever value is given to the parameter t. Equations (2) are parametric equations of the circle whose equation in rectangular coordinates is (1). 74 LOCI AND THEIR EQUATIONS [Chap. IV. Similarly, a pair of parametric equations of tlie ellipse x = a cos ^, 2/ = sm ^, (4) since these values of x and ?/ satisfy (3) for all values of t. A variety of pairs of parametric equations can be found ex- pressing the same relation between x and y. For example, a(l-f) T 2 at X — -^ ^ and ?/ = are parametric equations of the circle (1), since these values of X and y will satisfy (1) for all values of t. EXERCISES 1. Show that X = a sec t and y = b tant are parametric equations of an hyperbola. 2 . Show that x — — \-t and y = b + mt satisfy the slope form of the m equation of a straight line for all values of t. 3. Eliminate t from the equations x = t- and y = 2 t and thus show that these equations are parametric equations of a parabola. 4. Write a pair of parametric equations for the standard form y'^ = 4px. 5. Show that x — at and y = b(l — t) are parametric equations of a straight line. 6. Show that the equations cV2(l + fi)t 1 + «* ' y = !^ — 1 + t* are parametric equations of the lemniscate (Art. 54). 7. Write the parametric equations of the rectangular hyperbola x^ — y'^ = a^. 61. Geometrical construction of the ellipse and the hyperbola. The ellipse and the hyperbola can be constructed easily by means of parametric equations (Figs. 47 and 48). Draw the concentric circles whose radii are the semiaxes a Art. 61] GEOMETRICAL CONSTRUCTION 75 and h. These circles are called the major and minor auxiliary circles, respectively. For the ellipse, with any value of t, construct OD = a cos t and EP' = b sin t. - These are the coordinates of the point P on the ellipse. / fv^ ■— ^ ? / >^ /\ / \ X. \ y / / k ^ \ \ \ \ \^^ A \ [X. \y X 1 \| 3J V Fig. 47 Fig. 48 Similarly, for the hyperbola, OD = a sec ^ and EM=h tan ^ are the coordinates of the point P on the curve. The points P, P, and P" are called corresponding points. As the radius OP revolves about 0, the points P and P" move on their respective auxiliary circles, and P describes the ellipse in Fig. 47 and the hyperbola in Fig. 48. EXERCISES 1. Write the parametric equations and coustruct the ellipse whose semi- axes are 3 and 4. 2. "Write the parametric equations and construct the hyperbola whose semiaxes are 3 and 4. 3. Construct the following loci by assigning arbitrary values to the parameter t and tabulating the corresponding values of x and ?/ : (a) x = «-l, ?/ = 4-<2; (J)-) x = ~, y = -; (c) z = St, y -3f^ - t^. 4. With e as the parameter, construct the locus a; =: 6 cos 61 + 3 cos 2 6, 2/ = 6 sin 6* — 3 sin 2 0. This locus is called the three-cusped hypocycloid. 76 LOCI AND THEIR EQUATIONS [Chap. IV. 62. Recapitulation. Locus OR Curve Polar Equation Parametric Equations The circle. r = 2 a cos ^. X = a COS t, y = a sin t. The ellipse. «^ - c-^ _ ^ a> c. X = a COS t, y = b sin t. a — c cos ^ The hyperbola. ff — c cos 6 a<_c. x = a sec t, y = b tan t. The parabola. 2;j _,. 4p 1 — cos ^ EXERCISES 1. Find the lengths of the axes, the distance between the foci, and the eccentricity of each of the following cnrves. (a) 9 2/2 + 4 a;2 = 36. (c) 100 2/2 -25x2 ==-2500. (e) 64 2/2 + 25 x^ = 1600. 2. Show that the points (— 4, (2, — 1) lie upon two straight lines. (6) 7 x2 + 11 2/2 = 15. ((?) 17x2-25 2/2 = - 116. (/) 64 2/2 -25x2 =-1600. -2), (2, 1), (-6, 3), (0,0), and What are the equations of these lines ? 3. The semiaxes of an ellipse are 6 and 4. Find the length of the latus rectum. 4. Write the polar equation of the hyperbola, if the transverse axis is 6 and the distance between the foci is 10. For what values of 6 is r infinite ? 5. If the perpendicular to the major axis of an ellipse at the point D meets the major auxiliary circle in P and the ellipse in P', prove that DP : DP' -.-.a-.b, where a and b are the semiaxes. 6. In geometry it is shown that the areas of rectangles having the same width are to each other as their lengths. Combining this proposition with that in the preceding exercise, show that the area of the major auxiliary circle is to the area of the ellipse as a is to b, and hence the area ofthe ellipse is Trab. 7. If the major auxiliary circle is rotated around the major axis of the ellipse until its plane makes an angle whose cosine is — with the plane of a the ellipse, and if perpendiculars be dropped from every point of the circle upon the plane of the ellipse, show that the feet of these perpendiculars lie upon the ellipse. CHAPTER V EQUATIONS AND THEIR LOCI 63. Locus of an equation. The curve which passes through all the points whose coordinates satisfy a given equation, and through no other points, is called the locus of the given equation. 64. A second fundamental problem. In the last chapter we have found the equations of a number of important loci from given laws. There now arises a second fundamental problem of analytic geometry ; namely, given an equation connecting the varia- bles X and y, to construct the locus of the equation and to discover the important properties of the locus. In simple cases the general form of the locus can be determined by plotting (Art. 27). But this method alone often fails to reveal the important properties of the locus, and, at best, leaves wholly undetermined the form of the locus between consecutive points plotted. A discussion of the given equation, however, will reveal certain properties of the locus which, together with a few plotted points, will determine frequently the form and nature of the locus. 65. Discussion of an equation. The method to be followed must depend upon the particular equation under discussion, but the following outline will serve to indicate what to look for in any given case. (a) Symmetry (cf. Art. 29). (1) If the given equation contains only even powers of y, the locus is symmetrical with respect to the X-axis. For then, if P(a, b) is any point on the locus, Q(a, — b) is also on the locus. (2) If the given equation contains only even powers of x, the locus is symmetrical with respect to the I^axis. For then, if P(a, b) is any point on the locus, Q(—a, b) is also on the locus. (3) If the given equation contains only even powers of x and of 78 EQUATIONS AND THEIR LOCI [Chap. V. y, the locus is symmetrical with respect to the origin. For then, if P(a, 6) is any point on the locus, Q(— a, —6) is also on the locus. (4) If the given equation is unaltered when x and y are inter- changed, the locus is symmetrical with respect to the line bisect- ing the first and third quadrants. For then, if P{a, h) is any point on the locus, Q(h, a) is also on the locus (cf. Art. 39). (6) Intercepts (cf. Art. 30). The determination of the inter- cepts furnishes a good point from which to begin the construction of the locus. (c) Limits of the locus. It frequently happens that to certain values of either variable there correspond no real values of the other. There is no corresponding real point in such a case. Hence, tJie locus is confined to those prxrts of the plane such that to each value of either variable there corresponds a real vahie of the other. For example, the locus of the equation y^ = 2 x (Art. 28) is confined to the part of the plane to the right of the T'-axis. Whenever to any value of either variable there corresponds no real value of the other, the locus is said to be limited. The equa- tion of a limited locus, if it is algebraic, must establish at least one of the variables as a multiple-valued function of the other. For in no other way can imaginary values appear. The converse does not hold, for an equation which establishes one variable as a multiple-valued function of the other does not necessarily have a limited locus. Thus, the equation y- — x"^ establishes ?/ as a multiple-valued function of x, but for no value of either variable is the other ever imaginary. (rf) Change of one variable due to a given variation of the other (Art. 27). It is important to determine from the equation how increasing or decreasing one variable will affect the other. For example, if x is allowed to increase in value, will y increase or decrease in value ? In other words, to determine whether ?/ is a monotone function or not ; and, if not, for what values of x it is increasing, for what values it is decreasing, and, if possible, for what values of x it has turning points. (e) Behavior of the locus at great distances from the origin. It is also important to determine from the equation how in- creasing either variable indefinitely will affect the other. Art. 66] EXAMPLE I 79 The discussion of an equation according to the foregoing out- line will be illustrated in the following examples. 66. Example I. Discuss the equation x'^ + 4 ?/"' = 4 and construct tlie locus. (a) Assume any point P{a, b) whose coordinates satisfy the given equa- tion (Fig. 49). Then the coordinates of the points Q(a, — b), S{~ a, 6), Fig. 49 and i?(— ff, — b) also satisfy the equation. Hence the locus is symmetrical with respect to both axes, and also with respect to the origin. (b) The x-intercepts are ± 2 and the 2/-intercepts are ± 1. (c) Solving the equation for y, we have y V4 — x^ (1) from which it is seen that x is limited to the range of values extending from — 2 to + 2 in order that y may have real values. The range of values for x is indicated by writing -2g.>-^ + 2. The locus is thus limited to lie between the lines x = lines AB and CD in the figure. Again, solving the equation for x, we obtain a;=±2Vl r — 2 and x = + 2, or the (2) Hence y is limited to the range -l^y x > 2 :^ "J../ . . ;^: . : ; :~ - '\ l^' ■-■■--. -^' i — : , : : M i _j_._^... - -^S5^:M^ -— -;'- _X UL 1 M-^^ _ ^_ ! " ; """"^ ^ ~X ---'i'f'iiw LJ-1^ =-=;— ^ — - — K_p!r Wn ^ —/ III I^^T^^^ -■=^ i^v^--- "^^ \ ====^^i --H-+t — -^1 >~v^^^ ' '"'""r"""Tli " .": "' ^ ' - :-■; ■ 1 j- ^" — -+ -|--|--+-|--^-r|-|-|--|-|-| H l- -H y: I-4---I4-I-- Fig. 50 In order that rj may have real values. The locus, therefore, lies outside the strip bounded by the lines x = — 2 and x = + 2 (Fig. 50). Solving the equation for x, we have X = ± 2 Vl + 2/-. Arts. 67, 68] EXAMPLE III 81 Hence x is real for all values of y. The locus is therefore unlimited in the y-direction. {d) From equation (1), as x increases from 2, the positive value of ij increases continually and %Yithout limit, but as x increases from a large negative value to — 2, the positive value of y continually decreases to zero. Combining these facts with the symmetry in («)i "we conclude that the locus spreads out as it recedes from the origin in either direction. (e) As X increases indefinitely, the values of y approach nearer and nearer to ± -. For the radical Vx- — i is clearly always less than x in value, but for very great values of x, the difference _ ^yO. _ 4 =: . X + Vx- — 4 is very small and can be made as small as we please by choosing x sufficiently great. Therefore, at great distances from the origin, the locus lies close to the straight lines y=±ry These lines are called asymptotes. The X-axis bisects one of the angles between the asymj^totes and the locus lies within this angle, one branch on each side of the origin. The curve is an hyperbola. EXERCISES 1. Discuss the following ecpiations and draw the corresponding loci : (a) 4 x2 + 9 y2 = 36. (6) 4 a;2 - 9 2/2 = 36. (c) y^ = 16 x. {d) x^ = dy. (e) a;2 -1/2^4. (/) a;2 + ?/2 = 4. (g) ?/ = 4 x^. 2. Find the lengths of the axes, the distance between the foci and the eccentricity of the ellipse in exercise 1. 3. Find the lengths of the axes, the distance between the foci, and the eccentricity of the two hyperbolas in exercise 1. 4. Find the equations of the asymptotes for each of the hyperbolas in ex- ercise 1. 5. Find the coordinates of the focus for each of the parabolas in exer- cise 1. 68. Example III. Discuss the equation xy — x — y = and find the form and propeities of the locus. (a) The locus is obviously not symmetrical with respect to either axis nor with respect to the origin. It is, however, symmetrical with respect to a line bisecting the first and third quadrants, since if P(«, 6) is any point whose coordinates satisfy the equation, then the coordinates of the point ^(6, «) 82 EQUATIONS AND THEIR LOCI [Chap. V. also satisfy the equation. The points P and Q are symmetrically situated with respect to the line OA (Fig. 51). (5) The locus crosses the coordinate axes only at the origin. The inter- cept on each axis is therefore zero. Fig. 51 (c) The locus is not limited in either direction, since each variable is real for all values of the other. (d) Solving the equation for y, we have 2/ = (1) Hence, as x increases from a large negative value, y continually decreases from a value less than 1, through zero, becoming — co for x equal to 1. As X passes the value 1, y changes suddenly to a very great positive value and then continually decreases, approaching nearer and nearer to 1. The func- tion y is therefore monotone. It has a discontinuity at a; = 1. (e) From equation (1) we see that y approaches nearer and nearer to 1 as X increases or decreases indefinitely. For very great values of x, there- fore, the locus lies close to the line y = 1. This line is an asymptote to the curve. Similarly, solving the equation for ,r, the line a: = 1 is found to be an asymptote ; that is, for very great values of y the locus lies close to this line. The above discussion enables us to form a fairly accurate idea of the locus before any plotting has been done. 69. Example IV. Discuss the equation y and properties of the locus. l + a;^ and find the form Arts. 69, 70] EXAMPLE V 83 (a) The locus is not symmetrical with respect to either axis, but it is symmetrical with respect to the origin, since if P(a, b) is any point on the locus, so also is the point Q(— a, — b) on the locus (Fig. 52). ..i\7- j_ [ 1 1 ■ 1 1 1 i • i ' I 1 1 1 : 1 ; ■ > , ■ 1 ' ■ L_i_^ LL^3_ "] 1 - -L ^_ /■ 1 1 / ^ \ -." --f-J L_U_Lj.-aJJ ^ Q^ ^'--1 L_LJ__L^__Jj -J-- — -^ 1 ' 1 1 1 1 ; ' X^ ' : , 1 1 1 ■ [ I .1 1 M 1 1 i Fig. 52 (6) The locus crosses the axes only at the origin. (c) The locus is unlimited in the a:-direction, since y is real for all values of X. But if we solve the equation for x, we obtain 2y and therefore y is limited to the range in order that x may have real values. Consequently the locus lies within the strip bounded by the lines ?/ = — | and ?/ = + i. (d) and (e). As x increases from zero to 1, y increases from zero to ^ ; and as x increases indefinitely from 1, y decreases slowly from i towards zero. Hence the function y has a turning point at x = 1 and its value there is |. Also for very great positive values of x, the locus lies close to the X-axis. Since the locus is symmetrical with respect to the origin, its form to the left of the origin is known as soon as its form to the right has been determined. We conclude, therefore, that the function has a turning point at X = — 1 and that the X-axis is an asymptote to the curve. 70, Example V. Discuss the equation y^ — ^ — ^'^ and find the form and properties of the locus. 3 a; -f 6 84 EQUATIONS AND THEIR LOCI [Chap. V. (a) The locus is clearly symmetrical witli respect to the X-axis ; it is not symmetrical with respect to the F-axis, since the equation contains odd powers of X. (b) For the purpose of discussion we will suppose that & is a positive number (Fig. 53). The locus crosses both axes at the origin and also crosses the X-axis at x = ft. 1 1 1 1 1 1 1 1 1 iijj ;:i: y ' ' ■ ■ i_ — • . : - y +hr r : t H ^::\ ;: -ii Kv. ^ -_ fi^^ :::? =::■ ;:;■ t. J: :::::::::::;:::::::::: " 1 1 " 1 -a^m- ,::. mf llfl::::;::::::::::::::::::::: Fig. 53 (c) From the equation we see that x is limited to the range -^^CW0<^^ Fig. 57 (cl) As 6 increases from 0° to 45°, the absolute value of r decreases from a to 0. Again, as 9 increases from — 45° to 0° the absolute value of r increases from to a. The locus consists, therefore, of- two loops as shown IB Fig. 57. Arts. 76, 77] TRANSFORMATION OF THE AXES 91 EXERCISES 1. Discuss the following equations and construct the corresponding loci ; (a) r — a cos 3 6. (6) r = a sin 3 0. (c) r = 1 + cos 6. (d) r=- 1 (e) r = 1 — cos 6. 1 + cos d 2. Discuss the equations r = a cos nO and r = a sin nd for n an even integer ; for n an odd integer. What is the difference in the form of the curve ? 3. Discuss the equation r = a tan and draw the corresponding locus. 4. Change the equation in Example X, Art. 76, to rectangular coordi- nates and compare with Art. 54. What is the locus ? 62 5. Discuss the equation r - , first when c > a and then when a — c cos c <,a. What are the loci ? 6. Discuss the equation i- = 2 a sin tan and draw the locus. The carve is called the cissoid of Diodes. 7. Discuss the equation f- = — . The locus is the lituus. e 8. Discuss the equation r = a^. The locus is the logarithmic spiral. TRANSFORMATION OF COORDINATES 77. Transformation of the axes. The equation of a given locus can be simplified often by changing the axes to a new position in ;s jgF ra's: ^^=h -i=7^=t-D- -mi m ah 00: Fig. 58 the plane, and then finding the equation which the new coordinates of the points on the locus satisfy. The operation of changing the 92 EQUATIONS AND THEIR LOCI [Chap. V. axes is called transformation of coordinates, and the process of finding the new equation from the old is called transformation of the equation. When the new axes O'X' and 0' Y' are respectively parallel to the old axes OX and OY (Fig. 58), the transformation is called translation of the axes. Let the coordinates of the new origin 0', referred to the old axes, be h and k ; the coordinates of any point P, referred to the old axes, be x and y ; and the coordi- nates of P, referred to the new axes, be x' and y'. Then, from either of the positions of 0' shown in Fig. 58 (cf. Art. 3), we ^^^^® 0D= OE + ED and DP = DD' + D'P = h + x' =Jc + y'. Hence, x = h -\- x' and y = k -\- y'. (1) It is clear that equations (1) hold wherever the point 0' may be situated, provided the new axes have the same direction as the old. Example. Transform the equation y = 2x +3 by translating the axes so that the new origin shall be the point (1, 5). Here, for any point (x, «/) in the plane, and hence for any point on the locus of the given equation, x= 1 + x' and y = 5 + y'. Therefore, in terms of x' and y', the given equation becomes 5 -f- ?/' = 2(1 + x'), or y' = 2x'. (2) The given equation y = 2 x + 3 and the new equation y' = 2 x' repre- sent the same locus ; namely, the straight line shown in Fig. 59. The origin is, in the one case, at 0, and in the other, at 0'. 78. Rotation of the axes. When the origin is not moved, but the axes are each rotated through a given angle, the transforma- tion is called rotation of the axes. To obtain the equations for rotating the axes, let P be any point in the plane (Fig. 60) whose coordinates referred to the old Fig. 59 Art. 78] ROTATION OF THE AXES 93 axes are (x, y), and referred to the new axes are (x', y'). Let the angle XOX', through which the axes are rotated, be denoted by Fig. 60 e, the angle XOP by <^, and the angle X'OP by «. If OP = r, then X — r cos <^ = r cos (d -\- a) = r cos 9 cos a — r sin 6 sin a, y = r sin <^ = r sin (^ + «) = r sin 9 cos « + r cos ^ sin «. But r cos a — x' and r sin « = ?/' ; and therefore x = oc' cos 6—2/' sill 6, y = x' sin 8 + z/' cos 0. (1) These equations express the old coordinates of any point in terms of the new coordinates. To obtain the new coordinates in terms of the old, we can solve equations (1) for x' and y', or we can derive x' and y' directly from the figure. In either way we find , /I ■ /, X = X cos 9 + y sm 9, y' — y cos 9 — X sin 9. (2) Example. Transform the equation 24 xy — 7 y^ = 144 by rotating the axes through the acute angle whose tangent is f . 94 EQUATIONS AND THEIR LOCI [Chap. V. Here sin ^ = | and cos d= f , hence the equations for rotating the axes are X = f X' — y'. Substituting in the given equation and reducing, we have 9 a;'2 _ 16 m'2 = 144, or — - ^^ = 1. ^ 1(3 9 The given equation, therefore, represents an liyperbola whose semiaxes are 4 and 3. EXERCISES 1. Transform the equation 3 x + 7 y = 8 to a new set of axes parallel to the old set, and having the point (4, — 2) as origin. 2. Show that the equation x^ + y'^ = a^ ig unaltered by rotating the axes through any angle 0. What is the geometrical interpretation of this fact ? 3. Transform the equation x'^ — y^ = 10 by rotating the axes through an angle of 45°. 4. Transform the equation x^ + y^ = a^ by first translating the axes parallel to themselves, the new origin being at the point ( - , - | , and then rotating the new axes through the angle 45°. What is the locus of the resulting equation ? What is the locus of the original equation ? 79. Removal of terms of first degree. When the given equation is an algebraic equation of the second degree in the variables and contains terms of the first degree, the latter can often be removed by translating the axes to a new origin, as il- lustrated by the follow- ing example. Example. Given the equation 4 x^ + 9 ?/'- — 16 x — 18 y = 11. Translate the axes so as to remove the terms of first degree. Substituting for x and y their values in terms of x' and ?/', equations (1), Art. 77, we have 4(x' + ^0^ + 9(2/'+ *)^ - 1'^ + 9 ?/'- = 36. This equation represents an ellipse whose semiaxes are 3 and 2, hence the given equation represents this ellipse. Figure 61 shows the curve and both sets of coordinate axes. 80. Removal of the term in ocij. The term in xy can be removed from an equation of the second degree by rotating the axes through the proper angle. This is illustrated in Art. 78. As another example, we will remove the term in xy from the equation x2 -f 2 xy + 2 2/2 - 4 = 0. Substituting the values of x and y from equations (1), Art. 78, and collecting terms, the given equation becomes (cos2 ^ + 2 sin2 ^ + 2 cos 6* sin e)x'- + (2 cos 6 sin d + 2 cos^ — 2 sin^ e)x'y' + (sin2 Q ji^2 cos'- e — 2 sin 6 cos e)y''^ —4=0. (1) Putting the coefficient of x'y' equal to zero, we have the equation 2 cos d sin + 2 (cos2 6 — sia? 6) = 0, from which to determine 6. But this equation is equivalent to sin 2 ^ + 2 cos 2 ^ = or tan 2 ^ = — 2. Therefore 2 6* = arc tan (— 2) = 116"' 34', nearly, or ^ = 58° 17', nearly. Since tan 2 fi" = - 2, 2 — 1 we have sin 2 ^ = , cos 2 ^ = - Hence, cos^ d V5 V5 1 + cos 2 g _ \/.5— 1 ^ 2 ~ 2V5 1 — cos 2 g _ V'5 + 1 _ V2 2\/5 sin 2 ^ 1 and sin 6 cos 9 = 2 V5 Substituting these values in (1), it reduces to 3 _ V5 3 + V5 Hence the given equation represents an ellipse. 96 EQUATIONS AND THEIR LOCI [Chap. V. EXERCISES 1. Remove the terms of first degree and then the term in xy from the equations ^^^ xy-x-y = 0; (ft) xy-x + y = 0. What are the loci which these equations represent ? 2. Show that the terms of first degree cannot be removed from the equation ^g ^^ - 24:xy + 9y^ - 20x- 110 y + 225 = 0. Try to generalize this result so as to tell at a glance whether the terms of first degree can be removed or not from any equation of the second degree. 3. Given the equation x^ — 3 axy + y^ = 0. Rotate the axes through the angle 45° and compare the resulting equation with Example V, Art. 70. What locus does the given equation represent ? 4. In exercise 2, rotate the axes through the angle 6 = arc tan | and then translate the axes, taking for new origin the point (2, 1). What locus does the equation represent ? 81. Classification of algebraic curves. Theorem. The degree of an algebraic equation in the variables x and y is unaltered by transformation of coordinates. For, in transforming the equation by translation or rotation of the axes we replace x and y by expressions of the first degree in x' and y' and therefore the degree of the equation cannot be raised by this process. Neither can it be lowered, for then it would be necessary to raise the degree in transforming back to the original axes, and Ave have just seen that the degree cannot be raised by a transformation of coordinates. Since the degree cannot be raised or lowered by a transformation of coordinates, it must remain unaltered. Because of the theorem just proved, algebraic curves can be classified conveniently according to the degree of their equations, since we now know that the degree of the equation is independent of the position of the axes with reference to the curve. If the degree of a given algebraic equation is any integer n, the corresponding curve is said to be of order n. The straight line is the only locus of order 1 (Art. 47). The circle, the ellipse, the hyperbola, and the parabola are all loci of order 2. The folium of Descartes is a locus of order 3 (Art. 70). The ovals of Cassini are loci of order 4 (Art. 54). Art. 81] CLASSIFICATION OF ALGEBRAIC CURVES 97 The succeeding chapters will be devoted to a special study of loci of orders 1 and 2. MISCELLANEOUS EXERCISES 1. Show that the triangle whose vertices are (3, 2), (—1, —3), and (— 6, 1) is a right triangle. 2. On the line ?/ — 5 = a segment is laid off, having for abscissas of its extremities 2 and 5, and upon this segment an equilateral triangle is con- structed. What are the coordinates of its third vertex '? 3. Find the coordinates of the point dividing the segment (5, 2) to (4^ _ 7) in the ratio 4 : 7. 4. Change the polar equations = a -| to one in rectangular coordi- cos^ nates. Plot the locus. 5. Remove the terms of first degree from the equation4a;--|-9 j/ — 8 y— 6=0 and plot the resulting equation. 6. Find the rectangular equations of the asymptotes of the hyperbola whose polar equation is ~ 4 cos 61 — 3 7. Discuss the equation y- = 4:x:^ — x^ and plot the locus. Write the parametric equations of the locus if y = tx. t being the parameter. 2 4- s^ 8. Discuss the equation y- =:x- and plot the locus. Write the para- 2 — X metric equations if y = tx. Ill 9. Write the parametric equations of the locus ofx- + y^ = a'^, assuming X — a COS'* d. 10. OB is the crank of an engine and AB the connecting rod, A being the piston. A moves in a straight line passing through 0. Find the equa- tion of the locus of any point P on the connecting rod. Let P be a units from A and 6 units from B, and let r be the length of the crank, OB. Dis- cuss the equation and plot the locus. Write the parametric equations of the locus, assuming y= a sin d. What is the locus when r = a + b? 11. Discuss the equation y-(a — x)— oc^(a + x)= 0, and plot the locus. The curve is called the strophoid. 12. Construct the locus of w = sin 2 x H ; of ?/ = e-^ + 4 x'-^ : of y = - ■ cos X. X 13. Discuss the equation r^ = a^ sin 2 6 and plot the locus. 14. Discuss the equation r^ = o?- tan Q and plot the locus. CHAPTER YI LOCI OF FIRST ORDER 82. Linear equations. We have seen (Art. 47) that the equa- tion of every straight line is of the first degree in x and y ; and conversely, that every equation of first degree in x and y is the equation of a straight line. For this reason, an equation of the first degree in x and y is called a linear equation. Every linear equation is of the form Ax + 5?/ + C = 0, (1) where A, B, and C are constants. These constants can have any values, except that A and B cannot both be zero, for then the equa- tion would contain neither variable. If A is zero, (1) is the equation C of a line parallel to the X-axis, for then y has the value for all values of x. If B is zero, (1) is the equation of a line parallel C to the F-axis, for then x has the value for all values of y. A -^ Finally, if C is zero, (1) is the equation of a line passing through the origin, for then the equation is satisfied when x = and y = 0. For A, B, C different from zero, the slope of the line is given by the formula j and the intercepts a, b on the X- and I'-axes by «= — -T and ^—~^' respectively. EXERCISES 1. Find the slopes and intercepts of the lines whose equations are the following : (a) x + VSy + 10 = 0. {b)y = x-6. (c) 5 X - 12 2/ = 13. (d) 2 X - 3 y = 4. 98 Arts. 82, 83] INTERSECTION OF TWO LINES 99 (e) a; - rt = 0. (/) 4 ?/ + 3 x = 24. (g) 5x + 4:y=20. (h) 2x ~ iy + 9 =0. (i) 2x + 3y=0. ij) 2/ =4. (k) Ax + By+ C = 0. (I) (a^ - b-^)x =(a + b)y + c. 2. If a and 6 represent the intercepts on the J"- and I^'-axes respectively, and m the slope, determine the equations of the lines for which (l)a = 2, &=-3. (2)a=-l.m = 4. (3) 6 =3, wi = -2. (4) m=-5, a=-2. (5) a =2, and passing through the point (4, — 3). (6) Passing through the points (— 1, 2) and (5, — 4). ^ ^ C C 83. Intersection of two lines. The coordinates of the point of intersection of two lines must satisfy the equation of each line, since the point lies on each line. To find the coordinates of the point of intersection it is only necessary, therefore, to solve the equations simultaneously for x and y. For example, x = 3 and 2/ = 4 is the common solution of the two equations x — y -\- 1 = and Ax + y — 16 = 0; and these are the equations of two straight lines which intersect in the point (3, 4). In general, two straight lines intersect in one and only one point. But they may be : (1) Parallel to each other. (2) Coincident, In the first case the slopes of the lines are equal and their equations have no common solution. For example, the equations 2 a; - 3 !/ = 4 and 2 a- - 3y = 7 (1) are the equations of a pair of parallel lines, since the slope of each is -|. The equations have no common solution. The equa- tions of a pair of parallel lines are called incompatible or inconsist- ent* Thus equations (1) are incompatible. Obviously 2 ;i;— 3 2/ cannot be 4 and 7 at the same time for any values of x and y. In the second case the slopes of the lines are also equal, but their equations have an indefinite number of common solutions, since any pair of values of x and y that satisfies one equation must also satisfy the other. The equations, therefore, can differ only * See Rietz and Crathorne, College Algebra, p. 49. 100 LOCI OF FIRST ORDER [Chap. VI. by a constaut factor. The equations of a pair of coincident lines are called dependent. Thus, the equations '2x — 3y = 4: and 4 a; — 6 ?/ = 8 are dependent and are the equations of a pair of coincident lines. EXERCISES 1. Find the intersections of the lines represented by the following pairs of equations. Tell which are inconsistent and which are dependent equations, (a) 2x + 3?/ = 12, 4x- ?/ = 5. (6) 3x + 5 ?/ = 1, 6a; + 10?/ + 7 = 0. (c) 5 X - 2 ?/ + 4 = 0, a; - .4 ?/ = - .8. (d) x + 3 ?/ = 15, 3 x - ?/ = 5. Draw the lines in each case. 2. Write an equation representing the same straight line as 5x+4i/— 20=0 in which the sum of the coefScients shall be 22 ; in which the product of the first and third coefficients shall be equal to the second. 3. Change the equation 3x — 4?/ + 12 = into another representing the same straight hne and having the square of the second coefficient equal to twice the third minus four times the first ; having the product of all three coefficients equal to minus three times the last. 4. The equations 5x — 2?/ — 3 = and Ax + By + C = are dependent and B^-\-2(A + C) = 24. Find A, B, and C. 84. The pencil of lines. Let m = and v = he the equations of two straight lines, then the equation u + fcv = (1) is the equatio7i of a straight line passing through the intersection of ic = and v = 0, ivhatever value is given to Tc. Here u and -v are each expressions of the first degree in x and y and therefore u -{■ hv = is the equation of some straight line (Art. 47). Moreover, u + A;v = is satisfied by the coordinates of the point of intersection of u = and v = and therefore it is the equation of a straight line passing through the intersection of u = and v = 0. If k is allowed to vary, a series of lines is obtained each pass- ing through the intersection of m = and v = 0. The totality of lines so obtained is called a pencil of lines (Fig. 62). The constant k can be determined so that the line u -\- kv = shall satisfy any single condition, such as passing through a given point, having a given slope, etc. Art. 84] THE PENCIL OF LINES 101 Example. Find the equation of the line passing through the intersection of 2x + 3y - 4: = and x + 2?/— 5 = and also through the point (2, 3). The line whose equation is required is one of the pencil 2x + Sy -4: + k{x + 2y-5)=0. (2) Since tlie line is to pass through the point (2, 3), these coordinates must satisfy the equation. Hence k=—3. Substituting this value of k in (2), we have the required equation, x + Sy-ll =0. This result may be verified by solving the given equations simultaneously and then finding the equation of the line passing through the common point and the point (2, 3) in the usual way.* Fig. 62 EXERCISES 1. What is the equation of the line passing through the origin and through the intersection of the lines x + Sy — 8 = and 4 x — 5 ?/ = 10 ? 2. The equations of the sides, of a triangle are 5 a;— 6?/ = 16, 4x + 5 2/ = 20, and z + 2y = 0. Find the equations of the lines passing through the ver- tices and parallel to the opposite sides. 3. Find tlie equation of the line which passes through the intersection of the lines 2 x — 3 y + 1 = and a; + 5 ?/ + 6 = and is perpendicular to the first of these lines. Which is parallel to the line 6x — y + 10 =0. 4. What is the equation of the line which passes through the intersection of the lines ?/ = 7 x — 4 and y =— 2 x + 5 and makes an angle of 60° with the positive end of the x-axis ? 5. Find the equation of the line which passes through the intersection of the lines by — 2x— 10 = and ?/-!-4x — 3 = 0, and also through the inter- section of the lines 10 2/-fx-f-2l = and 3y — 6x + 1 = 0. Suggestion. The equations 6y — 2x- 10 + k{y + 4x - 3) = and 10?/ -1- a; -|- 21 -|- k'(3y — 5x + 1) = must be dependent (Art. 83). * The theorem of this article holds when u and v are expressions of any degree in X and y. 2( + kv =0 is then the equation of a pencil of curves. 102 LOCI OF FIRST ORDER [Chap. VI. 85. The pair of lines. Let u = and v = be the equations of two straight lines, then the locus of the equation is the jKtir of lines m = and v = taken together. For the equation u ■ v = is satisfied only when one of the factors, u or v, is equal to zero or when both factors are equal to zero. But the two straight lines pass through all the points whose coordinates satisfy the equations u = and v = 0, and through no others. Consequently these lines, taken together, form the locus of the equation u • v—0 (Art. 63).* Example. The locus of the equation x- — y'^ = is the pair of lines X + ?/ = and x — y = 0. EXERCISES 1. Draw the pairs of lines whose equations are the following : (a) x^ + xy = Q. (b) 2x^ + 5xy - 3y^ = 0. (c) x^ — 5x + Q = 0. {d} 2 if - xy + 4:X - 9 y + i = 0. (e) x'^ - ?/ - 2 ?/ - 1 = 0. 2. Write the equation of the pair of lines each of which passes through the origin and whose slopes are respectively V3 and — VS. 3. Write the equation of the pair of lines each of which passes through the point (1, 2) and whose slopes are respectively 2 and — |. 86. The normal form. We have seen (Art. 57) that the polar equation of a straight line is r cos (6 — a) =p, where j9 is the length of the perpendicular from the origin on the line and « is the inclination of this perpendicular to the X-axis. Expanding cos {9 — a), the equation becomes r (cos 6 cos a + sin 6 sin a)=p. Since r cos 6 = x and r sin 6 = y, the equation in rectangular coordinates is . ... X cos a + 1/ sma = p. ■ (1) * The theorem of this article holds when %(■ and r are expressions of any degree in X and y. The locus oi u- v = is then called a composite curve. Arts. 85-87] REDUCTION OF Ax + By + C = 103 This is called the normal form of the equation of a straight line (Fig. 63). 87. Reduction of Ajc+Bi/ + C=0 to the normal form. The problem before us consists in reducing the given equation Ax + Bi/+C = (1) to the form xcosa + y sina—p = 0. (2) Since (2) is to be the equation of the same line as (1), the two equations can differ only by a constant factor (Art. 83). Hence we must have cos a = kA, sin « = JcB, and — 2> = kC, (3) where Jc is the constant factor. F.rom the first two of these equa- tions, by squaring and adding, Ave get Fig. (i3 l = k^(A-' + B'-), or k = Therefore ± VA^ + £2 cos a = , SlU a = B :, and — p C . . (4) In order to determine which sign shall be given to the radical in any numerical example, we shall assume that |> is always a positive number and then, from (4), the sign of the radical must he opposite to the sign of C. For example, to reduce 3a;+4?/ + 10 = to the normal form, we divide both members of the equation by — V9 + 1" = — 5 and obtain Fig. 6i Therefore p = 2, cos a = — 1^, sin « = — A, and a = 233° 8' (Fig. 64). 104 LOCI OF FIRST ORDER [Chap. VI. EXERCISES 1. Write the equations of the lines for which ; («) p = 5, a = 60°. (d) p = 0, a = 225°. (6) p = 5, a = 120°. (e) p = l, a = 45°. (c) p=-6, a = 330°. (/) p = 6, a =-60°. 2. Reduce the following equations to the normal form and plot the lines of which they are the equations : (a) 4x-3y -25. (b) :«• + 4 = 0. {c) x + 2ij -- 8. • (fZ)52/-3=0. {e)2x-y-0. (/) x - 3?/ + 4 = 0. 3. What system of lines is given by xcos a + y sin a —p = when a is constant and p varies ? When p is constant and a varies ? 4. Two lines can be drawn through the point (2, 5) and tangent to the circle x'^ + y^ = 25. Find the equation of each line. Draw the figure. 88. Distance from a line to a point. With the help of the normal form of the equation of a straight line, it is easy to find the length of the perpen- dicular DP drawn from a given line AB to a given point P(x„ y,) (Fig. 65). Thus, let the equation of AB, reduced to the normal form, be yS a^cos « + ?/sin a — ^:> =0. (1) Through P draw BS paral- lel to AB, and let 0M = jh be the length of the per- pendicular from to ES. Then the normal form of the equation of PS is ^(^vVi) Fig. 65 X COS u-\-y sin « — 2h — 0- (2) Since P is on RS, the coordinates Xj^ and y^ must satisfy (2). Hence, . Xi cos « -f ?/i sm « = 2h- ( m m (9) is a tangent to the parabola y^ = ijxv. As an example of the use of the foregoiug formulas, we shall find the equa- tions of the tangents to the hyperbola x2 _ 4 2/2 = 36 which have the slope f . Here a = 6, 6 = 3, and m = |. Substituting in (7), we find the required equations are 6 y = 5 X ± 2i. Again, to find the equations of the tangents to the ellipse 4 x2 + 9 2/2 := 36, which pass through the point (2, 3), use equation fo). Here a = 3, 6 = 2, and we are to find m so that the tangents pass through the given point. Hence we must have 3 = 2m± V9m2+4, from which we find - 6 ± V61 m = — 5 The required equations are therefore ,/_3 = ^l±^(.r-2). 5 The formulas (A), (B), (C), and (D) are of frequent use in what follows. Note that properties of the hyperbola, expressed by equations involving the semiaxes, can be derived from the corresponding properties of the ellipse by changing the sign of b^. Thus, equa- tions (B) and (C) differ only in the sign of b^. EXERCISES 1. Find the equations of the tangents to the following conies : («) 2/2 — 4x, slope = J. (d) x^ — 4?/ 2 = 36, passing through (&) x^ + y'^ — 16, slope = — f - the point (3, 4). (c) 9 a:2 + 16 y'^ = 144, slope — — \. (e) a;2 + 4y2 = 36, perpendicular to 6x — 4?/ + 9 = 0. 114 LOCI OF SECOND ORDER [Chap. VII. 2. Eind the equations of the common tangents to the follomng pairs of conies. Construct the figures. (a) 2/2 = 5 X and 9 x^ + 9 y^ = 16. (6) 9 x2 + 16 ?/2 = 144 and 7 x'^ -32y^ = 224. (c) x2 + 2/2 = 49 and 13 x^ + 50 2/2 = 650. Suggestion. Find the equations of tangents to each conic in terms of the slope and then determine the slope so that the two equations shall be depend- ent (Art. 83). 3. Prove that two tangents to the parabola y^ = 4px which are perpendic- ular to each other intersect on the directrix. Suggestion. The slopes of the tangents are negative reciprocals of each other. Hence their equations are y = mx + £- and y = pm. in m But these lines intersect in a point whose abscissa is — p, whatever the value of in. Construct a figure illustrating this exercise. 4. Prove that two tangents to the ellipse 1- ^ = 1 which are perpendic- ular to each other intersect upon the circle x"^ -\- y"^ = a^ -{- b^. Suggestion. The equations of the two tangents are y =zmx+ Vahn^ -\- b^ and y = - ~ + /:^/«i±E™!\ _ in \ in / The point of intersection must satisfy both these equations ; hence the equa- tion of its locus is found by eliminating m from these equations. To do this, remove the radicals from each equation by transposition and squaring. Thus, (y — inx)'^ = a^m'^ + &2 and (my -|-x)2 = ^2 _|_ 52,j^2_ Add these equations, member to member, and divide by the common factor 1 + in^. The circle x- + y^ = a'^ -i- 62 ig called the director circle for the ellipse. Construct a figure illustrating this exercise. 5. The locus of the intersection of a pair of perpendicular tangents to the hyperbola is called the director circle for the hjrperbola. Find its equation and show that it is a real circle only when a > & ; it reduces to a point for the equilateral hyperbola, a = b ; and is imaginary for acoordinate of the point of contact x= . A," Since the point of contact is on the line y = mx + k, we have y = mx-^k=—^ + k= = -. Therefore, the poUds of contact on the tangents having the slope m ahn b'^^ are T, j, ivhere k has the values ± ^ahn^ + &^. Similarly, for the parabola, making use of equation (D) and the corresponding value of k, the .c-coordinate of the point of con- tact is given by the equation m'^x^ — 2px + — = 0, 116 LOCI OF SECOND ORDER [Chap. VII. from which x=—. Substituting in y = mx -{-!<:, we find that y — . Therefore, the point of contact on the tangent having the slope m is [—, — \m? m EXERCISES 1. Show that the coordinates of the points of contact of the tangents to the hyperbola = 1, having the slope m, are( , ), where k has a?- y'^ \ k k I the values ± \'a'^n^ — b'^. 2. Show that the coordinates of the points of contact of the tangents to the circle x^ + y^ = a^, having the slope m, are | ~ ^ "^ , — ], where k has the values ±aVin^ + 1. 3. Find the coordinates of the points of contact of the tangents to the conies in exercise 1, Art. 95. 4. rind the coordinates of the points of contact of the tangents to the pairs of conies in exercise 2, Art. 95. 97. Equation of a tangent in terms of the coordinates of the point of contact. First method. Let P(x^, i/i) be the point of contact of a tangent to the ellipse ^ + ^ = 1. a^ b^ Then, by the preceding article, — a'^m J b^ X. = and v. = — . k ^ k Eliminating k by division, we have Vi _ — b"^ . _ y^x^ Xi a^m ' a'^yi Since the tangent passes through the point of contact, its equation is 2/ - 2/1 = m(^ - ^i) = ~ ^ (•^' ~ ^^y (^) Art. 97] COORDINATES OF POINT OF CONTACT 117 Clearing of fractions and remembering that IP-x-^ -\- a?yi = a^b^, since the point of contact is on the ellipse, equation (1) reduces to ^^1 I UUl 1. (2) For the parabola, if P{Xi, y{) is the point of contact of a tangent, we have seen that ?/i = -^. Therefore the slope of the tangent m 2p 2p is -~^, and its equation is y -yi = m{x — X,) (3) Qcxi^h, 2/^+t) Fig. 71 Clearing of fractions and remember- ing that yi^ = 4 j^x^, we have 2/!/i = 2i>(a? + a?i). (4) Second method. Let a secant meet a curve in the points P(x^, ?/i) and Q(% + h, ?/i + k) (Fig. 71), so that the projections of the segment PQ upon the X- and F-axes are respectively h and k. The slope of PQ is then -. The coordinates of P and Q satisfy the equation of the curve. Hence, if the curve is an ellipse as in the figure we have ^ + lll = l and (^i+M + (li±J^ = i, (5) Subtracting the first equation from the second, member from member, we obtain the equation from vs^hich we get 2 hx, + h' , 2 ky, + k^ ^ ^ a^ b^ k^ b\2x, + h) h a\2y, + k) (6) As the secant rotates about P (cf . Art. 95), the point Q approaches P along the curve and in the limit coincides with it, and then the secant becomes the tangent at P. But the slope of the secant is constantly equal to the right-hand member of (6). When Q coin- 118 LOCI OF SECOND ORDER [Chap. VII. cides with P, both h aud k are zero, and the right-hand member of (6) gives the slope of the tangent at P; that is, m = -. The equation of the tangent is then found as in tlie first method. EXERCISES 1. Write the equation of the tangent to the ellipse 3 x'-^ + 4y" = 19 at the point (1, 2). 2. Show that the equation of the tatfgent to the hyperbola - — ^—1 Or' b- at the point (xi, ?/i) is ?^_Mi= i. 3. Write the equation of the tangent to the hyperbola 2 x^ — ?/2 — 14 at the point (.3, — 2). 4. Find the equations of the tangents to the ellipse 16 x^ + 25 y- = 400 which pass through the point (3, 4). 5. Write the equation of the tangent to the parabola y^ = 6 x at the point (6, - 6). 6. Find the angle which the ellipse 4 x- + 2/'- = 5 makes with the parabola y'^ — 8 x at ?i point of intersection. Suggestion. Find the equation of the tangent to each curve at a point of intersection and then find the angle which one tangent makes with the other. 7. Show that the equation of the tangent to the circle x^ -fy^ = a- at the point (xi, yi) is xxi + yyi = a^. 8. Show that the length of a tangent to the circle x- + y^ = a-, included between the point of contact and the point (xo, 2/2), is Vx-r^ f- yz^ — a'^. 9. Prove that the circles whose equations are x'^ + y^ — 8 x + 4 ?/ + 7 = and x^ + 2/2 — 10 X — 6 2/ + 21 = intersect at right angles. Suggestion. Show that the square of the distance between the centers is equal to the sura of the squares of the radii. 10. Using the second method of Art. 97, find the equations of the tan- gents to the following curves at the points designated : (a) 2/2 = x3, at (4, 8). (6) y = x2(x- 1), at (2, 4). (c) ys = x% at (8, 4). (d) y^=x(x -l){x - 2), at (8, V6). Draw each curve, Arts. 98, 99] TANGENT LENGTH, NORMAL LENGTH 119 98. Normals. Given any curve, the line drawn perpendicular to a tangent at the point of contact is called the normal to the curve at the point of contact. Let P(.x"i, 2/i) ^6 the point of contact and m the slope of the tan- gent at P, then the equation of the normal is y -yi = {x-x,). (1) For example, the slope of the tangent to the ellipse ^-4-^ = 1 at the point Cx^, Wj) is — (Art. 97). Hence the equation of the normal at (.v,, y^) is (2) 99. Tangent length, normal length, subtangent, subnormal. Con- nected Avith every point on a curve there is a special triangle whose sides are respec- tively the tangent at the point, the normal at the point, and the X-axis. In Fig. 72, PTN'\^ such a triangle, where FT is the tan- gent at P, PN is the normal at P, and TN is the X-axis. PT is called the tangent length and PN the normal length. DP is the ordinate of P, PT is the projection of PT on the X-axis and is called the subtangent, 1)1^ is the projection of PiVon the X-axis and is called the subnormal. Let the coordinates of P be ccj, y^ and the inclination of the tangent be ^, then DP — y^ and the angle DPN — angle DTP. Therefore, dot i , i • PT = I yi cosec ^ \, PX= |?/i sec ^1, Fig. 72 120 LOCI OF SECOND ORDER [Chap. VII. Z>r=|2/iC0tc/>I, where the bars indicate positive, or absolute, value. EXERCISES 1. Write the equation of the normal to the circle at the point (a:i, yi). Note that the normal passes through the center of the circle. 2. Write the eqiiation of the normal to the parabola at point (xi, yi). The equation of the normal to the hyperbola at (xi, yi). 3. The point (3, 2) lies on the ellipse ofi + 4?/2 = 25. Find the tangent length, the normal length, the subtangent, and the subnormal, at this point. Construct the figure. 4. Find the equation of the normal to the parabola ?/- = 8 x which is parallel to the line 2a; + 3?/ = 10. 5. Prove that the normal to the ellipse or the hyperbola at the point (xi, 2/1) meets the X-axis at a distance e-xi from the center. 6. Show that the subnormal to the parabola y'^ — 4:px is constant and equal to 2 p. 7. The line 3 x + 8 j/ = 25 is tangent to the ellipse x'^ + 4 ?/2 =25. Find the coordinates of the point of contact and write the equation of the normal at this point. 8. The line mx — iy = 1 is tangent to the hyperbola ^ = 1. Find 9 4 m and compute the subtangent and subnormal for the point of contact. 100. Reflection properties. The three theorems that follow express what are known as reflection properties. Theorem I. The angle formed by the focal radii drawn to any point of an ellipse is bisected by the normal at that point. The equation of the normal at the point (x^, y{) is given in (2), Art. 98. From this equation we find the intercept on the X-axis is (Fig. 73) (a^-b^)x, c^x, a^e^x, „ a^ a^ a- But FO = OFi = ae, and therefore FN = ae + e^x^ and NF^^ = ae — e\. The ratio of FN to NF^ is, therefore, FN _a-\-ex^_ FP NF^ a - ex^ F,P (Art. 50). Art. 100] REFLECTION PROPERTIES 121 Hence, we have sliowu that the normal at P divides the base of the triangle PFF^ into two segments which are proportional to the adjacent sides. Therefore the normal bisects the angle FPF^ Fig. 73 A ray of light from either focus of an ellipse is reflected from the curve to the other focus. Theorem II. The angle formed by the focal radii drawn to any point of an hyperbola is bisected by the tangent at that point. The equation of the tangent at {x-^, y^) is a" b- Hence the intercept on the X- axis is (Fig. 74), 0T=-- We can now show that FT ^ PF TF, ~ PF, ' For, and (Art. 52) Fig. 74 FT = ae + - X, TF, = ae- PF = ex- 1 + a , PFi = ex^ 122 LOCI OF SECOND ORDER [Chap. VII. Hence the tangent at P divides the base of the triangle PFF^^ into two segments which are proportional to the adjacent sides, and therefore bisects the angle FPF^. Theorem III. Any tangent to the parabola if = 4: iix bisects tJie angle formed by the focal radius drawn to the point of contact and a line through the p)oint of contact jKtrallel to the X-axis. The equation of the tangent to the parabola at the point P(x„ y,) (Fig. 75) is (Art. 97) Hence, the tangent meets the X-axis at the point T = ( - x„ 0). Therefore TO = OD, and, if EG is the directrix and F the focus, E0= OF. Hence, TF = ED. But ED = FP (defi- nition of the parabola), and, consequently, TFP is an isosceles triangle. The angles PTF and FPT are therefore equal. If PG is parallel to the X-axis, the angles PTF and TPG are equal. Consequently, the tangent PT bisects the ana:le FPG. Fig. 75 EXERCISES 1. Two parabolas have a common axis and a common focus, and extend in opposite directions. Show that tliey intersect at right angles. 2. Given the focus and the vertex of a parabola, but not tlie constructed curve, show how to draw the two tangents through a given point P. Suggestion. Let F be the focus and A the vertex. Draw the line AF and construct the directrix. With P as center and PF as radius, draw a circle meeting the directrix in D and D\. The perpendicular bisectors of DF and D\F are the required tangents. The tangents thus constructed meet the perpendiculars to the directrix at D and Z>i in the points of contact. 3. Show tliat an ellipse and an hyperbola having the same foci intersect at riglit angles. 4. Having given the length of the transverse axis and the distance between the foci of an ellipse, or an hyperbola, show how to construct the tangents to the curve from a given point P. Arts. 101, 102] CONJUGATE DIAMETERS 123 Suggestion. Let AB be the given transverse axis. Locate tlie foci on AB at the points F and Fi. With P as center draw a circle through the nearer focus Fi, and witla F as center and AB as radius, a second circle meeting the first in the points D and Di. The perpendicular bisectors of DFi and DiFi are the required tangents. These tangents meet the lines FD and FDi in the points of contact. 5. Why is light emanating from the focus of a parabolic mirror reflected in parallel rays ? What use is made of this fact ? ^(•ri,y,) DIAMETERS 101. Definition. Any line through the center of a circle, an ellipse, or an hyperbola is called a diameter of the curve. Any line perpendicular to the directrix of a parab- ola is called a diameter of the curve. That di- ameter of the parabola which passes through the focus is the axis. The circle, ellipse, and hyperbola are called cen- tral conies ; the parabola has no center, and is there- fore called noncentral. Fig. 76 102. Conjugate diameters. If P (.^\, ,?/,) is any point on a central conic and PT is the tangent at P, then the diameter through P and the diameter parallel to PT are called con- jugate diameters. Thus, PO and QO are conjugate diameters (Figs. 76 and 77). Theorem. If m and m' are the slopes of a pair of conjugate diameters. then mm' = T-^, Fig. 77 according as the conic is an ellipse or an hypierhola. 124 LOCI OF SECOND ORDER [Chap. VII. In either Fig. 76 or Fig. 77, the slope of PO is V\ a-i ^^^ the slope of QO, = the slope of PT, = m' = T^\ according as the conic is an ellipse or an hyperbola. Consequently, ' -r b' mm = T —' cr Since the product of the slopes is independent of the coordinates of P, it follows, in case of the ellipse, that the tangent at Q is parallel to the diameter PO. EXERCISES 1. Given any diameter of an ellipse or an hyperbola, construct its con- jugate diameter. 2. The point ( "^-^ , 1 ) lies on the ellipse 4 x^ + 9 y^ = 36. Find the equa- tions of the diameter through the point and its conjugate diameter. 3. The point (xi, yi) lies on the ellipse — + ^ = 1. Show that the equa- a^ 52 tions of the diameter through the point and its conjugate diameter are, re- spectively, 2/ix-xi2/ = Oand^ + M = o. 4. The point (xi, wi) lies on the hyperbola^ — ^ = 1. Find the equa- tions of the diameter through the point and its conjugate diameter. 5. If a diameter of the ellipse — -\-^^ = 1 meets the curve in the point (xi, 2/1), show that the conjugate diameter meets the curve in the points _Ml,MUndf^,-^V b a I \b a ) 6. Prove that the sum of the squares of any two conjugate semidiameters of an ellipse is equal to the sum of the squares of the semiaxes. 7. Show that conjugate diameters of a circle are always at right angles to each other. 8. What is the relation between the slopes of conjugate diameters of the equilateral hyperbola (5 = a) ? Art. 103] THE LOCUS OF MIDDLE POINTS 125 9. Two chords are drawn from any point of an ellipse or an hyperbola to the extremities of a diameter. Show that the diameters bisecting these chords are conjugate diameters. Suggestion. Let the coordinates of the point be Xz and yn, and the coor- dinates of one extremity of the diameter be Xi and ?/i. The coordinates of the other extremity are then — xi and — y-^. Show that the product of the slopes of the diameters bisecting the chords is for the hyperbola. — , for the ellipse, and -| d^ d^ 10. Show that the area of a parallelogram inscribed in the ellipse 1-^=1 whose diagonals are conjugate diameters is 2 ah. or b'^ Suggestion. Let be the center of the ellipse, P(xi, yi) and 'Q, two adja- cent vertices of the parallelogram. The coordinates of Q may then be taken as ( — , —] (exercise 6). The ai-ea of the parallelogram is four times \ b a I the area of the triangle POQ. 11. Prove that the axes of an ellipse or an hyperbola form a pair of con- jugate diameters. Suggestion. As the slope of one diameter approaches zero, what does the slope of the conjugate diameter approach ? 12. Can a pair of conjugate diameters of an ellipse or an hyperbola ever be at right angles unless they are the axes ? Why ? 103. The locus of the middle points of a system of parallel chords. Theorem. The locus of the middle points of a system ofjmrallel chords of any conic is a diameter of the conic. Let y = mx + A." be the equation of any line meeting the conic in the points Pi and P.,- If the conic is an el- lipse, as in Fig. 78, the a^coordinates of the points Pi and P2 are the roots of equation (B), ^ig 78 Art. 95, and if x^ and X2 represent these roots, then the ic-coordinate of the middle point 126 . LOCI OF SECOND ORDER. [Chap. VII. of the chord P1P2 is given by the equation „j ^1 ~l~ ^2 But the sum of the roots of (B) is Hence, and since the middle point is on the chord, y' = mx -\- k = ahn"^ + 6^ Therefore, whatever value is given to Tc, the coordinates of the middle point, x' and y' , satisfy the equation y = -^~^- (1) But (1) is the equation of a straight line passing through the center and is therefore a diameter of the curve. The line PO represents this diameter. If li = 0, the line P1P2 assumes the position QS, which is also a diameter of the curve. Since the product of the slopes of PO and QS is — — , these diameters are conjugate to each other. a- Combining this result with the theorem in the preceding article, we conclude that all the chords parallel to PO are bisected by QS. The theorem is proved for the other conies in a similar way, making use of equations (A), (C), and (D) of Art. 95. EXERCISES 1. Find the equation of the diameter of the hyperbola x^ — 8 i/^ = 96 bisecting all the chords parallel to the line 3 a; — 8 2/ = 10. 2. Find the equation of the diameter of the parabola y- =.Qx bisecting all the chords parallel to the line x + 3 ;/ = 8. 3. What is the equation of the chord of the ellipse 9 x^ + 36 ?/2 = 324 which is bisected by the point (4, 2) ? 4. Find the equation of the chord of the ellipse 13 x^ + 11 ifi = 113 which passes through the point (1, 3) and is bisected by the diameter 2 y = 3 x. Art. 104] DEFINITIONS 127 5. What is the equation of the chord of the parabola y^ = 6 x which is bisected by the point (4, 3) ? 6. Find the equation of the diameter of the hyperbola — _ ^ = 1 which bisects all the chords of slope m. ^ 7. Prove that the diagonals of any circumscribing parallelogram to an ellipse form a pair of conjugate diameters of the curve. ScGGESTiox. Let m and n represent the slopes of a pair of adjacent sides of the parallelogram. Show that the slopes of the diagonals are mki — nk ■, mk] + nk k\ — k A'l + k where k'^ = d-m- + i>- and k{^ = aP-n- + h-. The product of the slopes is therefore . The diagonals pass through the center of the ellipse and are therefore diameters. 8. Prove exercise 9 of the preceding article by showing that the diameter bisecting one chord is parallel to the other. POLES AND POLAR LINES 104. Definitions. It has been shown that the equation of a tangent to a conic can be expressed in terms of the coordinates of the point of contact. For example, we saw in Art. 97 that the equation of the tangent to the parabola at the point P(x^, y^) is yi/, = 22y(x-{x,). (1) This equation is the equation of a straight line, whatever values are given to x^ and yi, and is the equation of a tangent to the parabola only when the point P{x^, yx) is on the curve. In general (1) is the equation of a straight line called the polar line of P{xi, yi) with respect to the parabola y'^ = 4,px. The point Pix^, y-^ is called the pole. If the pole is on the curve, the polar line is tangent to the curve at the pole. Similarly, the equation of the polar line of any point with respect to any conic can be written at once. For example, the equation of the polar line of P{X], y^) with respect to the ellipse is We are led, then, to the following definition : 128 LOCI OF SECOND ORDER [Chap. VII. Tlie polar line of P {x^^, y^) with respect to a given conic is that line ivhose equation has the same form as the equation of the tangent to this conic when P (x^^, ?/i) is thep)oint of tangency. As an example, we will write the equation of the polar line of (1, 3) with respect to the circle x'^ + y^ = 4. Here the equation of the polar line of any point (xi, ?/i) is ^xi + 2/2/1 = 4. Hence, the polar line of (1, 3) is X 4- 3 2/ = 4. The student should draw the figure illustrating this example. As a second example, find the coordinates of the pole of 5 x — 4 ?/ + 20 = with respect to the ellipse + yA X^ 25 16 Here the polar line of any point (•^■i, 2/1 ) is x:/;i 25 16 Fig. 79 ^ = 5^^ y^ = - 25 16 If this equation and the given equa- tion are the equations of the same straight line, we must have (Art. 83) 4 k, and - 1 = 20 k, where k is the common ratio. From these equations, we find that Xi =— ^^ and 2/1 = V- Hence the required pole is the point ( — -\5.^ J/) (Fig. 79). EXERCISES 1. Write the equation of the polar line of each of the following points : 1. (1,-2) with respect to x^ + 4 y^ = 16. 2. (6, — 4) with respect to j/^ = 4 x. 3. (— 2, 2) with respect to 5 x^ — 8 y^ = 24. 4. (2, — 3) with respect to 5 x2 + 4 ?/2 = 10. 2. Find the coordinates of the pole of the line 3 x — 2 y = 5 with respect to the circle x'-^ + 2/^ = 25. 3. What are the coordinates of the pole of 5 x + 4 y = 7 with respect to the ellipse x^ + 2 ?/2 = 10 ? 4. Find the coordinates of the pole of the line x — ?/ = 10 with respect to the parabola 2/^ = 8 x. Akt. 105] GEOMETRIC PROPERTIES OF POLES 129 5. What are the coordinates of the pole of the line Ax + By + C = with respect to the hyperbola - — ^ = 1 ? 6. Through the point (.ri, yi) a line is drawn parallel to the polar line of the point with respect to the ellipse ^ + ?^ = 1. What are the coordinates of the pole of this parallel ? a b^ 105. Geometric properties of poles and polar lines. A point is outside a conic when two tangents can be drawn from the point to the conic. A point is inside a conic when no tangents can be drawn from it to the conic. Theorem I. If the jjoint (x-^, 2/i) is outside a conic, its polar line with respect to the conic passes through the points of contcLct of the tangents draivn from the point. Let the conic be the ellipse; the proof is similar for the other conies. Let Bix^, 1/2) a^^cl C(xs, y^) be the points of contact of tangents drawn from A(x^, y^ (Fig- 80)- The equations of the tangents at B and C are, respectively, ^ + m = l and^ + # = l. a^ Since the tangents pass through A, the coordinates of A satisfy each of the above equations. Hence, ^2 I .M? ce ¥ 1 and ^j^3 ^ M? = 1. a" 62 But these same equations result from substituting the coordinates of the points B and C in the equation of the polar line of A with respect to the ellipse. Consequently the polar line of A passes through B and C. 130 LOCI OF SECOND ORDER [Chap. VII. Exercise. Prove Theorem I for each of the other three conies. Theorem II. If P and Q are two points in the plane such that the polar line of P icith respect to a given conic passes through Q, then the polar line of Q jJctsses through P. Take the parabola y^ = 4pa; (Fig. 81) for the given conic. A similar proof establishes the theorem for the other conies. Let the coordinates of P be Xi, i/i and the coordinates of Q, X2, 2/2- The equation of the polar line of P is, then, Fig. 81 Since this line passes through Q, we 2/22/1 = 2p(a;2 + a-i). But this same equation results from substituting the coordinates of P in the equation of the polar line of Q. Hence, the polar line of Q passes through P. Exercise. Prove Theorem II for each of the other three conies. Theorem III. If a line through the pole A meets the p>olar line in C and the conic in B and D, then AB^ AD BC DC Let the coordinates of A (Fig. 82) be x^, y^ and the coordinates of G be x^, 2/2. The coordinates of the point B, dividing the seg- ment AC in the ratio AB : BC= r, are Xi + 7U*2 l + r But B lies on the conic, the figure, and y Ih + 'ilh l + r (Art. IT). Hence, if the conic is an ellipse as in (x-i + rx^y- , (yi + ry.y ^ ^ a\l + rf h\l + rf Expanding and arranging according to the powers of r, we have 1 2/2^ _ -J^ V.2 _j_ 2!^^ _|_ 2/1 ?/2 iV + + 1 Wo. Art. 105] GEOMETRIC PROPERTIES OF POLES 131 We should have been led to the same equation had we taken AB:DC=r. Hence the roots of this equation are the ratios in which the curve points divide the segment AC. But, since C is Fig. 82 on the polar line of A, the coefficient of r vanishes. Hence the roots are equal but opposite in. sign ; that is, AB^ AD BC ~ DC Four points A, B, C, and D situated on a straight line and such constitute a harmonic range. The segment AC that^ = -^i^ BC DC is said to be divided harmonically by B and D. Exercise. Prove Theorem III for the parabola and for the circle. Theorem IV. The polar line of a focus of a conic is the corre- sponding directrix. We shall establish the theorem for the case of an ellipse, leav- ing the remaining cases as an exercise for the student. The coordinates of the right-hand focus of an ellipse are x = ae and y = 0. Substituting these for x^ and ?/i in the general equation of the polar line with respect to the ellipse, we have, as the polar line of the focus, a X — -• 6 132 LOCI OF SECOND ORDER [Chap. VII. But this is the equation of the right-hand directrix. Exercise. Prove that the directrix of a parabola is the polar line of the focus ; that either directrix of an hyperbola is the polar line of the corre- sponding focus. Theorem V. Tlie line joinmg a focus of a conic to the intersec- tion of any tivo tangents, bisects one of the angles formed by the focal radii draivn to the points of contact of the tangents. In Fig. 83, let B and D be the points of contact of tangents from Q, F being a focus. Draw the line BD and let it meet the directrix corresponding to F in A, and the line QF in C. Draw the lines BE and DG perpen- dicular to the directrix. Since F is the pole of the directrix (Theorem IV) and Q is the pole of the line BD (Theorem I), it follows that QF is the polar line of A (Theorem II). Therefore we have the following equations : (Theorem III, CD = - DC), . Fig. 83 AB BC EB AB BF EB AD CD GD AD DF GD (Similar triangles), (Eccentricity, property A, Art. 94). Equating the product of the right-hand members of these equa- tions to the product of their left-hand members, we obtain BF^DF BC CD' Hence the point C divides the side BD of the triangle BFD in the ratio BF : DF, and consequently the line QF bisects the angle BFD. Art. 106] ASYMPTOTES OF HYPERBOLA 133 EXERCISES 1. Show how Theorem II can be used to construct the pole of any line with respect to a given conic. SuGGESTiox. Construct the polar line of any two points on the given line. "Where do these intersect ? 2. Two lines are drawn through a point P. The poles of these lines with respect to any conic are the points R and Q. Show that BQia the polar line of P with respect to the same conic. 3. Given any two lines in the plane such that the first passes through the pole of the second with respect to any conic. Show that the second passes through the pole of the first. 4. Prove that the intereection of any two tangents to an ellipse or an hyperbola is equidistant from the four focal radii that can be drawn to the points of contact. 5. Show that the intersection of any two tangents to a parabola is equi- distant from the focal radii to the points of contact and the diameters through the points of contact. 6. In Fig. 75, Art. 100, let PO meet the directrix in J/, and PT meet the vertical tangent in Q. Show that QF '\s the polar line of M. SYSTEMS OF CONICS 106. The asymptotes of the hyperbola. It has already been noticed that the hyperbola has asymptotes (Art. 67). This is a characteristic property of the hyperbola. No other conic has asymptotes. Solving the standard form of the equation of an hyperbola for y, we have y=±- ■\/x^ — al a Hence, as x increases indefinitely, y approaches nearer and nearer to the values ± — • Therefore a !/=±~ (1) are the equations of the asymptotes. The equations of the asymptotes can- be written ^-^=0, (2) 134 LOCI OF SECOND ORDER [Chap. VII. since the coordinates of any point on either asymptote will satisfy equation (2) (cf. Art. 85). The equation of a tangent to the hyperbola in terms of the slope is (Art. 95, Eq. 7) y z= mx + -yjahii^ — h"^. (3) If, in this equation, m is taken as the slope of either asymptote ; namely, ± -, the equation becomes y = ± — For this' reason, a « the asymptotes are often spoken of as tangents to the Jiyperbola, the points of contact being infinitely distant. Since an asymptote passes through the center, it is a diameter of the hyperbola (Art. 101). The product of the slopes of a pair of conjugate diameters of the hyperbola is — (Art. 102), and therefore each asymptote is its own conjugate diameter, or in other words, an asymptote is a self-conjugate diameter of the hyperbola. 107. Conjugate hyperbolas. The two hyperbolas a2 -t'^1, and^-r = _l ' ¥ ce b- are called conjugate hyperbolas. The transverse and conjugate axis of the one are respectively the conjugate and transverse axis of the other. Either hyperbola is conjugate to the other, but it Fig. 84 Arts. 107, 108] CONCENTRIC HYPERBOLAS 135 is convenient to speak of the first as the primary, and the second as the conjugate, liyperbola. The foci of the conjugate hyperbola are on the J'-axis, and, since c = Va^ + h"^ is the same for each hyperbola, the four foci lie on a circle of radius c and center at the origin (Fig. 84). The eccentricity of the conjugate hyperbola differs from the eccentricity of the primary hyperbola. The former is - and the h latter is -• a The asymptotes of the conjugate hyperbola coincide with the asymptotes of the primary hyperbola. For, from the equation of the conjugate hyperbola, we have ^ = ±-V.^•2 + Hence, as x increases indefinitely, the curve approaches nearer and nearer to the lines y of the primary hyperbola. and nearer to the lines y = ±-^' But these are the asymptotes EXERCISES 1. Show that the foot of the perpendicular from a focus of an hyperbola on either asymptote is at a distance a from the center and h from the focus. 2. Show that the circle of radius, h, whose center is at a focus of an hyperbola, is tangent to the asymptotes at the points where they cut the corresponding directrix. 3. Show that the product of the perpendiculars let fall from any point of an hyperbola on the asymptotes is constant. 4. Write the equation of the hyperbola conjugate to 9 x"^ — ?/2 = 9, and find the lengths of its semiaxes, its eccentricity, the coordinates of its foci, and the equations of its directrices. 5. If e and e\ are the eccentricities of two conjugate hyperbolas, show that — I = 1. Also ae = bei. 6. What is the eccentricity of the equilateral hyperbola ? Of the con- jugate to the equilateral hyperbola ? 108. The system of concentric hyperbolas. The equation. 136 LOCI OP SECOND ORDER [Chap. VII. where Ji is a variable parameter, is the equation of a system of concentric hyperbolas (Fig. 85). All the hyperbolas contained in the system have the same asymptotes, as can be shown by solving (1) for y and then allowing x to increase indefinitely. If A; is a negative number, (1) is the equation of a conjugate hyper- bola. As A' increases in value, the hyperbola approaches the asymptotes closer and closer, and coincides with them when Jc is Fig. 85 zero (Art. 106). When k is an increasing positive number, the hyperbolas are all primary and recede farther and farther from the asymptotes. The contour lines about two contiguous mountain peaks form a rough approximation to such a system of hyperbolas. 109. The system of confocal conies. foci are called confocal. The equation Conies having the same + y a'-k b'^-k 1, (1) where A; is a variable parameter, is the equation of a system of con- focal conies (Fig. 86). For, suppose a > b, then for every value of k a^, equation (1) is satisfied for no 'real values of x and y. In this case, (1) is said to be the equation of an imaginary ellipse. Each hyperbola of the system cuts every ellipse at right angles and vice versa (Exercises, Art. 100). If two sets of curves are so related that each curve of either set intersects all the curves of the other set at right angles, the two sets of curves are said to be orthogonal to each other. Thus the hyperbolas and ellipses of the system of confocal conies form two sets of curves orthogonal to each other. Sets of orthogonal 138 LOCI OF SECOND ORDER [Chap. VII. curves are of great importance in mathematical physics, since they represent fields of force. EXERCISES 1. What are the equations of the two conies of the system (9 _ A) (4 - ^•) which pass through the point {^2L^, X^\ ? 2. Show that the equation — + ^ = ^- is the equation of a system of con- centric ellipses, k being a variable parameter. Discuss the equation for various values of k. 3. Show that the equation y'^ = 4 k(x + k) is the equation of a system of confocal parabolas. Discuss the equation for various values of A;. 4. What system of conies is given by the equation — + ^ = 1, k being a variable parameter ? Show that the x-intercept of a tangent to any one of these conies is independent of k. How can this fact be utilized to construct the tangent at any point on one of the conies of the system ? 5. Discuss the system of circles given by the equation x^ + ?/2 — ffl2 _ 2 A;y = 0, k being a variable parameter. 6. Discuss the system of circles given by the equation x2 + 2/2_^(5[2_ 2 mx=0, m being a variable parameter. 7. Show that the two systems of circles in exercises 5 and 6 form two sets of circles orthogonal to each other. Draw a figure illustrating this exercise. MISCELLANEOUS EXERCISES 1. Find the equations of the tangents to the ellipse x^ + 4 y- = 16 which pass through the point (2, 3). 2. Find the equations of the tangents to the hyperbola 2ofi — Zy'^ — 18 which pass through the point (4, — Vo). 3. Find the coordinates of the points of contact of the tangents in ex- ercises 1 and 2. 4. For what value of k \s y = 2x ■\- k a tangent to the hyperbola x2 - 4 2/2 := 4 ? 5. For what value of m isy = m:c + 2 a tangent to the ellipse x^ + 4 j/^ = 1 ? Art. 1091 SYSTEM OF CONFOCAL CONICS 139 6. What relation connects A, B, and C, if Ax + By + C = is a tangent to the parabola ?/-^ = 4 x ? 7. Are the following points on, inside, or outside the hyperbola 4 x2 - 2/2 = 4 y («) (1, 3), (5) (2, 1), (c) (3.25, 3). 8. The coordinates of one extremity of a diameter of the ellipse — + ^ = 1 are xi — a cos di and yi-h sin ^1. Show that the coordinates of a- &'^ one extremity of the conjugate diameter are given by the equations X2 = — a sin di and y2 — h cos di. 9. Show that the segments of any line contained between an hyperbola and its asymptotes are equal in length. 10. Find the equation of the tangent to the parabola ij"^ = 4:px which has equal intercepts. 11. The earth's orbit is an ellipse whose eccentricity is .01677 and whose major semiaxis is 93 million miles, the sun being at one focus. Find the greatest and the least distance from the earth to the sun. 12. Find the angle which one diameter of an ellipse makes with its conjugate diameter. 13. A comet moves in a parabolic orbit with the sun at the focus. If the comet is 2 million miles from the sun when the line from sun to comet makes an angle of 60° with the axis of the orbit, find the least distance from sun to comet. 14. Show that the bisector of the angle formed by lines joining any point of an equilateral hyperbola to the vertices is parallel to an asymptote. 15. Find the equation of the locus of the mid-points of chords drawn from one end of the major axis of an ellipse. 16. Show that the ordinate of the intersection of any two tangents to the parabola y'^ = 4px is the arithmetic mean of the ordinates of the points of contact, and the abscissa is the geometric mean of the abscissas of the points of contact. 17. Find the equation of the locus of the intersection of two tangents to the parabola ?/- = 4 px, if the sum of the slopes of the tangents is constant. 18. Show that the angle formed by any two tangents to the parabola is half the angle formed by the focal radii to the points of contact. 19. Any two perpendicular lines are drawn from the vertex of a parabola. Show that the line joining their other points of intersection with the parab- ola cuts the axis at a fixed point. 20. Show that the tangents to the parabola at the extremities of any chord intersect on the diameter bisectins; the chord. 140 LOCI OF SECOND ORDER [Chap. VII. 21. Show that the eccentricity of an hyperbola is equal to the secant of half the angle between the asymptotes. 22. Show that the tangents at the vertices of an hyperbola intersect the asymptotes at points on the circle about the center and passing through the foci. 23. Show that the product of the distances from the center of an hyper- bola to the intersections of any tangent with the asymptotes is constant. CHAPTER Vlir LOCI OF THE SECOND ORDER EQUATIONS NOT IN STANDARD FORM 110. Translation of the coordinate axes. If the coordinate axes are translated by means of equations (1), Art. 77, and then the primes are dropped, the standard forms of the equations of the several conies become : Circle: (x + Ti)"-^ + (// + A;)^ = r^, (1) Ellipse: (^±^+(^±_M^' = l, (2) Primary hyperbola : (^±^ _ UL^ = i, (3) Conjugate hyperbola : (^±^ _ iJlA^ = _ 1, (4) Parabola, X-axis parallel to the axis of the curve : (y + hyi = 4iJ(x + h), (5) Parabola, T^axis parallel to the axis of the curve : (.r + /i)2 = 4i>(»/ + A;), (6) On the other hand, if an equation of the second degree can be reduced to one or the other of the above forms, the locus is the corresponding conic. The center of the circle, the ellipse, or the hyperbola is then the point (— /*, — k), and the vertex of the parabola is the point (— h, — k), Figs. 87 and 88. As an example, take the equation 9 x^ + 4 2/2 + .54 X- I6y + 61 =0. Completing the squares of the terms in x and the terms in 2/ separately, the equation can he written f,, , .^.„ , ., -.xo 00 ^ 9(x + S)^ + 4c{y — 2)2 z= 36. Comparing with (2), we see that the locus is an ellipse whose center is the point (— 3, 2) and whose semiaxes are 2 and 3. 141 142 EQUATIONS NOT IN STANDARD FORM [Chap. VIII.. If any one of the equations (1) to (6) is expanded and cleared of fractions, it is seen to be a special case of the equation rta?2 + 6t/2 + 2sr£c + 2/2/ + c = 0, (7) where a, b, g, f, and c depend upon h and k. We are led, then, to the following Theorem. The equation of a conic referred to coordinate axes parallel to the axes of the curve (the axis and tangent at vertex in case of the parabola) has the form (7). Fig. 87 Fig. 88 111. Discussion of the equation ax^ + by^ + 2gx + 2fy + c = 0. The question now arises : Is ax' + by'-\-2yx + 2fy + c = (1) the equation of a conic for any given set of values of the coeffi- cients ? We shall answer this question by a discussion of the equation (cf. Art. 46). General case. Let us first suppose that none of the coefficients is equal to zero, and we may farther suppose without loss of generality, that a is a positive number. For, if a is a negative number in any particular case, we can change the signs of all the terms in the equation. Equation (1) can then be written in the f°^'^ - No / ^\2 2 « a x-\- +*(^' + 5J='^+%-'- (2) Denote the right-hand member of (2) by D, then the nature of the locus will depend upon the signs of b and D. Thus, if b is nega- tive, the locus is an hyperbola which is primary or conjugate Art. Ill] DISCUSSION OF ax2 + &2/2 + 2 ffx + 2/2/ + c=0 143 according as D is positive or negative (cf. Art. 107). If h and Z) are positive, the locus is an ellipse. There is no locus if h is positive and D is negative, for the sum of two positive numbers can never be negative. The locus is then said to be imaginary. If either a or 6 is zero, the above method fails. But if a is zero, while 6 and g are different from zero, (1) can be written and if & is zero, while a and / are different from zero, (1) can be written In either case, the locus is a parabola, as we see on comparison with equations (5) and (6) of the preceding article. Special cases. If D is zero and b is negative, the left-hand member of (2) is the product of two linear expressions, and there- fore the locus of (2), and consequently the locus of (1), is a pair of intersecting straight lines (cf. Art. 85). If b is positive, the locus consists of the single point, ( — -» — V ), since this is the only- point whose coordinates will then satisfy (2). Finally, if a and g, or b and /, are each equal to zero, equation (1) contains but one of the variables. For example, if b and / are each equal to zero, (1) becomes ax'' -f 2 gx +c = 0. (5) If the roots of (5) are real and distinct, the locus consists of a pair of lines parallel to the F-axis. If the roots of (5) are equal, the locus consists of a single line parallel to the y-axis. If the roots of (5) are imaginary, there is no locus, but we shall say, in this case, that the locus consists of a pair of imaginary lines. In these special cases, the locus is said to be degenerate. We shall find it convenient to say that the locus of (1) is a conic, but that in certain cases, the conic is imaginary, or degenerates into a single line, or into a pair of real or imaginary lines, or consists of a single point. 144 EQUATIONS NOT IN STANDARD FORM [Chap. VIII. As an example of the foregoing analysis, consider the equation 9 x^ - 16 y-^ - 30 X + 9(3 ?/ - 108 = 0. Here we find that a is positive, D is zero, and b is negative. Therefore the locus consists of two intersecting lines. The equations of these lines are 3(x-2)i4(j/-3) = 0, and they intersect in the point (2, 3). As a second example, consider the equation 9 x2 + 2 ?/ - 18 X + 8 ?/ + 17 = 0. In this case, a is positive, D is zero, and h is positive. Hence the locus con- sists of a single point. Completing the squares of the terms in x and the terms in y separately, the equation becomes 9(x-l)2 + 2(?/ + 2)-^ = 0. Therefore the point (1, — 2) is the only point whose coordinates satisfy the given equation. A summary of the possible loci of the equation ax^ + hi/ + 2 gx + 2/y + c = is given in the following table, where D = -^ -{--^ c. a b a ^ 0, 5^0 a=0 (or & = 0) a>0 b>0 ffl>0 6<0 a = 0,g^O,b^O (or 6 = 0, /^O, rt 9^0) a = g = 0, b=^0- (or &=/=0, a^O) D>0 Ellipse D>0 Hyperbola (priraary) Parabolas A pair of real par- allel lines, a single line, or a pair of imag- inary lines ; according as the roots of by^ + 2fy + c = (or a.x2 + 2 gj- + c = 0) are D = Point Z> = Intersecting lines D Imaginary D<0 Hyperbola (conjugate) real and distinct, equal, or imaginary. In the following exercises, use is to be made of this table. Art. 112] EQUATION OF SECOND DEGREE 145 EXERCISES 1. Determine the nature of the loci of the following equations. Find the coordinates of the center and the coordinates of the foci of each ellipse or hyperbola ; the coordinates of the vertex and the coordinates of the focus of each parabola. Make a sketch of each curve. (a) 2x^ + Sy^-6x + ^y = 10. (b) x^ + 2y^ - 6 x + ij = 10. (c) ix^-Sy^-4:X + 8 = 0. (d) .r^ + 4x - 2 ?/ = 15. (e) 3x^ -y2 + 0y^0. (/) y^ + 2x-'iy = 7. 2. Determine the nature of the loci of the follow^ing equations : (a) x^ + y^ — 4 X - 6 y + IS = 0. (b) x- — y^ — 4 x + 6 y + 5 = 0. (c) x2 - 5 r. + 6 = 0. (d) y^ - 6 ?/ + 9 = 0. (e) ?/2 - 6 ?/ + 10 = 0. (/) y2 _ 6 2/ + 8 = 0. 112. The general equation of second degree. The equation ax' + 2 hxu + hii- + 2 gx + 2./)/ + c = (1) is called the general equation of second degree, because it contains every term that can appear in an equation of the second degree. We will now prove the following theorem. Theorem. The term in xy can be removed from the general equation of second deg)-ee hy rotating the axes through a positive angle 0, less than 90°. Keplacing x and y in (1) by their values in terms of x' and y' ; namely, , „ , • n '' X = jc cos 6 — y sm 6, y = x' sin + y' cos d, (Art. 78), we obtain a'x'^ + 2 h'x'y' + b'y" + 2 g'x' + 2f'y' + c = 0, (2) a' = a cos2 ^ + 2 /i sin ^ cos 6> + 6 sin^ 0, (3) b' = a sin2 e-2h sin ^ cos ^ + 6 cos^ 0, (4) 2 h' = 2 h{cos^ 6 - sin2 d) - 2(a - b) sin cos ^2h cos 2 6-{a-b) sin 2 6, (5) 2(7' = 2f/cos^ + 2/sin^, (6) 2/' = 2/ cos ^ - 2 ^ sin ^. (7) If, now, we can choose so that h' shall equal zero, the term in x'y' will drop out of (2) and the general equation will be reduced to the form ^^,^, ^ ^,^,, ^ ^ ^,^, ^ ^_^,^, ^ ^^ ^ ^_ ^^^ where 146 EQUATIONS NOT IN STANDARD FORM [Chap. VIII. Putting h' equal to zero in (5), we have for the determination of 6, 2 h cos 2 ^ -(a - 6) sin 2^ = 0, from which tan 20 = ^^. (9) Since the tangent of an angle assumes all possible positive and negative values as the angle increases from 0° to 180°, it follows that it is always possible to find an angle 6, less than 90°, ivhicJi iviJl satisfy equatiori (9). If the axes are rotated through this angle, the term in x'y' drops out and the general equation is thus reduced to the form (8). Equation (8) has the same form as that discussed in the pre- ceding article. Therefore we can say that the locus of the general equation of second degree is a conic, hut that this conic may be imaginary, or may .consist of a single line, or of a pair of real or imaginary lines, or of a single 2'>oint. The values of the coefficients a' and b' can be found easily from equations (3), (4), and (5). Thu«, adding (3) and (4), we ^^^^® a' + h' =a + h, (10) and subtracting (4) from (3) gives a' -b' = 2 h sin 2 ^ + (a - b) cos 2 9. (11) Squaring (5) and (11) and then adding, we obtain 4 /i'2 + (a' - &')2 = 4 ¥ + (a - bf. (12) Subtracting (12) from the square of (10) gives a'b' -h'^ = ab-h-. (13) When the coordinate axes have been rotated through the angle given by (9), we have seen that h' becomes zero. Hence equa- tions (10) and (13) give respectively the sum and product of the required coefficients. These coefficients are then the roots of the quadratic equation X^ - (« + b)\ + ab- 7t- = 0. (14) The roots of this equation are always real, since the discriminant, (a + by — 4(a6 — h"^) = (a — 6)- + 4 h"^, is always positive. Art. 112] EQUATION OP SECOND DEGREE 147 In order to decide which of the roots of (14) to take for a', eliminate cos 2 6 between (9) and (11), thus obtaining 2 /i(a' - 6') = [4 W + (a - hf] sin 2 6. (15) For < 90°, sin 2 is positive. Tlierefore a' must be so chosen that a' — b' will have the same sign as h. The roots of (14) will be both different from zero if ab — h^ =fc 0, and will be alike in sign, or unlike in sign, according as their product ab — h^ is positive, or negative. The values of the coefficients g' and /' can be computed from equations (6) and (7), but the computation is often tedious, and can be avoided frequently by a translation of the axes (Art. 113). If a given equation of the second degree contains no terms of the first degree ; that is, if g and / are each equal to zero, then, by (6) and (7), g' and/' are also each equal to zero and the foregoing analysis serves to determine the nature and the position of the locus. For example, consider the equation X- + 2 xy + 2 y'^ — 4 = 0. Here tan 20 = -^-^ = -^— = - 2, from which we get 8 = 58° 17', nearly, a — b 1 — 2 If the axes are rotated through the angle 8, the term in x'y' will drop out. The coefficients of x'^^ and ?/''' are the roots of (14) which becomes, in this case, The roots are ^+^^ and ^ - V5 Since 2 2 h is positive, a' — b' must be positive, and therefore we choose h' — — '— and b' = — V5 2 2 The given equation, thus, reduces to 3 + V^.o , 3 - VS ,., . ~~2 — 2 — ^ ' ^^ 2(3 -V5) 2(3+\/y) The locus is, therefore, an ellipse whose semiaxes are V2(3 — VS) and V 2(3 +VE). The major axis coincides with the new F-axis (Fig. 89). 148 EQUATIONS NOT IN STANDARD FORM [Chap. VIII. EXERCISES 1. Determine the nature of the locus of the equation 5 X'+2xy + 5y^ = 12. Find the angle through which the coordinate axes must be rotated in order to remove the term in xy. Plot the curve and both sets of axes. Find the eccentricity of the curve. 2. What is the locus of each of the following equations ? (a) 3x^-2xy + y-^-6 = 0. (b) Sx^ - 2xy + y^ = -Q. (c) Sx^-2xy + if = 0. (d) 9x^-20xy + 11 y^ - [>0 = 0. (e) 25 a;2 _ 60 xy + 36 2/2 - 81 = 0. 3. Reduce the following equations to standard form. Draw the figure for each exercise. (a) x2 + xj/ + 2/2 _ 1 = 0. (6) x2 + 3 x?/ - 3 2/2 - 4 = 0. (c) 2 x2 - 12 xy - 3 2/2 + 14 = 0. (d) 43 x2 + 30 xy + 59 y^ - 68 = 0. (e) 8 x2 - 12 x?/ + 3 2/2 - 9 = 0. 4. The locus of the equation Ax"^ + 4:xy + y^ + k =0 is two straight lines for any value of k. Discuss the change in these lines as k varies. 5. Show that 3 x2 + 2 hxy + 12 2/2 = 3 is the equation of a system of concen- tric conies, h being a variable parameter. Discuss the change in the locus as h varies from a great negative number to a great positive number. 113. Removal of the terms of the first degree. If the terms of the first degree can be removed from the general equation of second degree, this can be done by translating the axes, as in Art. 79. Let m and n be the coordinates of the new origin. The equations for translating the axes are then x = x' -{- m and y = y' -\- n (Art. 77). Substituting in the general equation, (1), Art. 112, and arranging according to the powers of x' and y', the resulting equation can be written r , . , ,■. {am + an + g)X ax'^+21ix'y' -\-hy'' + 2\ [ + qim + hn^f)y'\ (m{am + lin + 9') 1 + n{hm-\-hn+f)\=0. (1) -\-{gm-irfn+c) \ If the terms of first degree can be removed, m and n must sat- isfy simultaneously the two equations am, + Tin -\- g -d, /9\ hm + hn + f = 0. ^^' Arts. 113, 114] FIRST CASE, 06 - /i^ ^ 149 The values of m and n derived from these equations are, ah — Ji ,n = M^zAu^ (3) ab — hr ^^ hg -af We now have three cases to consider (cf. Art. 83) : 1. Equations (2) are consistent and have but one common solution if, and only if, ab — h^ is not equal to zero. For then equations (3) give but a single pair of values for m. and n. 2. Equations (2) are inconsistent, and therefore have no com- mon solution, if ab — h'^ = and hf— hg ^ 0. For then hg—af^O and the numerators in (3) do not vanish while the denominator is equal to zero. 3. Equations (2) are dependent and therefore have an indefi- nite number of common solutions if ab — h^ = and hf— bg = 0. For then hg — af= and both the numerator and the denomina- tor of each fraction in (3) vanishes and any pair of values of m and n that satisfies one of the equations (2) also satisfies the other. We shall consider the three cases separately, and we assume that in no case is h equal to zero. For, if h is equal to zero, the general equation has the form discussed in Art. 111. 114. First case, ah — h^ ^0. Central conies. In this case the terms of first degree can be removed. Setting the values of m and n from (3), Art. 113, in (1) and dropping the primes, (1) becomes ax^^2hxy + bf + yMllM.,&lj:i^ + c = Q. (1) ab — 7r ab — /r The absolute term in this equation is dhc + 2fgh —hg^ — af^ — ch^ ,c, ^ ab-W ^"^ For simplicity, let A represent the numerator in C2) and C, the denominator. Equation (1) can then be written ax^ + 2 hocy + by^ +^ = 0. (3) 150 EQUATIONS NOT IN STANDARD FORM [Chap. VIII. A cau be expressed as a determinant. Thus, (4) a h ff h b f ff f c C is the cofactor corresponding to the element c, and the numerators in (3), Art. 113, are respectively the cofactors corre- sponding to the elements g and /. The determinant A is called the discriminant of the general equation of the second degree. If the axes are now rotated so as to remove the term in xy, (3) becomes aV^'+6y^ + ^=0, (5) where a' and b' are the roots of (14), Art. 112. If ab — h^ > 0, a' and b' are alike in sign (Art. 112) and (5) can be written „ , , ,„ , /^^ \a'\x'^+\b'\y'^ = c', (6) where 1 a' \ and | b' \ are positive numbers and c' is ± — . The locus is then an ellipse which is real or imaginary according as c' is positive or negative. If ab — li^ < 0, a' and 6' are unlike in sign and then (5) can be ''''■'^*®'' \a'\x"'-\b'\y'' = c'. (7) The locus is then an hyperbola which is primary or conjugate with respect to the axes X', Y' according as c' is positive or negative. Degenerate conies. If A = 0, then c' is zero, and the locus of (6) is a single point ; namely, the origin, and the locus of (7) is a pair of straight lines intersecting at the origin. Neither a' nor b' can- equal zero, since a' ■ b' = ab — li^ ^ 0. We conclude, therefore, that in this first case, the locus of the general equation of second degree is either an ellipse, real or imaginary ; an hyperbola ; a pair of real and intersecting lines ; or a single point. But the lomis is never a parabola, a pair of parallel lines, or a single line. We may further conclude, from equations (3), (6), or (7), that the locus, whatever it may be, is symmetrical with respect to the Art. 114] FIRST CASE, ab-h^^O 151 a h 9 h b f 9 f c origin, since if x^ and y^ satisfy any one of these equations, — x^ and — yi will also satisfy the equation. Therefore the locus of the general equation, in this first case, is symmetrical with respect to the point {m, n). Or, in other Avords, the point {m, n) is the center of the locus. Hence the condition ah — W^Q characterizes the central conies. As an example of the foregoing analysis, let us reduce the equation 8:B2_^4x2/ + 5y2-|.8u;— 16 2/-16 = to the standard form and thus determine the nature and position of the locus. u. (t y Here C = a6 - /t^ = 40 - 4 = 36, and ^ == h b /" = - 1296. Also from (3), Art. 113, we have m = — 1 and n = 2. Therefore the center of the conic is the point (— 1, 2). Again, from (9), A^t. 112, t^n 2^ = 1, from which we find d = 26° 34', nearly. Hence we conclude that when the axes are translated so that the new origin is the point (— 1, 2), the given equation wUl take the form (3), or 8x--]-ixy + 6y^ — SG = 0, and when the axes are rotated through the angle 26^^ 34', the equation will take the form (5), where «' and b' are the roots of X2_ 13X + 36 = 0; i.e. 9 and 4. Since h is positive, we choose a' = 9 and b' — 4. The given equa- tion then becomes a;2 , ?/2 _ . 9 x2 + 4 ?/2 - 36 = or - + i^ = 1. 4 9 Fig. iK) The locus is therefore an ellipse whose semiaxes are 2 and 3 Figure 90 shows the curve and the three sets of axes. As a second example consider the equation 2 a;2 - x?/ - 3 2/2 - 2 a: + 18 2/ - 24 = 0. Here we find A = and C = - ^-f. The roots of (14), Art. 112, are therefore unlike in sign, and the locus consists of two intersecting lines. The left-hand member of the given equation must be the product of two linear factors 152 EQUATIONS NOT IN STANDARD FORM [Chap. VIII. Fig. 91 (Art. 85) . Solving the given equation for one of the variables, considering the other as a knovni number, reveals the factors at once. Thus, solv- ing for X, we have ^^2M:_2 5y-U_ 4 4 Hence the locus consists of the two lines 2x-Sy + 6 = and »• + ?/ - 4 = (Fig. 91). EXERCISES 1. Reduce the following equations to standard form. Determine the coordinates of the center and the angle through which the axes must be rotated in order to remove the term in xy : (ffl) 5 x2 + 4 X2/ — ?/"2 + 24 X — 6 ?/ — 5 = 0. (ft) xy + y- -}- y + 1 = 0. (c) 4: xy + 4: y'^ — 2 X + 3 - 0. (d) x^ + xy +y^ + 3y =: 0. (e) x^ -2 xy + 5y'^ -8y = 0. (f)x^ + 2xy + 9y^ = 0. {g) 2 x^ - Q xy + 6 y^ + 6 X — 12 y + 9 = 0. 115. Second case, ab - h'^ =0 and hf-bg=p 0. Non-central con- ies. In this case the terms of first degree cannot be removed, since equations (2), Art. 113, are inconsistent and have no com- mon solution. We begin the discussion, therefore, by rotating the axes so as to remove the term in xy. The angle through which the axes must be rotated is determined from (9), Art. 112. After rotation, the general equation assumes the form (8), Art. 112, where a' and b' are the roots of (14) and g' and/' are deter- mined from (6) and (7). But in the case we are discussing, ah — /i^ = and equation (14) becomes X" - (a -\- b)X= whose roots are and a -\- b. We may assume that a is positive (Art. Ill) then b is also positive, since the product ab is positive ( = /i^). Hence we choose a' = a 4- b, or b' = a + b, according as h is positive or negative (Art. 112). The general equation is then reducible to one or the other of the forms (a + b)3c''^ + 2 g'xJ + 2 fy' + c = 0, or (a + 6)2/'2+2sr'a?' + 2/'*/' + c = 0, according as li is positive or negative. (1) (2) Art. 115] SECOND CASE, 06-/1^ = AND hf - bg ^ 153 We must now determine g' and/'. From (9), Art. 112, we have , 0/1 2tan^ 2/t tan 2 6 = = 1 — tan^ a — b Solving this quadratic for tan 6, we obtain tan 0^ — - ± H by h^ + 1. But since h^ = ab, the expression under the radical is a perfect square. Therefore, tan 6 = - or — -, from Avhich sin 6 and cos 9, h h and thence g' and /' can be calculated. But, since < 90°, tan 0, sin 0, and cos are all positive. Hence we have the following results, where the sign before the radical is positive : h positive h negative tan e = b h a h sin d = b a Vb-^ + /i2 Va2 + A2 cos^ = h -h y/b-^ + h^ Va^ + h^ ff' = hg + bf Vb-^ + li^ af- hg Vd^ + h^ /' = hf-bg - (hf + ag) V62 + ^2 (3) (4) Since neither hf— bg nor hg — af is zero, we see that if h is positive, and therefore the equation reducible to the form (1), /' cannot equal zero; and if h is negative, and the equation re- ducible to the form (2), g' cannot equal zero. Hence, on com- parison with equations (3) and (4), Art. Ill, we conclude that the locus of the general equation, in this case, is necessarily a parabola. 154 EQUATIONS NOT IN STANDARD FORM [Chap. VIII. As an example, let us reduce the equation x2 — 2 x?/ + 2/2 — 2 2/ - 1 = to the standard form and thus determine the position of the locus. Since ah — fi^ is here equal to zero and lifis not equal to hg, the locus is a parabola. Again, since h is negative, we choose the form (2). Computing g' and/' from (3) and (4), the given equation becomes 2 2/'2 - V'2 x' - ^2 ?/' - 1 = 0. Completing the sqnare of the terms in ?/', vre have 1 y'-- 2y/2 V2 ■ 4V2/ Comparing with (5), Art. 110, we see that the vertex of the parab- ola, referred to the new axes, is the point ( ^, ^^V If V 4\/2 2V2/ the axes are translated so that this point is the new origin, the equa- tion reduces to the standard form 9 v''2 y =-7-»^> Fig. 92 where the primes have been dropped. The angle through which the axes have been rotated is given by the equation tan 61=-- =1, h Therefore d = 45^. Figure 92 shows the locus and the three sets of axes. EXERCISES 1. Reduce the following equations to the standard form . Determine the angle through which it is necessary to rotate the axes in order to remove the term in xy^ and the coordinates of the vertex referred to the original axes : (a) x2 - 2 a;?/ -h 2/2 - 8 X + 16 = 0. (6) x2 — 2 x?/ + 2/^ + 2 X — 2/ - 1 = 0. (c) 4 x2 + 4 X2/ -I- 2/^ — 4 X = 0. (d) 9 x2 + 12 X2/ -h 4 2/2 - 2 2/ = 0. (e) x2 -h 4 X2/ + 4 2/2 - 6 X -I- 8 2/ -t- 1 = 0. Art. 116] THIRD CASE, afc - /i^ = AND hf - bg = 155 116. Third case, ab—h^=0 and hf—bg=0. Since ah — h^ is again eqnal to zero, the general equation is reducible to the form (1), or the form (2), of the preceding article, according as h is positive or negative. But here hf—bg = and af — hg=Q. Hence, if h is positive, /' = ; and if 7i is negative, g' = 0. Consequently equations (1) and (2) of the preceding article be- come respectively . , ^ ,., r. , r ^""^ (a + b)y" + 2f'y'+c = 0. ' (2) Each of these equations contains but a single variable. There- fore, in this case (cf. Art. Ill), the locus consists of a pair of parallel lines; a single line ; or ci pair of imaginary lines, according as the roots of the equation, (1) or (2), are real and distinct; equal; or imaginary. It is not necessary to calculate the coefficients g' and /' in order to determine the nature of the locus of an equation satisfy- ing the conditions of this third case. For, since a is not zero, the general equation of the second degree can be written arx- + 2 ahxy + aby^ + 2 agx + 2 afy + ac = 0, (3) and since ab = h- and af — hg, (3) becomes {ax + hyf -H 2 g{ax + hy) + ac = 0. (4) The locus then consists of a pair of parallel lines, a single line, or a pair of imaginary lines according as g^ is greater than, equal to, or less than ac. For example, the locus of the equation 4.T2-f 12a;?/ + 9/ + 4;^ + 6y + l =0 is a single line, since here g- = ac. The equation of the line is '^'^'^^ 2x+3y + l = 0. EXERCISES 1. Determine the nature of the loci of the following equations. Draw the locus when possible. («) x^ — 2 xy + y^ + 2 tj — 2 X + 1 = 0. (6) 4 x^ + 12 xy + 9 y^ + 4 X + 6 y + 2 = 0. (c) x^ + 2xy + y'^—1 = 0. Id) 9 a;2 - 12 x?/ + 4 2/2 + 1.5 X - 10 2/ + 6 = 0. 156 EQUATIONS NOT IN STANDARD FORM [Chap. VIII. 2. If a = 0, in the third case, show that the general equation is neces- sarily by'^ + 2fy -\- c = 0, and therefore the locus is a pair of parallel lines, a single line, or a pair of imaginary lines according as f'^ is greater than, equal to, or less than be. 117. Recapitulation. The results of the foregoing three arti- cles can be exhibited in tabular form as follows : Loci of the general equation of the second degree A = a h y h b f 9 f c C=ab-h\ First case, ab-h^^O Second case, ab-h'^=0, hf-bg=j^O Third case, ab-h-^ = 0, hf-bg = (7>0 C<0 Parabolas Parallel lines, a A^^O Ellipse, real or imag. Hyperbola single line, or imag- inary lines A = Point Intersecting lines EXERCISES 1. Analyze the following equations. What is the locus of each ? (a) x^ + Qxy + y'^- 4 X- 12 y + 10 = 0. (6) x^ - xy + y^ + 3 x = 0. (c) 9 ,x2 _ 30 xy -I- 25 ?/2 _ 10 X = 0. (d) 2oi^ — xy + bx -2y -\- Q =Q. 2. Analyze each of the following equations and draw the corresponding locus. (a) ,x2 - 2 x?/ -f- ?/2 - 10 X - 6 2/ + 25 = 0. (6) x^ - x?/ -|- 5 x - 2 ?/ -|- 6 = 0. (c) 2 x2 + xy -I- J/''^ - 5 X - 10 ?/ -f 18 = 0. (fZ) x2 -(- 3 xy + 2 2/2 _ ^ _ 2/ = 0. 3. The locus of the equation 3 x^ — 3 xy — y^ + 15 x + 10 y — 24 = is an hyperbola ; find the equations of its asymptotes. Suggestion. The center of the curve is found to be the point (0, 5). The standard form of the equation is 3 x'- — 7 y^ = 2. Hence the equations of the asymptotes, referred to the axes of the curve, are 3 x^ — 7 y^ = Arts. 117, 118] TANGENTS 157 (Art. 106). When the coordinate axes are transformed back to the original position, the equations of the asymptotes become 3 x2 -3x(2/ - 5) - {y- by- = 0, or 2/_5 = (-3±V2"l)| 4. For what value of k is the locus of x^ + 2 xj/ + 2 ?/2 + ,t + ^• = a pair of straight lines ? Are these lines real or imaginary ? 5. If the locus of the general equation of second degree in x and «/ is a central conic, show that the equation can be written in the form a{x — mY + 2 h{x — m) {y — n) + h{y — n)- = , where to and n are the coordinates of the center, and A and C have the meanings assigned in Art. 117. 6. Making use of the preceding exercise, show that the equations of the asymptotes of any hyperbola are a(x — to)- + 2 h,{x — m){y — n) + b{y — n)- = 0. Apply this method to find the equations of the asymptotes of the hyperbola x2+6 X2/ + 2/2 - 4 X — 12 ?/ -f- 10 = 0. 7. Find the coordinates of the vertex, the coordinates of the focus, and the equation of the directrix of the parabola x2_4x?/ + 4?/2-4x-22/ + 8 = 0. TANGENTS AND DIAMETERS 118. Tangents. It is often convenient to write the equation of a tangent to a conic at a given point without first having to re- duce the equation to the standard form. Suppose the equation has the general form ax' + 2 hxy + hy'' + 2gx + 2fy + c = 0. (1) Let Pi{xi, ?/i) and P-iix^, y^) he any two points on the curve. Then we must have ax,^ + 2 hx,y, + hy,^ + 2 gx, + 2fy, + c = 0, (2) axi + 2 hx^y^ + fty^' + 2 gx^ + 2fy^ + c = 0. (3) Subtracting (3) from (2), we obtain a{xy^ - X.J) + 2 li{x{y^ - x,y^) + h{y^~ - y.2^)+2 g{x, - x,) + 2f{y,-y,) = Q. (4) 158 EQUATIONS NOT IN STANDARD FORM [Chap. VIII. Dividing (4) by x^ — X2, we have a(x, + x,) + 2h ^'^y' - ^^^^ + 6 ^y^ + y-^^^'\- y^^ + 2g+2fy^^^^ = 0. (5) Now -'^ ~ ■ - is the slope of the secant PiPj. Let this slope be represented by m. The term ^'"'{^ ~ ^'f' can be written a^i.Vi - ^-1^/2 + ^xVi - ^-iD-i «V1 ~~' t^l/o and is therefore equal to mx^ + 2/2. Hence (5) becomes a{x^ + 0^2) 4- 2 hiinfix^ + ^/j) + 6(^1 + y^'m -\-2g^ 2fm = 0. (6) Solving for m, we obtain ^ ^ _ a(xi + a;^) + 2 %2 + 2 .g / ^s 2^a;i + 6(?/i + ^2) + 2/ ^ Equations (2) to (7) hold as long as Pj and P2 are on the curve. When P2 approaches Pj along the curve, the secant approaches the position of the tangent at Pi, and in the limit coincides with it (Art. 97, second method). Hence, making x^ = x^ and 2/2 = Ui ir^ (7), we have the slope of the tangent at Pj ; that is, ax +hy +(j m = — — i i . (8) hXj^ + hij^ +f The equation of the tangent at P, is, therefore, ax, + hy, -\- a X . hx, + byi +f Clearing of fractions and reducing by means of equation (2), we have ax,x + h (x,y + y^x) + hy,y + g{x, + x) + f{y^ + y) + c = 0. (9) This equation is easily remembered, since if the subscripts are removed, it returns to the original form (1). A convenient way of writing (9) is the following : X {ax^ + %i + 9') + y(hx, + hy,+f) (10) + (^^i+./^i + c) = 0. Arts. 118, 119] DIAMETERS 159 Either (9) or (10) is the equation of a straight line whether the point Pi is on the conic or not. If P^ is not on the conic, then (9) or (10) is, by definition (Art. 104), the equation of the polar line of Pi with respect to the conic whose equation is (1). EXERCISES 1. Find the equations of the tangents to the following, at the points indicated. (a) a;"-^ + 4 2/2 + 5 X = 0, at the points whose ordinate is 1. (6) xy = 4, at the point whose abscissa is 2. (c) a;2 _|_ xy + 4 = 0, at the point whose abscissa is 2. {d) y~ + 2 a-?/ — 3 = 0, at the point whose ordinate is — 1. (e) x^ — 3 xy — 4 y^ + 9 = 0, at the points whose ordinate is 2. 2. Find the equation of the polar line of the point (1, 2) with respect to the conic x^ — 3 xy + y- = 4. Draw the figiire to illustrate the problem, 3. In exercise 7, Art. 117, show that the directrix is the polar line of the focus with respect to the given laarabola. 119. Diameters. In case of a central conic, we have found that the coordinates of the center are the values of m and n which satisfy equations (2), Art. 113. This amounts to saying that the center of the conic is the point of intersection of the lines whose equations are hx + by+f = 0. ^ ^ Hence these two lines are diameters of the conic (Art. 101). The equation of any diameter is, therefore (Art. 84), (ax + hy +g)+k (Jix + by +/) = (2) where fc is a variable parameter. Let P(a;i, y^) be any point on the conic. When the diameter (2) passes through P, A; has the value o.i'i+7^?/i +g hXi+by^-{-f But this is the slope of the tangent at P, (8), Art. 118, and hence, also the slope of the diameter conjugate to (2), Art. 102. There- 160 EQUATIONS NOT IN STANDARD FORM [Chap. VIII. fore, the parameter k is the slope of the diameter conjugate to (2). The slope of (2) is h + hk Hence, k and — — are the slopes of a pair of conjugate /I — |- ufC diameters of the conic tvhose equation has the general form (1) of the preceding article. The two diameters will be perpendicular to each other, and therefore the axes (Art. 102, exercises 12 and 13), when the product of their slopes is — 1 ; that is, when — ~^ = 1, or h + bk hk^ +{a-b)k-7i = 0. (3) If ki and ^2 ^-i"© the roots of (3), the equations of the axes are (aoc + hy + g)+Ti.-^{hx + by + f) =0, . (aac + hij + g)+ Ji:..2(hx + by +f) = 0. The roots of (3) are always real, since the discriminant, 4 h^ + (a - by, is necessarily positive. We have seen (Art. 106) that an asymptote of an hyperbola is a self -con jugate diameter. But if the slope of any diameter of a conic is equal to the slope of its conjugate diameter, we must have 7. __a-{-hk ~~ h + bk' or bk^ + 2hk + a = 0. (5) If k' and k" are the roots of (5), the equations of the asymptotes (ax + hy + g)+ Jc'ihac + by +f)=^Of ,g and (ax + hy + g) + k"(hx+ by + f)-0. ^^ The roots of (5) are real and unequal only if a& — A^ < ; that is, only if the conic is an hyperbola or a pair of intersecting lines (Art. 117). As an example, consider the equation 2 a-2 -f 4 x?/ — 2/2 + 4 a; - 2 ?/ + 3 = 0. Here equation (8) is ^ ,,.2 , 3 i- _ 9 — Art. 119] DIAMETERS 161 whose roots are | and — 2. Hence, from (4), the equations of the axes are 2 X + y + I = 0, and X — 2y — 2 = 0. Equation (5) becomes in this case ]c-2-4k-2 = 0, whose roots are 2 ± V6. Hence the equations of the asymptotes are (2x + 22/ + 2) + (2±V6)(2x-y-l)=0. The student should draw a figure illustrating this example. The diameters of a parabola are perpendicular to the directrix (Art. 101) and therefore perpendicular to the tangent to the curve at the vertex. We have seen (Art. 115) that the equation of a parabola is reducible to one or the other of two forms by a rotation of the axes through an angle ^<90°. But if h is positive, the new X-axis is parallel to the tangent at the vertex (Eq. 1, Art. 115) ; and if h is negative, the new X-axis is parallel to the axis of the curve (Eq. 2, Art. 115). Hence, from the values of tan 9 in these two cases, we conclude that the slope of a diameter to the jyarabola is or according as h is positive or negative. But - = - • Consequently, is the slope of any diameter. b h ^ b The slope of the tangent to the parabola at the point P(x, y) is, (8), Art. 118, m= ax + hy-\-g hx + by+f This tangent is perpendicular to the diameters of the curve if — = 1, or in other words, if the coordinates of the point of con- b tact satisfy the equation (ax + hy + g)7i ___^ ,j. {hx + by+f)b ■ ^ ^ Clearing of fractions and remembering that h^ = ab, (7) becomes (a + b)(hac + bij)+hg + fb = 0. (8) 162 EQUATIONS NOT IN STANDARD FORM [Chap. VIII. Now (8) is the equation of the axis of the parabola. For it is a diameter of the curve since it has the slope As an example, consider the equation a:2 _ 2 xy + y"- — 2 ?/ - 1 = 0. (cf . Art. 115. ) From (8), we get tlie equation of the axis, 2{x-y)+ 1 = 0. If we solve the equation of the curve and the equation of the axis simulta- neously, we get the coordinates of the vertex. In this example, the vertex is the point ( — |, — f ) • The equation of the tangent at the vertex is, therefore, y + I ^_(.x + 1), or X + ?/ + f = 0. As a second example, consider the equation 9 a;2 + 24 xi/ + 16 2/2 — 62 X + 14 2/ - 6 = 0. Here we find the equation of the axis is 3x + 4?/-2 = 0. Solving simultaneously with the given equation, we find that the vertex is the point (.08, .44). The equation of the tangent at the vertex is, therefore, (2/ -.44) =1(0; -.08) which reduces to 4 a; _ 3 ?/ + 1 = 0. The student should construct a figure to illustrate this example. EXERCISES 1. Find the equations of the axes of the ellipse a;2 - 2 xi/ + 4 2/2 + 2 X + 10 1/ + 10 = 0. 2. Find the equations of the axes and the equations of the asymptotes of the hyperbola a;^ _ 7 xy + 2/2 + 12 x + 3 2/ + 171 = 0. 3. Find the equation of the axis, of the tangent at the vertex, and of the directrix of the parabola x2 - 2 X2/ + 2/"^ - 10 X - 6 2/ + 25 = 0. 4. In the general equation of a conic, show that the line gx + fy + c = is the polar line of the origin with respect to the conic. * That the tangent at the vertex is the only tangent perpendicular to the diameter through its point of contact follows from Art. 96. We there saw that the coordi- nates of the point of contact are -^ and — , where hi is the slope of the tangent. Hence, as hi increases indefinitely, the point of contact approaches the origin. Arts. 120, 121] THE SYSTEM OF CIRCLES 163 SYSTEMS OF CONICS 120. The pencil of conies. If U and V denote expressions of the second degree in x and y and Jc is any constant, then U-\-kV=0 is the equation of a conic passing through the p)oints common to U= and V=0. For U'-{-kV= is of second degree in x and y and is, therefore, the equation of a conic. This conic passes through the points common to U= and V= 0, since its equation is satisfied by the coordinates of these points. When k is allowed to vary, we obtain a series of conies, each passing through the common points. This series, or system, of conies is called a pencil of conies. The parameter k can be chosen so that the conic U+kV=0 shall satisfy some additional condition, for example, that it shall pass through a given point in the plane. 121. The system of circles with a common radical axis. Suppose that U= and V= are the equations of two circles ; that is, U= x^ + if + Ax + By + C= 0, (1) and F= x2 + 7/2 + yli.i- -f 5i?/ -f- Ci = 0, then (x' + f + Ax + By +C)A- k{x- + f + A-->^' + B,y + C,)=0 (2) is in general the equation of a circle passing through the common points of the two given circles. But if A: = — 1, the terms of sec- ond degree drop out, and (2) becomes (A-A,)x+{B-B,)y + (C-C,) = 0, (3) which is of first degree in x and y, and therefore the equation of a straight line. This line is called the radical axis of the system of circles U+kV=0. The radical axis is a real line, whether the circles intersect in real points or not. If the circles intersect in real points, the radical axis is the common chord. EXERCISES 1. Find the equation of the conic whicli passes through the points common to the conies a;^ — Sxy + y^ — Qx = and 4 x^ — y'^ + 3 — 0, and also through the point (3, — 2) . 164 EQUATIONS NOT IN STANDARD FORM [Chap. VIII. 2. Find the equation of the radical axis of each of the following pairs of circles : (a) (X - 2)2 + (2/ - 3)2 - 10 = 0, (x + 3)2 + {y + 2)2 -6 = 0. (6) x2 + 2/2 _ 4 y = 0, (X - 3)2 + ?/ - 9 = . (c) (x + 3)2 + 2/2 _ 2 2/ _ 8 = 0, x2 + 2/2 - 2 ?/ = 0. (d) x2 + (2/ - a)2 = c2, (x - 2)2 + 2/^ = cV. 3. Three circles, taken in pairs, have three radical axes. Show that these radical axes intersect in one and the same point. This point is called the radical center of the three circles. 4. Find the coordinates of the radical center of the three circles (oc _ 3)2 ^ 2/2 = 16, x2 + 2/2 - 9, and x2 + («/ - 2)2 = 25. Construct the figure illustrating this exercise. 5. Show that the length of a tangent from the point (xi, y{} to the point of contact on the circle x2 + 2/2 + Z>x + ^(/ + -^ = is Vxi2 + 2/i2 + Dxi +Eiji + F. Suggestion. The triangle whose vertices are the center of the circle, the point of contact, and the point (xi, 2/1) is right-angled at the point of contact. 6. Prove that the locus of a point, the lengths of tangents from which to two fixed circles are equal, is the radical axis of the two circles. 7. Show that the radical axis of two circles is perpendicular to the line joining the centers of the circlee. 122. The parabolas in the pencil U+JcV=0. If the constant ]c is chosen so that the terms of second degree in U'+JcV=0 form a perfect square, the corresponding conic is in general a parabola. But it may be a pair of parallel lines, a pair of imaginary lines, or a single line (Arts. 115 and 116). Since the condition for the parabola is of second degree in the coefficients of x^, xy, and /, there are in general two parabolas in every pencil of conies. For example, consider the pencil of conies determined by the circle U=x' + y"" -l^x-d>y + U = 0, and the hyperbola 7-== ^^2 _ ^2_ §0; + 12 = 0. Here the equation (aj2 _|. ^2 _ 16 a; _ 8 2/ + 44) + A:(a;2 _ ^/^ - 8 .t + 12) = can be the equation of a parabola only if k = ± 1. Therefore the pencil contains the two parabolas aj2_ i2a;-4?/ + 28=0 and ?/2-4a;-4?/ + 16 = 0. Arts. 122, 123] STRAIGHT LINES IN U + kV = 165 Fig. 93 illustrates tlie ex- ample. The circle and hy- perbola have but two real points in common and the two x^arabolas pass through these points. EXERCISES 1. Find the equations of the parabolas which pass through the points common to the circle x'^ + y^ — x— 9 = and the hy- perbola oi'y —1 = 0. 2. Find the equations of the two parabolas which pass through the points where the ellipse x!^ — Sxy + iy^ — X — 2 =0 cuts the coordinate axes. (The equation of the coordinate axes is xy = 0.) Construct the figure illustrating this ex- ercise. 123. Straight lines in the pencil JJ+ kV=0. AVhen k is chosen so that the discriminant of U+lcV=0 vanishes, the correspond- ing conic is in general a pair of lines (Art. 114). For example, consider the pencil of conies determined by the ellipses l/= 2x^ + xy + 6 2/2 + 3; - 6 = and V= 3x^ + 6xy + 10 y-2 + x - 10 = 0. Here the pencil U+ kV = is (2 -h 3 k)x^ H- (1 + 5 k)xy -|- (6 -h 10 k)y^ +(i + k')x-(6 + 10 k) = 0. Forming the discriminant, 1 + 5k FiCx. 93 A = (2 -h 3 k) l + 5k 2 1 + k 2 + 10 k) 1 +k 2 (6 + 10 k) we find that it reduces to - 12(6+ 10 k){k + l){2 k + I). Hence, if k is — -|, — 1, or — J, the discriminant is zero and the correspond- ing conic consists of a pair of lines. If A = — |, U + kV = becomes x^—lOxy + 2?; = 0, 166 EQUATIONS NOT IN STANDARD FORM [Chap. VIII. which is the equation of the pair of lines X = and x — lOy + 2 =0. These are the lines BC and AD in Fig. 94. It k = —1, U+ kV=0 be- comes a;2 + 4 3:;/ + 4 2/2 - 4 = 0, which is the equation of the pair of parallel lines x+ 2y -2 = and x + 2y + 2 = 0, 1 1 /V' ^^ !S, ,^: _ : =■; - >,i^- :> -- = '"T'Ti^ ^ ^'"^ :^, : / •n -■^ ? '*•«.,//'' ^^^. "-' - 2V^ ,^... ^t s^ 7 ■*; j^ 7 "^-.^ -7 "^^ : '=^. Fig. 94 or BD and ^Cin the figure. If ^• =— i, Z7+ A;F= becomes a;2 _ 3 x?/ + 2 2/2 + X — 2 = 0, or (.r - 2?/ + 2)(x— ?/ - 1)= 0, which is the equation of the pair of lines AB and 01). The coordinates of the points A, B., C, and B are now easily found, represent the common solutions of the equations 17= and Y= 0. They EXERCISES 1. Find the equations of the straight lines which join in pairs the points common to the following pair of conies : (ff) X? + 2/'- -25 = 0, 5 x2 + 14 2/ + 3 x - 110 = 0. (6) 4 x2 + 9 2/ - 36 = 0, x2 + 4 2/ = 0. (c) x2 + 2 X2/ + 7 2/2 - 24 = 0, 2x^ -xy-f--^= 0, 2. Find the coordinates of the points common to each pair of conies in exercise 1. Art. 124] PENCIL OF CONICS 167 124. The pencil of conies through four given points. In the pre- ceding article we have seen how tlie coordinates of the points common to two conies U= and V^^ may be found. On the other hand, if we are given the coordinates of four points, we can determine tlie pencil of conies which has these points in com- mon. For example, let the four given points be ^(1, 0), 5(2, 1), C(l, 2), andZ>(0, 1) (Fig. 95). The equation of the pair of lines AB and CD is then {x-xj-\){x-y + 1)= 0, and the equation of the pair ^IC and Therefore the equation of the pencil of conies having the four given points in common is , ,,, , -,\ , i , is , i\ n. (x-y -l)(x-y + l)+k(x- l)(y- 1)=0. Clearly, the parameter k can be determined so that the corresponding conic shall pass through any fifth point in the plane. Thus, if we wish the conic of the pencil wliich passes througli tlie origin, we must determine k so that the above equation shall be satisfied by the coordinates of the origin. But the equation is satisfied for x = and y — H kis 1. Therefore the ellipse x^ — xy + y'^ — X — y = belongs to the pencil and also passes through the origin. EXERCISES 1. Find the equations of the conies which pass through the following sets of points (a) (0, 0), (1, 0), (2, 1), (1, 3), and (- 1, -4). (b) (1, 1), (3,2), (0,4), (-4,0), and (-2, -2). MISCELLANEOUS EXERCISES 1. Show that the line x — 2y = touches the circle x^ + y^ — 4:X + 8y = 0. 2. The line ?/ = 3 x — 9 is tangent to the circle x"^ + y^ + 2 X + 4: y — 5 = 0. Find the coordinates of tlie point of contact. 168 EQUATIONS NOT IN STANDARD FORM [Chap. VIIT. 3. Prove that the distances of two points from the center of a circle are proportional to the distances of each from the polar line of the other. 4. Find the equations of the circles which pass through the intersections of x^ + 2/2 = 9 and x"^ + y- + x+ 2y — li and touch the X-axis. 5. Find the coordinates of two points whose polar lines with respect to the circles x'^ + y^ _ 2 x — 3 = and x'^ + y'^ + 2x— 17 =0 coincide. 6. Find the coordinates of the radical center of the three circles x2 + 2/2 _ 4 a; _ 8 2/ - 5 = 0, x2 + z/2 - 8 X - 10 2/ + 25 = 0, and ft;2 + 2/2 + 8 X + 11 2/ - 10 = 0. 7. Reduce the following equations to a standard form : (a) (4 2/-3x)2 + 4(4x+32/) = 0. (5) 4x2-24x2/+ 112/2- 16x- 2 2/- 89 =0. (o) 5 x2 - 4 X2/ + 8 2/2 - 22 X + 16 2/ - 10 = 0. (d) 9 x2 - 12 X2/ + 4 2/2 = 10(2 x + 3 2/ + 5). (e) 3 x2 - 2 X2/ + 2 2/2 - 16 X - 8 2/ + 8 = 0. (/) 6 x2 + 24 x?/ - 2/^ + 50 2/ - 55 = 0. (g) x2 — 2 X2/ + 2/2 — 5 X — 2/ — 2 = 0. (/i)4x2 + 4x2/ + 2/^ + 4x — 32/-|-4 = 0. (i) 25 x2 - 20 X2/ + 4 2/2 + 5 X - 2 2/ - 6 = 0. ( j) x2 - 6 x?/ + 9 2/2 - 2 X + 6 2/ + 1 = 0. (yfc) x2 — 2xy-y-- 20. {I) xy + 3 X - 5 2/ + 5 = 0. (m) x2 + 2 X2/ + 2/2 + 1 = 0. (n) (5?/ + 12x)2 = 102x. (o) x2 — x?/ — 2 2/2 — X — 4 2/ — 2 = 0. 8. Wliat curve must be used as a pattern for cutting elbo-ws of stovepipes from sheet iron ? CHAPTER IX LOCI OF HIGHER ORDER AND OTHER LOCI 125. Certain loci of higher order, as well as certain transcen- dental loci, are of importance either because they are useful in mechanics or because of their historical interest. The more im- portant of these loci are considered in the following articles. ALGEBRAIC LOCI 126. The Cissoid of Diodes. Let G be the center of a circle of radius a, and OCA, any diameter of it. Through draw any chord OR and produce it until it meets the tangent at A in the point Q. If P is so chosen that PQ is equal to OR, then the locus of P is a curve called the Cissoid of Diodes. To find the equation of the cissoid, let be the origin, OCA the X-axis, and the tangent at the Y-axis. Let 6 denote the angle AOQ. Then OQ = 2 a sec ^ and OR = 2 a cos 6. Hence, OP=OQ- PQ^ Oq- 0R = 2a (sec d - cos 6). (1) But OP = Vic^ -f- ''f and d = arc tan ^ • Therefore sec = ^^-=^, X and cos^ = — • Substituting in (1) and reducing, we get the equation sought, y^ = 7T-^ — 2 a — X Either from the definition, or the equation, the curve is seen to have the form indicated in the figure. 169 170 LOCI OF HIGHER ORDER [Chap. IX. EXERCISES 1. Show that the line 2 « — x = is an asymptote to the cissoid. 2. Using the method of Art. 118, show that the tangent to the cissoid at the point (xi, yi) is 2(2 a — Xi)ijiy — (3 Xi^ + yi^)x + 2 ay{^ = 0. 3. In Fig. 96 let C3I be taken twice the length of CB. Draw 3IA and let it meet the cissoid in the point F whose ordinate is FE. Prove that FF^ = 2 ■ 0E\ If CM is n times CB, show that FE^ = n- OE^. Note. The cissoid was invented by Dio- des for the purpose of duplicating the cube. Thus, in Fig. 96, Avhen CM is twice CB, and OF is the edge of a given cube, FE is the edge of a cube of twice the volume. The duplication of the cube is one of the famous problems of antiquity. Diodes lived about 150 B.C. 127. The Conchoid of Nicomedes. Let XX' be any straight line and any point not on XX'. Throngh. draw a series of straight lines forming a pencil, and on each of these lines lay off a constant length a on each side of XX'. The locus of the points so determined is called the conchoid of Nicomedes. To find the equation of the conchoid, let XX be the X-axis and the perpendicular through 0, the y-axis. The point of intersec- tion A is the origin. Let OA = 6, and P, any point on the con- choid. Construct the right triangle POD, PO and PD meeting XX' in i^and £", respectively. From similar triangles, we have EP:FE::DP: OD. Fig. 96 Now, EP=y, DP = 11^ h, and OD = x. and hence FE Va^ — y' also for the point P', where P'F=FP=a and reducing, we have the equation sought ; namely, By construction, PF=a, It is clear that these relations hold Substituting in (1) x^y^ = (y -\- h)-{a? — y"^). Arts. 127, 128] THE WITCH OF AGNESI 171 EXERCISES 1. Construct conchoids for which a > 6, a = & and a < 6. Note the differ- ence in form. 2. Show that the X-axis is an asymptote of tlie conchoid. 3. In Fig. 97, let AB be twice the length of OF. Draw the perpendicu- lar XX' at F and let it meet the conchoid at K. Draw OK and take KB = OF. Show that KB = RF = OF, and con- sequently the angle KOB is one third the angle FOB. Show how this construction enables one to trisect any given angle. Note. The conchoid was invented by Nieome- des for the purj)ose of tri- secting a given angle. This is another famous Fio. 07 problem of antiquity. Neither the duplication of the cube nor the trisection of an angle can be effected by means of the circle and straight line alone, hence the ancients were forced to the invention of other curves for these purposes. Nicomedes was a contemporary of Dio.cles. 128. The Witch of Agnesi. Let G be the center of a circle whose radius is a ; and OB any diameter. Draw the tangents at ^- and B ; and let OR be any chord through which, produced, meets the tangent at B in N. Through R draw the parallel to OD, the X-axis, and through N, the parallel to OB, the Y-axis. The locus of the point P, where these parallels meet, is called the witch of Agnesi. Fig. 98 172 LOCI OF HIGHER ORDER [Chap. IX. EXERCISES 1. Show that the equation of the witch, referred to the lines OX and OY 8a3 as coordinate axes, is y + 4a2 Suggestion. Use the similar triangles NBP and NOD to find the re- lation between the coordinates of P. 2. Shpw that the X-axis is an asymptote of the witch. Note. Donna Maria Agnesi, who invented the witch, was horn at Milan, 1718, and died there, 1799. She was appointed Professor of Mathematics at the University of Bologna, 1750. 129. The Limacon of Pascal. Let C be the center of a circle whose radius is a ; and OD, any diameter. Through draw a series of lines, and on each of these lay off a distance b on each side of the circle. The locus of the points thus determined is called the limacon. To find the equation of the limaqon, let be the pole and OD the polar axis. The length of the chord within the circle is 2 a cos 6. Hence the radii of the points P and P' on this chord are given by the equation r = 2 a cos ± b, which in rectangular coordinates reduces to (a;2 + / - 2 axf = b' (x' + /). EXERCISES 1. Construct the limagons for which 6 > 2 a, & = 2 a, and 6 < 2 a. Note the difference in form. When & =2 a, the limacon is called the car- dioid from its heart-shaped form. 2. When b = a, the limagon fur- nishes a neat curve for trisecting a given angle. In Fig. 99, let PCB be the given angle. Show that PM = MC = CO and, therefore, the angle POB is I the angle PCB. Note. Pascal (1623-1662) was a celebrated French mathematician and philosopher. Fig. 99 Arts. 129, 130] THE CYCLOID 173 MISCELLANEOUS EXERCISES 1. Show that the locus of the intersection of a tangent to the parabola 2/2 =— 8 ax and a line drawn through the origin perpendicular to this tangent is the cissoid. SuGGESTiox. The equation of a tangent in terms of the slope is (Art. 95, 2 a Eq. 9) 2/ = inx — , and the equation of the line through the origin per- pendicular to this tangent is, y — '-■ The locus of the intersection of these m two lines is found by eliminating m from the two equations. 2. A tangent is drawn to the parabola 2/^ = 4 px at a point T. The per- pendicular to this tangent through the origin meets the ordinate of T, pro- duced, at P. Find the equation of the locus of P as T moves along the curve. The locus is called the semicubical parabola. 3. The two parabolas j/^ = 2 ax and x^ = ay meet at the origin and also at another point P. Find the coordinates of P. If a is the edge of a given cube, show how the construction of the two parabolas solves the problem of the duphcation of the cube. 4. Show that the conchoid is the locus of the points of intersection of the line y =- — b with the circle (x — bky'+ y'^ = a-, k being a variable ^■ parameter. 5. A tangent is drawn to the equilateral hyperbola ^" — y'^ — cfi at the point T. The perpendicular to this tangent through the origin meets the tangent in the point P. Show that the locus of P, as T moves along the curve, is the lemniscate (Art. 54). 6. Find the locus of the intersection of the two straight lines X -\- ky + a{k'^ — 3)= and y = kx, k being a variable parameter. The locus is called the trisectrix of Maclaurin. Discuss its equation and draw the locus. TRANSCENDENTAL LOCI 130. The cycloid. The locus of a point in the circumference of a circle which rolls (without sliding) along a fixed straight line is called the cycloid. The circle is called the generating circle. To find the equation of the cycloid, take the line on which the circle rolls as the X-axis, the radius of the rolling circle equal to a, and one of the positions in which the tracing point is on this line as the origin 0. Let C be the center of the generating circle when the tracing point has the position P. Join P and C, and 174 LOCI OF HIGHER ORDER [Chap. IX. draw CT perpendicular to the X-axis. Let 9 represent the angle PGT. Now or = arc Pr=a^. Hence, x^OD=OT-PE = ad-a sin d, y = DP= TC - EC = a- a cos B. ^^^ Equations (1) are the parametric equations of the cycloid, being the parameter. When varies from to 2 tt, P traces out one arch of the ciirve. The entire curve consists of this arch and repetitions of it to the right and left corresponding to values of 9 outside the limits 0, 2 tt. EXERCISES 1. In Fig. 100, OB is tlie span of one arcii of the cycloid and F is its middle point. Show that the area of the triangle OAB is twice the area of the generating circle. 2. When the point T bisects OF, show that the area of the triangle OPA is half the area of the square inscribed in the generating circle. Fig. 100 3. Prove that the tangent to the cycloid at P passes through H, the upper extremity of the diameter of the generating circle which is perpendicular to the base OB. Suggestion. At any instant of the motion of the generating circle, T (its lowest point) is at rest, and the motion of every point of the generating circle is for the moment the same as if it described a circle about T. Hence the normal to the cycloid at P must pass through T. 4. If ^ = -, write the equation of the tangent to the cycloid. Note. The cycloid was much studied by the most eminent mathema- ticians of Europe during the first half of the 17th century. In particular Galileo and Pascal discovered many of its properties. Its area was found to be three times the area of the generating circle by Roberval in 1634. The method of drawing tangents (exercise 3) was discovered by Descartes. Art. 131] THE HYPOCYCLOID 175 131. The Hypocycloid. This locus is the curve traced by a fixed point on the circumference of a circle which rolls internally along the circumference of a fixed circle. To derive the parametric equations of the hypocycloid, let the radii of the fixed and rolling circles be a and b respectively. Let A be one of the positions in which ^ the tracing point lies on the fixed circle. Take the center of the fixed circle as origin, and the line OA as X-axis. Let C be the center of the rolling circle when the tracing point has arrived at the position P(;x, y), 6 the angle througli which the line of Fig. 101 centers OCB has turned, and ^ the angle through which the radius CP of the rolling circle has rotated since P left the position A. The coordinates of P are the coordinates of C plus the projections of CP upon the X- and F-axes, respectively. Hence (Fig. 101), we have X = 0M= 0H+ NP= OCcos ^ + CP cos <^ = (a - 6)cos ^ + & cos c^, y = MP = nC- NC = (a -b)s\nO-b sin <^. But arc PB = arc AB, and therefore we have b{6 + 0) pass through the origin, folloio the diagonals of the square ABCD more or less closely, and pass out of the square through A and one other vertex. III. If n is a positive number represented by - (a and b p)Time to each other), then (1) when a, is even and b is odd, the curves pass through (2) when a is odd and b is odd, the curves pass through (7 = (— 1, — 1) ; and (3) wheti a is odd and b is even, the curves pass through D^{1,-1). IV. The parabolic curves of the system fill the square ABCD and the infinite regions of the plane which corner on this square ; i.e. the shaded regions in Fig. 108. V. When n is a positive even integer, the curves touch the X-axis more and more closely the larger n is taken ; i.e. the curvature at the ori- — ^ gin becomes less and less as the value of n is increased. When ri- is a positive odd integer, the curves touch the X-axis at the origin, but the curvature changes from concave downward on the left of the F-axis to concave upward on the right. Each curve has a point of inflexion at the origin. When n is fractional, with neither numerator nor denominator equal to unity, each curve has a cusp at the origin. B ir Fig. 108 Art. 136.] LOCI OF TYPICAL EQUATIONS 185 VI. Tlie hyperbolic curves of the system (n < 0) Jill the regions of the plcme outside the square ABCD and the infinite regions corner- ing on this square ; i.e. the un- shaded regions in Fig. 108. The axes are asymptotes to each hy- perbola of the system. Types 3 and 5 (Art. 135) do not differ in form from types 2 and 4. Type 6 is illustrated in Fig. 109. Each curve of the system passes through the point (0, 1) (a is assumed to be unity in drawing these curves), the curvature depending upon the value of b. The curves illus- trate phenomena that follow the "compound interest law." Type 7 is shown in Fig. 110. If a^ = &', the curve is the witch (Fig. 98). This curve is of special importance in representing phenomena where the ob- served value (the function) gradually decreases, from a maximum Fig. 109 y=bTT. ;«=4,6=1 Fig 110 186 LOCI OF HIGHER ORDER [Chap. IX. at X — 0, as the variable . increases in value. Other curves ap- plicable to phenomena of this character are the probability curve, Fig. Ill A; y= — =6"'^'"'^', Fi^. Ill, and the curve y = a = 1, the latter is the hyperbolic secant curve. -, Fig. 112. If Fig. 112 137. Selection of type curve and determination of constants. Frequently the law which experimental data must follow is known beforehand, and then it is only necessary to determine the constants in the equation. For example, in experiments on falling bodies, the law is known to be of the form y = Cx^, where y represents the distance fallen during the time x. In this case, Arts. 137, 138] TEST BY LINEAR EQUATIONS 187 one pair of values of x and y will determine the value of the con- stant C. Experimentally determined values of any function are never absolutely exact, so that plotted points, determined from experi- ment, never all lie exactly on the curve representing the known law. The values of the constants, determined as above, are there- fore more or less approximate. The aim is to find such values for the constants as will give the best average curve to represent the observed values of the function. In case the law is not known, the curve which best represents the observed values of the function must be selected by trial. The procedure is as follows : (a) Plot the observed values carefully; (h) From the known forms of curves discussed in Art. 136, or elsewhere, select that one which resembles the plotted curve most closely ; (c) Determine the constants in the equation of the selected curve so that it will fit the observed values most closely. To fulfill the requirements (6) and (c) satisfactorily requires good judgment and a good eye as well as some knowledge of the forms of various types of curves. The results obtained are often quite as serviceable as though more intricate analysis had been employed to find them. 138. Test by means of linear equations. After a trial curve has been selected, it is often rather difficult to determine whether this curve actually represents the observed values of the function with sufficient accuracy for the purposes of the problem or not. The following device is of great assistance in determining whether to retain or reject the trial curve. The typical equations in Art. 136 can be transformed into linear equations as follows : 2. ?/ = Cx" can be written log y — log C -f n log x. 3. (y — k)=C{x — hy can be written log (2/ — A;) — log C + n log (x — h). Types 4 and 5 are included in the above. 6. y = ab^ can be written log y = log a -f- x log b. Ct 1 & 'V^ 7. y = , can be written - = - -f — . b + x"' y a a 188 LOCI OF HIGHER ORDER [Chap. IX. If the observed values satisfy sufficiently accurately an equa- tion of the form y = Cx^, say, then the logarithms of the observed values must satisfy the equation log y = log C -{- n log x. Hence, the points plotted from the logarithms of the observed values must lie closely upon a straight line. If they do not lie upon a 3 4 5 6 7 891 Fig. 113 4 5 6 7 6 91 straight line, or nearly so, the required equation is not of the form y = Cx". Similarly, if the observed values are to satisfy an equation of the type y = alf, then the values of log y and x must satisfy the equation log y = log a -\- x log h. Again, if the observed values are to satisfy an equation of the Art. 139] EXAMPLES AND EXERCISES 189 type 7, the values of - and x^ must satisfy the equation y y « o Instead of looking up the logarithms of the numbers in a given table, logarithmic paper may be used. The horizontal and vertical scales on this paper represent the logarithms of numbers. Figure 113 shows a sheet of this paper on which has been plotted the table in Example II, Art. 139. Because the plotted points lie very closely upon a straight line, we may assume the equa- tion y = Cx'^. 139. Examples and exercises. The foregoing statements will be better understood from the following illustrative examples and exercises. Example I. Find an equation which is satisfied by the following pairs of values of x and y -. a; = 12 3 4 5 6 y = -i 1.5 1.22 1.125 1.08 LOG Plotting the given pairs of values, we have the curve in Fig. 114. We now note that this cui've resembles one of the hyperbolic curves belonging to Fig. 114 the system y = Cj;", but with this difference ; instead of approaching the X-axis as an asymptote, it apparently approaches the line ?/ = 1 as an asymp- tote. We therefore assume, as a trial equation, y — 1 — Cx". If the given 190 LOCI OP HIGHER ORDER [Chap. IX. values of x and y satisfy this equation, then the values of \og{y — 1) and log X must satisfy the equation log (y - 1) = log C 4- n log x, From the given values of x and y, we or the equation of a straight line. obtain, logx =0 .301 .477 log (y - 1) = .301 - .301 - .658 .602 .903 .70 .80 1.10 -1.22 Plotting these values, we obtain Fig. 115. We see that the straight line passing through the first and next the last of these points very nearly passes through the others. The slope of this line is — 2 and the intercept on the F-axis is .301 = log 2. Hence, log (y — 1) = log 2 — 2 log X, or 2/ = l+ V a;- By actual substitution, this equation is seen to be very closely satisfied by the given pairs of values of x and y. Logarithmic paper may be used in this example, as explained in the preceding article. Example II. For water flow- ing in pipes, the loss of pressure due to friction is ' approximately propoi'tional to the square of the velocity. If y is the loss of pres- sure per 1000 feet and x is the velocity in feet per second, then (approximately) y = ax'^. Find the value of the constant a so that the following experimental data will fit the given equation closely : Fig. 115 1.9 2.8 3.6 4.3 4.8 6.1 2 4 6 8 10 15 7.2 8.2 9.1 20 25 30 Example III. Show that the observed data in the preceding example will more closely satisfy an equation of the type y = ax". See Fig. 113. Example IV. The following observations were made of wind pressure on inclined surfaces. Art. 139] EXAMPLES AND EXERCISES 191 Inclination from vertical : 30^ 40°, 50°, 60°, 70°. Pressure (pounds per square foot) : 5.5, 5.3, 4.4, 3.5, 2.1. DeteiTQine the curve representing the pressure as a function of the angle. Suggestion. Assume the equation k — y = ax'K Plot the observed values. Estimate k = 6. Plot the values of the logarithms of Q — y and x, and fit a straight line to the plotted points. jX l^'llililjl[l ':^^V:^—z" :4^§ \ :-Ll-E -r::r E£. - _.;:.^:::::: -4tfo ] --~ — r : fcfej- — -Li±+ \±- - . iw- — -^ : = ^-4 :: --++ ^— — - — ---ti -4++::::} --^50- H ^ __J lJ --4-^ m -- _ — \ , ■-' • .^ ■-■10--£ - 1 ^3T-;-" ^:: 1 I III ' 'j--^:-- \ \~ 4 ::x:;::::;::: ::::::::::::: -^ \ N ■ ■■:::- -.'_■ rf::: "10-------|3 ^ wH Srfe^ II III , --- :::&+T^3)'— 30±f90yit264-r56^E 0^2] [0-2-i0-2T0~o00-3^ 0^3G0r:390n420:^ Fig. 116 Example V. In a certain investigation upon the strains in railway bridges due to the passage of trains, the following data were found : x= 30 60 90 120 150 180 210 240 270 300 .330 360 390 420 y^lOO 95 84 72 58 45 40 30 25 22 18 16 15 12 10 192 LOCI OF HIGHER ORDER [Chap. IX. Find an empirical equation which these observations will satisfy with close approximation. Suggestion. Plot the given pairs of values carefully. Note that the curve obtained seems to approach the X-axis as an asymptote. Since the function begins with a maximum value at x = and steadily decreases in value as x increases, choose type 7 as a trial equation. Plot the values of i and x^ and fit a straight line to the plotted points. y Figure 116 shows the points plotted from the given pairs of values of x and y and also the locus of the equation y = — ~ — in which a = 2,000,000 and b + x^ b — 20,000. The scales are indicated on the figure. Example VI. In a series of experiments on the adiabatic expansion for air, the following data were obtained, where v stands for volume and p for the corresponding pressure. v= 3 4 5.2 G.O 7.3 8.5 10.0 j; = 107.3 71.5 49.5 40.5 30.8 24.9 19.8 Find the empirical equation connecting p and v. Suggestion. Since the curve obtained from the given pairs of values of p and V resembles one of the hyperbolas of the system y = Cx», plot the values of log p and log v and fit a straight line to the plotted points. The equation sought ispv^-'^ = 497.7. EXERCISES 1. If I represents the length of a steel bar and t represents temperature, find the equation connecting I and t from the following observations : / = 1 1.0004 1.0008 1.0012 1.0016 1.0024 1.0040 t = 20 40 60 80 120 200 2. Find the equation connecting Q and h from the following set of obser- vations : h= .583 .667 .750 .834 .876 .958 Q = 7.000 7.600 7.940 8.420 8.680 9.040 3. Show that the following set of corresponding values satisfies an equa- tion of the form y = a6^. Find the values of a and b. x = 2.000 3.20 4.70 8.5 10.3 12.6 ?/ = 7.086 12.64 125.07 163.0 388.4 1178.0 4. The following set of observations represents the deflection d of a beam of length L. Find the equation connecting d and L. L= 12 16 20 24 28 32 36 40 d=.17 .043 .085 .145 .220 .342 .512 .718 Art. 140] TYPE y = a + bx + ex'- + dx^ + --- + kx"- 193 5. Find the equation connecting u and v from tlie following set of corre- sponding values : 10= .5 1.1 1.70 2.30 5.10 6.40 V = 13.6 4.0 2.37 1.84 1 33 1.28 140. Type tj = a + bx + cx'^ + dx^ + ■■• + kx^. When a given set of corresponding values will not satisfy, in a satisfactory manner, any of the type-equations 1 to 7 (Art. 137), the general equation y = a-\- hx -f c.^•2 + d.v^ -\ + l-x" may be assumed. By substituting pairs of corresponding values in this equation, the values of the constants o, b, c, •••A; can be determined and may often be so adjusted that the locus of the resulting equation will represent the function to a fair degree of approximation within the limits of observation. EXERCISES 1. Show that the following set of corresponding values satisfy an equation of the form y = a + bx + cx'^. Pind the values of a, b, c: x= 8 23 39 53 63 y=10 10 27 33 36 2. Find an equation of the form y = a + bx + cx^ which will be satisfied by the corresponding values of angle and wind pressure in Example IV, Art. 139. Why is the equation found in this way not as satisfactory as the equation found in Example IV ? PAET II SOLID ANALYTIC GEOMETRY ^x CHAPTER X SYSTEMS OF COORDINATES 141. Rectangular and oblique coordinates. As has been said, it requires one number to locate a point on a line and two numbers to locate a point in a plane (Art. 4). To locate a point in space it requires three numbers, called the coordinates of the point. These coordinates may be chosen iii several different ways ; any par- ticular way of choosing them gives rise to a system of coordinates. Thus (Fig. 117), let OX, OY, and OZ be three linear scales hav- ing a common origin and not lying in the same plane. They determine in pairs three planes XOT, YOZ, and XOZ, called the coor- dinate planes. If through any point in space, as P, three planes are drawn parallel to the coordinate planes, they intersect the linear scales in the points D, E, and F. The distances x = OD, y = OJE, and z = OF are the Cartesian coordinates of the point P. The linear scales OX, OY, and OZ are called the X-, Y-, and Z- axes, respectively. The coordinate planes XO Y, XOZ, and YOZ are called the XY-, XZ-, and FZ-planes, respectively. The sys- 195 196 SYSTEMS OF COORDINATES [Chap. X. tern of coordinates thus set up is called the Cartesian system of coordinates. When the axes, OX, OY, and OZ are miitually perpendicular, the system of coordinates is called rectangular or orthogonal. If the axes are not mutually perpendicular, the system is called oblique. From the definition of the coordinates of a point, and the definition of a linear scale, it follows that, in the Cartesian system of coordinates, to each point in space there corresponds one set of values of x, y, z ; and to each set of values of x, y, z there corresponds one point in space. EXERCISES (In the following exercises, take the axes to be mutually perpendicular. Cross-section paper may be used.) 1. Plot to scale the following points, the coordinates being always written in the order (x, y, z): (1, 1, 1), (2, 0, 3), (- 4, - 1, ^ 3), (- 4, 2, 8), (0, 0, 2), (1, - 3, 0). 2. Find the distance between the points (1, — 2, 3) and (— 1, 2, — 2). 3. Where are the points located for which a; = 0? 2/ = 0? = 0? What are the equations of the coordinate planes ? Where are tlie points located for which x — a ; y —b ; z = c? What are the equations of the planes parallel to the coordinate planes ? 4. Where are the points located for which x = and y = ? for which X = a and y = b? for which x = y? for which x = y — z? 5. The points (2, 2, 3), (2, 4, 3), (4, 2, 3), and (3, 3, 2) are four of the vertices of a parallelopipedon. Find the coordinates of the remaining four vertices. Is there more than one solution to this problem ? 142. Spherical coordinates. Let OX, OY, OZ (Fig. 118) be a set of rectangular axes, and P any point in space. The distance OP = r, the angle ZOP = 6, and the angle which the plaiie ZOP makes with the fixed plane XOZ = <^ are the spherical coordinates of the point P. They are written in the order (r, 6, (f>). If the point P is on the surface of the earth, then 6 is the co- latitiide and <^ is the longitude of P. If P is on the celestial sphere, then is the co-declination and ^ the right ascension of P. If Z is the zenith, then 6 is the zenith distance and (^ is the azimnth of P. Arts. 142, 143] CYLINDRICAL COORDINATES 197 143. Cylindrical coordinates. In Fig. 118, let OD=r', the angle XOD = (f>, and DP = z ; (?•', <^, z) are the cylindrical coordi- nates of P. Again, let a, P, y denote the angles which 2 OP = r makes with the X-, Y-, and Z-axes, respec- tively ; then (r, «, /3, y) are the polar coordinates of P. Spherical coordinates and cylindrical coordinates are modifications of polar co- ordinates in space. Each is in common nse and each has its advantages. Spher- ical coordinates are espe- cially nseful in astronomy and in geodetic surveying. Fig. 118 EXERCISES 1. Using Fig. 118, show that the rectangular coordinates of P and the spherical coordinates of P are connected by the following formulas : x = r s\nd cos 0, y = r sin 6 sin 0, z = r cos 0. Conversely, show that r"^ = x^ + y- -\- z^, tan^g= ^' + y' , tan (p — 2. What are the formulas connecting the rectangular coordinates of P with the cylindrical coordinates of P ? 3. Will a given set of integral or fractional values of r, 6, 4> or of )-', <^, z locate one and only one point in space ? Does a given point in space have more than one set of polar coordinates ? 4. Locate the points whose spherical coordinates are : (3, 30°, 60°) , (2, -, TT ], (1, 45°, 45°). Find the rectangular coordinates of these points. 198 SYSTEMS OF COORDINATES [Chap. X. 5. Find the spherical coordinates and also the cylindrical coordinates of the following points: (2, 3, 4), (3, 3, - 2), (- 1, - 2, 1). 6. ^Vhere are the points located for which r — const. ? for which 9 — const. ? for which = const. ? for which >■' = const.? 7. Where are the points located for which — const, and ^ = const. ? for which — const, and r — const. ? for which r — const, and Q = const. ? for which r' — const, and z = const. ? CHAPTER XI E, A DIRECTED SEGMENTS IN SPACE 144. Projections upon the coordinate axes. As in plane geome- try, we shall call a segment of a straight line to which a direction has been attached, a directed line-segment, or simply, a directed segment. If P1-P2 is a directed segment, then Z Pj is called the initial point, and P,? the termi- nal point. If planes are drawn throngh the initial and terminal points of a di- rected segment and per- pendicular to each of the coordinate axes in turn, these planes will deter- mine upon each axis a segment called the pro- jection of the given di- rected segment upon that axis. In Fig. 119, let Pi=(xi, 2/1, z^) and P2 = (x'2, yz, z^) ; then we have projection of P1P2 upon the X-axis = x^ — x^. projection of P1P2 upon the I^axis = 2/2 — Vi- projection of P1P2 upon the Z-axis = Z2 — z^. 145. Length of segment. A segment P1P2 is the diagonal of a rectangular parallelopiped whose edges are the projections of the segment upon the coordinate axes. Hence, we have Fig. 119 P,P, = ^{x, - x,f+ {y, - y,y+{z, - z,y. 199 200 DIRECTED SEGMENTS IN SPACE [Chap. XI. EXERCISES 1. Find the lengths of the following segments and their projections upon the coordinate axes. (a) (1, 2, 3), C- 2, 1,1); (b) (0, 0, 0), (2, 0, 1) ; (c) (3, - 2, 0), (2, 3, 1) ; (d) (0,4,1), (-2, -1,-2); (e) (0,3,0), (3, -1,0). 2. A straight line five units in length has one extremity at the origin and is equally inclined to the coordinate axes. Find its projections upon the axes. 3. The initial point of a directed segment is at the point (3, 2,-1) and its projections upon the X-, F-, and Z-axes are respectively 4, — 6, and — 2. Find the coordinates of the terminal point and construct the figure. 4. If the terminal point of a directed segment is (—1, 3, 5) and its pro- jections upon the X-, Y-,. and Z-axes are respectively — 2, 3, and — 6, what are the coordinates of the initial point and the length of the directed segment? 146. Direction angles and direction cosines of a directed segment. The angles which a directed segment makes with the positive directions of the coordinate axes are called the direction angles of the segment. The cosines of the direction angles are called the direction cosines of the segment. Through Pi draAV lines paral- lel to the axes ; i.e. the lines P,X', P,Y', P,Z' (Fig. 120). The direction angles of P1P2 are then, ^^_^.,p^p^^ ^=Y'P,P„ y = Z'P^P^. If I is the length of P^P^, then the direction cosines are given by the equations: ^X COS « = I COS /3 _ ^2 - yi Fig. 120 COS y 147. Relation connecting the direction cosines of a segment. Theorem. The sum of the squares of the direction cosines of any segment is equal to unity. Arts. 146, 147] RELATION CONNECTING COSINES 201 For let I be the length of any segment. Then (Art. 145), we have Z2 = (a,'2 - a."i)2 + (?/2 - 2/i)2 + {z^ - z^f. Dividing by l"^, we obtain I J \ I J \ I or COS^ « + COS^ ^ + COS^ y = 1. EXERCISES 1. Find the length and the direction cosines of each of the following segments : Pi=(4, 3, -2). P2=(-2, 1,-5); Pi=(4, 7, -2), P2 = (3, 5, -4); Pi=(3, -8,6), Po=(6, -4,6). 2. Find the lengths and the direction cosines of each side of the triangle whose vertices are the points (.3, 2, 0), (— 2, 5, 7), and (1, — 3, — 5), the sides being taken in the order given. 3. Given the direction cosines of the segment P1P2 ; what are the direc- tion cosines of the segment P2P1 ? What is the direction of a segment when cos a =0 ? when cos j3 =0 ? when cos 7 = 0? when cos a — cos ^ — 0? when cos a = cos 7 = 0? when cos ^ = cos 7 = ? 4. A segment is five units long and its initial point is (— 2, 1, — .3). If cos « = J and cos /3 = |, find the coordinates of the terminal point and the projections upon the axes. There are two solutions, find each of them and construct the figure. 5. Show that the direction cosines of each of the lines joining the points (4, — 8, 6) and (— 2, 4, - 3) to the point (12, — 24, 18) are the same. How are the points situated ? 6. Find the direction angles of the segment drawn from the origin to the point (8, 6, 0). From the origin to the point (2, — 1, — 2). 7. Show by means of direction cosines that the three points (3, — 2, 7), (6, 4, — 2), and (5, 2, 1) lie on a straight line. 8. If two of the direction angles of a segment are - and - , what is the third ? ^ * 9. Show that the numbers 3, — 4, and — 2 are proportional to the di- rection cosines of the segment joining the origin to the point (3, — 4, — 2). 10. Show that any three real numbers a, b, and c are proportional to the direction cosines of the segment joining the origin to the point (a, b, c). 202 DIRECTED SEGMENTS IN SPACE [Chap. XI. Fig. 121 148. Projection of a segment upon any line. Let PiP^ be any segment, and AB any line in space. Through the extremities of the segment draw planes perpendicu- lar to AB. These planes determine a segment CD upon AB which is called the projection of PiPo upon AB. Through P^ draw a line parallel to AB, meeting the planes in the points Pj and Q. Let 6 represent the angle P2P1Q ; then CD = P,Q = P,P2 cose. 149. Projection of a- broken line. A. series of segments so ar- ranged that the terminal point of each is the initial point of the next following and the terminal point of the last is the initial point of the first, constitutes a closed line, or polygon, in space. The sum of the projections of the sides of a closed line upon any line in space is clearly equal to zero. It follows from the fore- going property that : Theorem. The sum of the i^Tojections of a series of segments joining the point A to the point B, upon any straight line in space, is equal to the projection of the segment AB upon that Z J'l line. For, the succession of segments AP^, P1P2, P.P. BA forms a closed line, and hence the sum of the projections of its sides upon any line is equal to zero; i.e. the sum of the projections of Pj^ ^22 the sides of the broken line joining J. to B is equal to the projection of the straight line joining A to B. Arts. 148-151] PERPENDICULAR SEGMENTS 203 150. The angle between two segments. When two segments do not intersect, the angle between them is defined to be the angle between two intersecting segments drawn parallel to, and agreeing in direction with, the given segments. To find the angle between two given segments in terms of their direction angles, let li and h be the lengths, and aj, ^i, yj ; Ojj /?2> 72? their respective direction angles. From the origin draw two segments, OP and OQ, having lengths and direc- tion angles equal respectively to the lengths and direction angles of the given segments (Fig. 123). By defi- nition, 6 = POQ is the angle to be found. Let the coordinates of P be X = OD, y = DE, and z = EP. Now, by the preceding article, the projection of the broken line ODEP upon OQ is equal to the projection of OP upon OQ. That is, ?j cos = X cos (u-\-y cos fSo + z cos y2. Dividing through by /j and remembering that - = cos a^, etc., we Fig. 123 have cos e = cos aj COS a.2 -|- COS Pj COS P2 + COS 7i cos 73. We will assume that the angle between the given segments is the smallest positive angle satisfying this equation ; that is < ^ < TT. 151. Perpendicular segments. Parallel segments. (a) Tivo segments are perpendicular to each other if . cos a, cos «2 + cos (3i cos jB^ + cos y^ cos y2 = 0. For then cos ^ = and therefore 6 = 90°. (6) Tioo segments are parallel and extend in the same direction if their direction angles are equal, each to each. 204 DIRECTED SEGMENTS IN SPACE [Chap. XL For then the expression cos a^ cos «2+cos /3i cos ^82 + 003 y^ cos 72 becomes cos^ a^ + cos^ /3i + cos^ yi = 1 (Art. 147). Therefore, in this case, cos 6=1, and 6 = 0°. (c) Two segments are parallel and in opposite directions if their angles differ by 180°, each from each. ^ovihen cos«i=-cos«2, cos /?! = — COS 182, cos yi = — cos y2. Hence the expression cos a^ cos a2 + cos fi^ cos IB2 + cos yi cos ya be- comes — (cos- «! + cos^ y8i + cos^ yj) = — 1. Therefore cos 6= — l, and 6* = 180°. EXERCISES 1. Find the angle between two segments whose direction cosines are as follows : r n\ 6 3 2 finfl 3 2 fi . (h'\ ^ 2 1 n-nA 3 6 2. /'/.N 2 _ 1 2 (^a) 7, y, y ailU y, — y, y, ^U ) 35—3,-3 di^U y, y, y, \^h ) 3, 3, 3- rrn/l 3_ _4_ 12 ""^ 13' 13' 13' 2. Show that the lines whose direction cosines are f , ^, f ; — |-, f , — f ; and — f , I, f are mutually perpendicular. 3. Show that the points having the coordinates (— 6, 3, 2), (3, — 2, 4), (5, 7, 3), and (— 13, 17, — 1) are the vertices of a trapezoid. 4. Show that the points (7, 3, 4), (1, 0, 6), and (4, 5, —2) are the ver- tices of a right triangle. 5. Show that the points (7, 2, 4), (4, - 4, 2), (9, - 1, 10), and (6, - 7, 8) are the vertices of a square. 6. Prove that if the direction angles of two segments are supplementary, each to each, the segments are parallel and in opposite directions. 7. Find the length of the projection of the segment Pi= (3, 2, — 6), P2 = (— 3, 5, - 4) upon the line drawn from (1, 2, 3) to (3, 3, 1). 8. Find the length of the projection of the segment Pi = (6, 3, 2), P2= (4, 2, 0) upon the line drawn from (7, — 6, 0) to (— 5, — 2, 3). 152. Point dividing a given segment in a given ratio. Let P be P P a point on the segment P^P^ situated so that -^~=r, a given -/-to P P r number. Then, by composition, — — = -. Tlirough P draw planes perpendicular to the coordinate axes. These planes divide Art. 152] POINT DIVIDING A GIVEN SEGMENT 205 .Z the projections upon the axes in exactly the same ratio as P divides the segment; that is P,P ^ D,D ^ E,E ^ F^F ^ r P,P,~ D1D2 E,E., F,F, r + 1 OD = X, on, = x„ DiD^ = X2 — .^'l. Substituting, we ^X have X^— U/1 ~|~ v*^2 1 y Fig. 124 '^1 "r t'^2 ~ r+1 * Similarly, v^e obtain r + 1 r + 1 r + 1 Z = Zi + (Zo — Zi) r ^ gj + rz2 r+1 r+1 EXERCISES 1. Fiud the coordinates of the point dividing the segment joining the following points in the given ratio r. (a) (3, 4, 2), (7, - 6, 4), r = 2. (b) (7, 3, 9), (2, 1, 2), r = 4. 2. Sliow that the coordinates of the point bisecting the segment (^.1, ?/i, ^i), (3-2, 2/2i •52) aie — - — , •- — —^-, — 3. Find tlie coordinates of the points which trisect the segment (1, — 2, 4), C-3, 4, 5). 4. Show that the medians of the triangle whose vertices are the points (1, 1, 0), (2, - 1, 1), and (3, 2, - 1) meet in the point (2, f, 0). 206 DIRECTED SEGMENTS IN SPACE [Chap. XI. 5. Show that the medians of any triangle meet in a point. Suggestion. Let the coordinates of the vertices be (xi, ?/i, Zi), (0:2, 2/2, ^2), and (X3, 2/3, Z2). The medians meet in the point a:i + a:2 + Xz ?/i + 2/2 + Vs ^i + z. + zs 3 ' 3 ' 3 ' This point is the center of gravity of the triangle. 6. Show that the lines joining the middle points of opposite edges of a tetrahedron pass through the same point and are bisected by that point. 7. Show that the lines joining the vertices of any tetrahedron to the point of intersection of the medians of the opposite face meet in a point which is three fourths of the distance from each vertex to the opposite face. This point is called the center of gravity of the tetrahedron. 8. Find the ratio in which the point (2, — 1, 5) divides the segment (4, 13, 3), (3, 6, 4); the point (2, —2, —6) divides the segment (4, - 5, - 12), (-2, 4, 6) ; the point (2, 1, 4) divides the seg- ment (-3, 4, 2), (7, -2, 6). CHAPTER XII LOCI AND THEIR EQUATIONS 153. Surfaces and curves. Iii space there are two kinds of loci to be considered. If a point moves according to a given law, it will, in general, describe a surface. Thns, if a point moves so as to be always at a given distance from a fixed point, it will describe a sphere whose center is the fixed point and whose radius is the given distance. If a point moves so as to satisfy simultaneously two independ- ent laws, it will, in general, describe a line, straight or curved. Thus, if a point moves so as to be at a fixed distance from the point A and at the same time at a fixed distance from the point B, it will describe the circle of intersection of the two spheres whose centers are at A and B and whose radii are the given fixed distances. 154. Equations of loci. AVhen the law governing the motion of a point is expressed in terms of the coordinates of the point, the resulting equation is called the equation of the surface de- scribed by the point. The surface is called the locus of the equation. Similarly, when a moving point is governed by two independ- ent laws and these laws are expressed in terms of the coordi- nates of the moving point, the resulting equations are called the equations of the curve described by the point. The curve is called the locus of the equations. As in plane geometry, two fundamental problems arise : First, given the law (or laws) governing the motion of a point, to find the equation (or equations) of the locus ; and second, given the equation (or equations), to find the properties of the locus. These problems will be illustrated in the succeeding pages. 207 208 LOCI AND THEIR EQUATIONS [Chap. XII. 155. The sphere. Let C= (a, b, c) be the center of a sphere whose radius is r, and P= (x, y, z), any point on the sphere. .The length of CP is then, r r = V(x - ay -\-{y- hf + {z - cf. Hence, the equation of the sphere is ^'^ Fxo. 125 (^^ - «)' + iV-bf+i^- of = r^^^ When the binominal squares are expanded, the equation has the form ^2 _^ ^2 ^ ^2 + ^4.^ + By + Cz + D = 0, (2) where A, B, C, and D are constants depending upon the coor- dinates of the center and the radius. Conversely, an equation of the form (2) represents a sphere. For it can be written in the form , AW f , BW f ^ CV A'B-'.C' r) A _B _C 2j \ 2j \ 2 J 4. 4: and hence represents a sphere whose center is JA"^ B^ C^ and whose radius is a h -r + ^; — ^^• \ 4 4 4 The sphere is real, so long as the expression under the radical is positive ; it will be a null-sphere, or a point, when the expres- sion under the radical is zero ; and it will be an imaginary sphere when the expression under the radical is negative. EXERCISES 1. Write the equation of a sphere whose center is (5, — 2, 3) and whose radius is 1 ; also of a sphere whose center is (2, — 3, — 6) and which passes through the origin. What is the equation of a sphere whose center is on the Z-axis, has the radius a, and passes through the origin ? 2. Which of the following spheres are real, which are null-spheres, and which are imaginary spheres ? Find the center and radius of the real spheres. Arts. 155, 156] SURFACES OF REVOLUTION 209 (a) x"^ + y^ + z^ -2x + 6y - 8z + 22 =0. (b) x^ + tf + z^ + 10 X - 4 y + 2 s + = 0. (c) x^ + y^ + z^ + 4x + 4y + 6 z + I = 0. (d) x^ + y^ + z^ + 6x = 0. (e) x^ + y"' + z^ + 4x + y + 6 z + 21 = 0. 3. Find the equation of the sphere passing through the four points (0, 0, 0), (2, 8, 0), (5, 0, 15), (- 3, 8, 1). Suggestion. Substitute the coordinates of the given points in equation (2) and solve the resulting equations for the unknown coefficients A, B, CD. 4. Find the equation of the sphere passing through the four points (2, 5, 14), (2, 10, 11), (2, 5, - 14), (2, - 10, -^11). 5. Find the equation of each of the two spheres whose center is at the origin and which touch the sphere x2 + y-2 + ^-2 _ 8a;- 6?/ + 242; + 48 = 0. 156. Surfaces of revolution. When a curve in tlie XZ-plane is rotated about the JY-axis, it describes a surface of revolution. Every point on the curve, as Q, describes a circle whose plane is perpendicular to the X-axis and whose radius is the ordinate DQ. Let the coordinates of Q be OD = X and DQ = z', and the equation of the curve MQR be f{x,z')=Q. Xow z' = DQ = DP = Vz^ + 9/. Hence, we have the following conclusion : Fig. 126 To find the equation of the sur- face described by MQR, replace z' in the equation f(x, z')= by its value ^/z^ + y-. By a similar consideration we may find the eqviation of a sur- face of revolution obtained by rotating a given curve about either of the other axes. 210 LOCI AND THEIR EQUATIONS [Chap. XII. EXERCISES 1. Find the equation of the ellipsoid obtained by rotating the ellipse — I = 1 about the X-axis. The ellip.soid of revolution is called the pro- late spheroid when a > &, and the oblate spheroid when a < 6. Explain, by familiar examples, the difference in form. 2. Find the equation of the paraboloid of revolution by rotating the parabola z'~ = ipx about the X-axis. 3. If the hyperbola - — — = 1 and its conjugate — —— 1 are ro- tated about the X-axis, how will the two surfaces obtained differ ? Find their equations. The first is called an hyperboloid of two sheets and the second, an hyperboloid of one sheet. 4. Show that if a curve in the XF-plane, whose equation is /(a;, y) = 0, is rotated about the X-axis, the equation of the resulting surface is found by replacing y by Vy'^ + z^ ; and if the curve is rotated about the F-axis, the equation of the resulting surface is obtained by replacing x by Vx"^ + z'^. 5. What is the equation of the surface obtained by rotating the parabola ?/2 =:4^x about the X-axis? about the F-axis? How do the two surfaces differ ? 157. Cylinders. If a straight line moves so as to be always parallel to one of the coordinate axes and, at the same time, intersects a curve lying in the plane of the other two axes, it describes a cyl- inder whose equation is the same as the equation of the curve. Eor, suppose the moving line is always parallel to the Z- axis and meets a curve in the XY-plane ; then the x- and ^/-coordinates of any point on tliis line will be the same as the X- and ^/-coordinates of the point where the line meets the curve, and will conse- quently satisfy the equation of the curve whatever be the value of z. Moreover, the x- and y-co- ordinates of a point not on the cylinder cannot satisfy the equa- tion of the curve. Therefore the equation of the curve, re- garded as the equation of a locus in space, represents a cylinder parallel to the Z-axis. Similarly we may obtain the equation ■^X Fig. 127 Arts. 157-159] PLANE SECTIONS OF RIGHT CONE 211 of a cylinder parallel to any other axis. Hence, we have the conclusion : Any equation in tivo of the three variables x, y, z represents a cylinder jiarallel to one of the coordinate axes. EXERCISES 1. The following equations represent loci in space. Interpret them and draw the figures, (a) — -f ^ = 1 ; (&) z^ = -^px; {c) s + 3 «/ = 6. a- t- 2. A point moves so as to satisfy simultaneously the two equations - + ^ = 1 and ^ + - = 1. Plot its locus in space. 2 3 4 5 3. A point moves so as to satisfy simultaneously the two equations x^ + ?/'^ = 4 and - + - = 1 . Plot its locus in space. 4. Show that a point can move so as to satisfy simultaneously the three equations 3 x + 2 ?/ = 6, 5 j/ + 4 = 20, and 8 2—15 x = 10. 158. The right circular cone. When the straight line z = mx is revolved about the X-axis, it generates a right circular cone whose vertex is at the origin and whose axis is the X-axis. Every generator of the cone makes an angle with the axis whose tangent is ni. By Art. 156, the equation of this cone is f/"2 + s^ = in-x'^. 159. Plane sections of a right circular cone. Let APB (Fig. 128) be the curve common to the cone and any plane, as AFPB. In- scribe a sphere in the cone touching the cutting plane at F, and the cone along the small circle LES. The cutting plane and the plane of the circle meet in the line I)D^. Through P, any point of the ciirve APB, draw the generator of the cone VP, meeting the small circle in E. From P drop the perpendicular PK upon the plane LES, and draw PR perpendicular to DD^. The angle PRK= a is the angle between the cutting plane and the plane LES and is therefore constant for all positions of P. The angle PEK = (3 is also constant for all positions of P. The lines PF and PE are equal in length, since they are tangents to the sphere from an external point. Hence, PF^^PE^PK^ PK ^ sina . PR PR PR ' PE sin/3' 212 LOCI AND THEIR EQUATIONS [Chap. XII. V Fig. 128 and the curve APB is therefore a conic having one focus at F, the corresponding directrix being Z)Z), (Art. 94, property A). The conic will be- an ellipse when a < y8, a parabola when a = 13; i.e. when the cutting plane is parallel to one of the generators of the cone, and an hyperbola when a > ft. In the figure, a is less than fi and the section of the cone is therefore an ellipse. EXERCISES 1. The equation of a right circulai- cone in spherical coordinates is ^=const. By means of the relations, Art. 143, exercise 1, transform this equation to rectangular coordinates. 2. Rotate the straight line - + - = 1 about the Z-axis and thus obtain the 2 3 equation of a right circular cone whose vertex is at the point (0, 0, 3). 3. From Fig. 128, show how to locate the second focus of the section of the cone and its corresponding directrix. 4. The cone in example 2 is cut by a plane parallel to the Y-axis and meeting the XZ-plane in the line - + ?= 1. Find the coordinates of the ^ 3 2 foci of the ellipse which this plane cuts from the cone. 5. The equation — + ^ — = represents a right circular cone. Write a^ (j2 j,2 the equation of the straight line which describes this cone and tell about which axis it is revolved. CHAPTER XIII THE PLANE AND THE STRAIGHT LINE IN SPACE 160. The normal form of the equation of a plane. Let p denote the length of the perpendicular from the origin to the plane, and a, (3, y, the direction angles of this perpendicular. If L is the foot of the perpendicular, then any point, as P, will be in the plane if the angle OLF is a right angle ; i.e. if the projection of the segment OF upon OL is equal to jk But the projection of OF upon OL is equal to the projection of the broken line ODEF upon OL (Art. 149). Leb the coordinates of F be OD = x, DE = y, and EF = z ; then a? cos a -h ?/ cos P + s cos 7 = 1? (1) is the equation sought. It is called the normal form of the equation be- cause it is expressed in terms of the perpendic- ular from the origin. It follows that the equation of a plane is of first degree in the vari- ables. We shall now show that, conversely, every equation of the first degree in the variables X, y, z, is the equation of a plane. For, let Ax + By + Cz + D = (2) be any equation of the first degree in x, y, and z. Now if the coordinates of a point F satisfy this equation, they will still sat- "•21?, Fig. 129 214 THE PLANE [Chap. XIII. isfy it after each of the coefficients A, B, C, D is multiplied by any constant k. The constant Ti can be chosen so that the equa- tion TcAx -\- JcBy -{- kCz = — kD will coincide with equation (1), term for term ; that is, so that kA = cos a, kB = cos ji, kC= cos y, kD = -2x Squaring and adding the first three of these equations, we have kXA" + £2 ^ (7-)= 1 ; (Art. 147) and therefore k = . In order that p may be positive, the sign of the radical must be opposite to the sign of D. With this value of k, equation (2) agrees in form with equation (1) But (1) is the equation of a plane; therefore (2) is the equa- tion of a plane for which A cos a = cos/3 = cos y — P = ± V^2 + B'+C^ B ±V.4' + 5'+C2' C ± V^2 + B^+ C^' -D The distances from the origin to the points where a plane meets the coordinate axes are called the intercepts. The lines in which a plane meets the coordinate planes are called the traces. EXERCISES 1. Construct the planes and find their equations, for which {a) tt =-i |3=-, 7=^,i>=4; (6) «= — , j8 = 5J:, ^ = Zi:,p=6; (c) cos « : cos i3 : cos 7 00 . o 4 o = 6 : — 2 : 3, j9 = 8 ; (d) cos 06 : cos ^ : cos 7 = — 2 : — 1 : — 2, p = 5. 2 rind the equation of the plane such that the foot of the perpendicular from the origin to the plane is the point (a) (3, —2, 6); (6) (2, —5, 1) ; (c) (3,4, -2). Arts. 161-162] EQUATION OF A PLANE 215 3. Reduce the following equations to normal form and find a, ^, 7, and p. (a) 6x-3y + 2 2-7=0. (6) x - V2 ?/ + ^ + 8 = 0. (c) X - 4 ?/ - 2 - 3 = 0. (fZ) X - 2 2/ - 3 = 0. 4. Find the intercepts and equations of the traces of the following planes, (ff) 2 X + 5 (/ - 3 g - 4 = 0. (6) X - ?/ - s + 10 = 0. (c) 3 x - ?/ + s = 0. 5. Find the area of the triangle which the coordinate planes cut from the plane 2 x + 2 ?/ + - 12 = 0. 161. Intercept form of equation. Let the x-, y-, and 2-intercepts of a plaue be a, b, and c respectively ; then (Fig. 126) the plane passes through the three points A =(a, 0, 0), B=(0, b, 0), and C = (0, 0, c). Since the equation of the plane is of the form Ax + By + Cz + D==0, this equation must be satisfied by the coordinates of the points A, B, and C. Hence, , ^ „ Bb + D = 0, Cc + D = 0, from which A = — , B = , and (7= . Substituting and a b c reducing, the required equation is ^ + i/ + ^=l. a b c 162. Equation of a plane through three given points. If a plane is required to pass through three fixed points, the coordinates of these points must satisfy the general equation Ax + By + Cz+D = 0, and there are thus three equations from which to determine three of the unknown coefficients A, B, C, D in terms of the fourth. Substituting the three coefficients thus determined in the general equation gives the equation of the plane through the three given points. EXERCISES 1. Write the equation of each of the planes having the following inter- cepts and find the length of the perpendicular from the origin upon each : («) 3, 1, 2. {h) -1,-2,3. (c)4, -2, 5. (d) - 5, 2, - 3. 216 THE PLANE [Chap. XIII. 2. Find the equation of the plane passing through the points (1, 0, 2), (0, 3, 4), and (— 1, 5, 0). Find the intercepts and the perpendicular from the origin. 3. Why will not the three points (1, 1, 2), (3, - 1, 3), and (5, - 3, 4) determine a plane ? What are the direction cosines of the segments which join the first point to each of the other two ? 4. From each of the points (2, 3, 0), (- 2, - 3, 4), and (0, 6, 0) drop perpendiculars to the XZ-plane. What are the coordinates of the feet of these perpendiculars ? What is the area of the triangle formed in the XZ- plane ? Drop perpendiculars to each of the other coordinate planes and compute the areas of the triangles formed in each. These triangles are called the projections of the space triangle upon the coordinate planes. 163. Determinant form of the equation. If a plane is required to pass through three given points (Xi, y^, Zi), (x^, y^, z^, and (.^•3, ?/3, z^, the general equation Ax + 5^/ + C^ + Z) = (1) must be satisfied by the coordinates of these points. Hence the following equations hold, Ax, + By, + Cz, + D = 0, (2) Ax^ + By^ + Cz, + D = 0, (3) Ax, + By,+ Cz,-\-D = 0. (4) But in order that the four equations (1) to (4) may be satisfied by other than zero values of A, B, C, and D, it is necessary and sufficient that the determinant of their coefficients shall vanish ; that is, we must have = 0. This equation is of the first degree in the variables x, y, z ; and it is clearly satisfied by the coordinates of the given points. Therefore it is the equation of the plane passing through these points. X y z 1 x^ Vi Zi 1 *^2 2/2 ^2 1 .^•3 3/3 h 1 Arts. 163, 164] PERPENDICULAR DISTANCE 217 EXERCISES 1. Using the determinant form, find the equation of the plane which passes through the points (2, 3, 0), (— 2, — 3, 4), and (0, 6, 0). 2. In the same way, find the equation of the plane passing through the points (1, 1, - 1), (- 2, - 2, 2), and (1, - 1, 2). 3. Show that the direction cosines of the normal to a plane passing through three given points are proportional to the cofactors corresponding to X, y, and z in the determinant form of its equation. 4. Show that the cofactors corresponding to x, ?/, and z are proportional to the areas of the projections of the triangle whose vertices are (xi, yi, zi), (X2, 2/2, ^2)5 and (x^, 2/3, 03), upon the coordinate planes. 164. Perpendicular distance from a plane to a point. Given the equation of the plane ABC and the coordinates of the point Fig. 130 Pj = (xi, yi, Zi), it is required to find the length of the perpendic- ular DPi, where Z) is a point in the plane ABC and P^ lies outside of this plane. If the equation of the plane is not in the normal 218 THE PLANE [Chap. XIII. form, reduce it to that form (Art. 160) so that the equation is xcos a + y cos (3 + z cosy —2:> = 0, (1) where the direction angles and the perpendicular are known. Let d be the length of the required perpendicular, so that the projection of OPi upon OJV is equal to jj + d. But this projection is equal to the projection of the broken line OEFP^ upon ON. Hence ' p-\-d = x'l cos cc + ^1 cos /3 -{-Zi cos y, or (I = if 1 cos a + yi cos p.+ si cos y—p. The length of the perpendicular is therefore equal to the result of substituting the coordinates of the given point in the left member of (1). The result of the substitution will be negative if the point lies on the same side of the plane as the origin; and positive if the point and the origin are on opposite sides of the plane. EXERCISES 1. Find the distance from the plane 6x — By + 2z — 10 = to the point (4, 2, 10). From the plane ix + 3ij + 12 z + 6 = to the point (9, -1,0). State if the point and the origin are on the same side, or on opposite sides, of the plane. 2. Find the length of the altitude of the tetrahedron from the vertex (2, 0, 1) to the plane of the vertices (0, 5, — 4), (0, 3, 1), and (2, - 7, 1). 3. The X- and y-intercepts of a plane are 3 and 4, respectively, and the plane touches a sphere whose center is at the origin and whose radius is 2. Find the equation of the plane. 4. Find the volume of the tetrahedron whose vertex is the point (5, 5, 6) and whose base is the triangle cut from the j)lane x + 2?/ + 5s— 10 = by the coordinate planes. 5. Find the volume of the tetrahedron whose vertices are (3, 4, 0), (4, — 1, 0), (1, 2, 0), and (6, — 1, 4). Of the tetrahedron whose vertices are (3, 0, 0), (0, 1, 0), (0, 0, 6), (5, - 2, 4). 6. Find the locus of points which are equally distant from the two planes x — 2y -\-3z -4^ = and 2 x + Stj — z - 5 = 0. 7. What is the equation of the locus of a point which is equally distant from the origin and from the plane x + y + s; — 1=0? 165. Angle between two planes. The angle between two planes is equivalent to the angle between the perpendiculars to these Arts. 165, 166] PENCIL OF PLANES 219 planes. Let the equations of two planes be A^x + B^y + CiZ + A = and A^x + B^y + C^z + D2 = 0. The direction cosines of the perpendiculars to these planes are given in Art. 160. Hence, Art. 150, we have A,A, + B,B, + aC, cos t^ = ±VA' + A' + Ci2- ±VA' + A'+C2=' the signs of the radicals being chosen as in Art. 160. It follows from this formula that : two planes will be perpen- dicular if, and only if, A,A, + B,B, + C\C, = 0. For only then can cos B be equal to zero, and consequently 9 = 90°. Two planes will be parallel if, and only if, A2 For only then will the perpendiculars to the planes be parallel to each other. EXERCISES 1. The three planes a; + 2/ + 2 — 2 = 0, .r — ?/ — 22: = 4, and 2a5 + ?/ — 5! = 2 meet in a point forming a trihedral angle. Find the vertex of the angle and the three dihedral angles. 2. Find the equation of the plane which passes through the points (0, 3, 0) and (4, 0, 0) and is perpendicular to the plane 4a: — 6x — s— 12 = 0. 3. Find the equation of the plane which passes through the point (1, 2, 4) and is perpendicular to each of the planes 2x — 3y — s + 2 = and a; — ?/ + 2s — 4 = 0. 4. Find the equation of the plane that is perpendicular to the segment joining (3, 4, — 1) to (— 3, 6, 1) at its middle point. 5. Find the equation of the plane which passes through the point (3, — 3, 0) and is parallel to the plane 3a; — jz + z — 6 = 0. 166. Pencil of planes with a common axis. The system of planes passing through the line of intersection of two given planes A^x + B{ij + (7iZ + Di = and A.x + B.jj + CaZ + i>2 = 220 THE PLANE [Chap. XIII. is called, a pencil of planes with a common axis, or a coaxial pencil. The pencil is represented by the equation AiX + J5i2/ +C,z-{-D, + k (Aox + Boy + C^z + D^) = 0, (1) where k is an arbitrary constant. Foi*, it is clear that every point whose coordinates satisfy both the given equations will be a point lying on the locus of (1); and, since (1) is of first degree in the variables, it is the equation of a plane. There- fore (1) is the equation of a plane passing through the line of intersection of the given planes, whatever value is given to Jc. 167. Pencil of planes with a common vertex. The system of planes passing through the point Pi = (x^, y^, z^ is called a pencil of planes with a common vertex, the point Pj being the vertex. It is represented by the equation A(x-x,)+B{y-y,)+C{z-z,)=0, where A, B, and G are arbitrary constants. For this equation is the equation of a plane, whatever be the values of A, B, and C, and it is clearly satisfied by the coordinates of Pi. Therefore it represents a plane passing through P^. EXERCISES 1. A plane passes through the point (3, 2, — 1) and is parallel to the plane 7x — y + z — li^O. Find its equation. 2. Determine k so that the plane x + ky — 2 z — 9 = shall pass through the point (1, 4, — 3). So that it shall be parallel to the plane 3 x — 4 ?/ + 2 — 5 = 0. So that it shall be perpendicular to the plane 5x — 3y — z — 2 = 0. 3. Find the equation of the plane which passes through the intersection of the planes 2x — Sy — z — 6 = and x + y + z = 5, and (a) passes through the point (3, — 2, 1); (6) is perpendicular to the plane x — y — s + 2 = 0. 4. Find the equations of the planes which pass through the intersection of the planes x — y — 3 z — 4c = and x + y + 5z — 6 = 0, and are perpen- dicular respectively to each of the coordinate planes. 5. Find the equations of the planes which are parallel to the plane 6x— 5y — 3z — 2 = and which touch a sphere of radius 3 whose center is at the origin. 6. Find the equation of the plane which is parallel to the plane 5x — 3y — 7z — 8 = and such that the point (5, — 1, 2) lies midway between the two planes. Arts. 167, 168] EQUATIONS OF A STRAIGHT LINE 221 7. Find the equation of a plane through the point (2, — 3, 0) and having the same trace upon the XZ-plane as the plane x — 3 y -r 7 z — 2 = 0. 8. Find the equation of the plane parallel to the plane 2x+y + 2z-{-5 = 0, and forming a tetrahedron of unit volume vdth the coordinate planes. 9. Find the equation of the plane parallel to the plane 5x + Sy + z-7 -0 such that the sum of its intei'cepts is 23. 10. Find the equation of the plane having the trace x + 3 y — 2 = 0, and forming a tetrahedron of volume f with the coordinate planes. 168. The equations of a straight line in space. If Pi = (x^, y^, z{) and P2 = (x2, 3/2, Z2) are any two points in space, then the coor- dinates of the point dividing the segment P1P2 in the ratio ^=r, are (Art. 152), r + 1 ' y = y^, (1) r + 1 When r is allowed to vary, these equations give the coordinates of a variable point on the line Pj^P., and are, therefore, the para- metric equations of the line P1P2, r being the parameter. From equations (1), or from the figure, Art. 152, it follows easily that X — Xi _ y — Vi z — z^ ^2 •^'1 Vi Vi ^2 (2) These equations are called the two-point form of the equations of the line P1P2. Since the direction cosines of P1P2 are proportional to the pro- jections upon the coordinate axes (Art. 146), we have, from equations (2), X - a^i ^ y-yi ^ z-Zi ^ .gx cos a cos /8 cos y These equations are called the symmetric form of the equations of the line P^Pi- 222 THE PLANE [Chap. XIII. As an example, tlie equations of the straight line through the points (3, — 2, 1) and (4, 5, — 6) are, in the parametric form, 3+4 r -2+5 r 1-6 r X = — , y = , z = r+1 ' ^ r+1 ' r+1 In the two-point form, they are 1 7 - 7 ' The direction cosines are cos a = — ^, cos jS = — ^, cos y = ~ ' V99 V99 V99 EXERCISES 1. Find the equations of- the lines joining the following pairs of points : (a) (0, 0, - 2) to (3, - 1, 0). (6) (- 1, 3, 2) to (2, - 2, 4). (c) (2, - 3, 1) to (2, - 3, - 1). 2. In the preceding exercise, iind the coordinates of the points where each line meets the coordinate planes. 3. rind the direction cosines of each of the lines in exercise 1. 4. Find the equations of the line through the point (— 1, 2, — 3) if (o) « = 60°, ^ = 60°, 7 = 45^ (6) a = 120°, /3 = 60°, 7 = 135°. (c) cos a = ^ , cos /3 = i cos 7 = 0. Show that the given values are possible in each case and plot the line, 5. Find the equations of the line through the origin and equally inclined to the axes. 169. The projecting planes of a line. The planes drawn through a given line and perpendicular to each of the coordinate planes in turn, are called the projecting planes of the line. The equation of a projecting plane can contain only two of the three variables x, y, z (Art. 157). Hence the equations of the projecting planes can be found from equation (2) or (3) of the preceding article, by neglecting one of the ratios involved. Thus, for example, the equation of the projecting plane perpendicular to the XF-plane is x — x^_y — y^ X^ - .Ti 2/2 - Vl Arts. 169, 170] INTERSECTION OF TWO PLANES 223 For this equation represents a plane parallel to the Z-axis, and it is satisfied by the coordinates of the points Pj and P^- Similarly, the equations of the other projecting planes can be found. The equation of any plane through the criven line is, = Jc Fig. 131 For, this equation is linear in x, y, z and is therefore the equation of a plane ; moreover it is satisfied by the coordinates of P^ and Pa irrespective of the value of the parameter k. 170. The intersection of two planes. If a line is given as the intersection of two planes, its equations are the equations of the two planes considered as simultaneous equations. Thus, the equa- tions A,x + B,ii + C,z + Z>i = and A^x + B.y + C^z + A = (1) are the equations of the line of intersection of the two planes whose equations are those just written. The direction cosines of the perpendiculars to these planes are respectively proportional to A^, Pi, Ci and A^, B^, C^ ; and, since the line of intersection is at right angles to both these perpendiculars, its direction cosines must satisfy the two equations A^ cos a + Pi cos ^ + Ci cos y = 0, An cos « + P2 cos ;8 + C2 cos y = 0. (Art. 151) Hence, we have cos « : cos ;8 : cos y = (P1C2 - P2C1) : (A.,C, - A^CV) : (^liP, - A.B,). Therefore the equations of the line of intersection are X — x^ _ y — 1/1 _ z —z-^ Pi C2 - PoCi A.C, - A, a A,B, - A.B, ' where (a-j, y^, z^ is any point on the line. 224 THE PLANE [Chap. XIII. To find a point on the line, put one of the variables equal to zero in equations (1) and solve the resulting equations for the other two. For example, consider the two planes 2x + iy + 2z -S =0, and 5 X + 6 ?/ + ,j - 16 = 0. The direction cosines of their line of intersection must satisfy the two equa- 2 cos « + 4 cos /3 + 2 cos 7 = 0, and 5 cos a + 6 cos /3 + cos 7 = 0. Hence, cos a : cos ;8 : cos 7 = — 8 : 8 : — 8 : = — 1 : 1 : - 1. To find the coordinates of a point on the line, put z =0 in the equations of the planes and solve the resulting equations for x and y. In this way we find that the point (2, 1, 0) lies on the line. Therefore the equations of the line are • 01 X — 2 _ y — I __ z ~- 1 ~ 1 ~ - 1' EXERCISES 1. What are the equations of the projecting planes of the line x — 2 _ y — 3 _gt) 6 ~ 3 ~ 2 ■ What is the equation of the plane passing through this line and through the origin ? Through this line and through the point (1, — 2, 5) ? 2. What are the equations of the line through the point (2, 5, 7) if cos a = |, cos/3= — , and cos 7=0 ? If cos « = ^, cos/3 = 0, and cos 7= — '- ? If cos « = 0, cos ^ — i, and cos 7 = 2 — - ? o 3. Find the equations of the projecting planes of the line 2x-Sy + z-6 = 0, x + y-3z-l = by eliminating x, y, and z in turn from these equations. 4. Find the direction cosines of the line 2x + 3y-^2z-lS = 0, 3x + 6?/-3.?-24 = 0. 5. Find the coordinates of the points in which the line 2x + 2y-Sz-2 = 0, 4x-?/-.s-6 = meets the coordinate planes. Arts. 170, 171] INTERSECTION OF LINE WITH PLANE 225 6. Reduce the equations x + 2y + 6z — ^ = 0, 3a;— 2y— 10 s — 7=0 to the symmetric form. Eliminating in turn x and y from the given equations, we find the equa- tions of two of the projecting planes to be 2y + 7 z — 2 = and X - s — S = 0. From the first, s — — — ^^^^JZ — 2 j and from the second, s = x — S. Hence we have, X — .^ _ y — 1 _ s 2 ~ - 7 ~2' from whicli the symmetric form follows at once. 7. In the same way, reduce the equations 2x + 2y-Ss-2 = 0, 4:X - y - s - 6 = 0, to the symmetric form. 8. Reduce the equations x = mz + a and y = nz + h to the symmetric form. 9. Find the direction cosines of the following lines : (a) 4x-52/ + 32; = 3, 4x-52/ + + 9 = O. ip) 2 .X + s: + 5 = 0, x + 3 2; - 5 = 0. (c) 3a;-?/-25; = 0, 6x-32/-4s + 9 = 0. 10. "What are the equations of the line through the point (2, 0, — 2) and perpendicular to the lines ^-^ ^y^ ^ + 1 and ^ = ^ + ^ = ^" + 2 9 2 12 3-1 2 ■ 11. What are the equations of a line through the point (2, 3, 4) if cos a = cos /3 = ? 171. Intersection of a line with a plane. The coordinates of the point of intersection of a line with a plane must satisfy the equations of the line and also the equation of the plane. Hence, to find these coordinates, solve the three equations simultaneously for X, y, and z. When the equations of the line are not given, but the coordi- nates of two points on the line are known, a more expeditious method is to write the equations of the line in parametric form (Ai't. 168, (1)), substitute these values of x, y, and z in the equa- tion of the plane, and solve for r, thus determining the ratio in which the required point divides the segment joining the given points. For example, to find the coordinates of the point in 226 THE PLANE [Chap. XIII. which the line joining the points (1, —2, 0) and (3, —4, 5) meets the plane x — y-}-4:Z-\-2 — 0, write the parametric equations of the line ; viz. : ^ .. 1+or x = y = and z = r + 1' - 2 - 4 r r + 1 ' 5r and substitute these values of x, y, and z in the equation of the plane. In this way we find r = — ^. Hence the point of inter- section is (i|, - If, - If). The line whose direction angles are a, ^, and y will be parallel to the plane Ax -{- By + Cz + D = if, and only if, A cos a -\- B cos /3 + C cos y = 0. For only then will the line be at right angles to every perpen- dicular to the plane. The line will be perpendicular to the plane if, and only if, ^ ^ B ^ C cos a cos ^ cos y For only then will the line be parallel to every perpendicular to the plane. EXERCISES 1. Find the coordinates of the point in which the line x + 'ky + 2 z = 0, y — S z — 7 =0 meets the plane 3x — 2y + s + 4:=0; the coordinates of the point m which the line = « = - — - meets the plane x -\- v 3 2-4 f -r y + ^ — 2 = ; the coordinates of the point in which the line joining the points (2, - 3, 1), (2, — 2, 4) meets the plane x - y — z — b=0. 2. Show that the line ^^ = l^^ = 5 is parallel to the plane 2 -7 3 ^ ix + 2y + 2z =9. 3. Show that the line ^ = 1 = ^ ig perpendicular to the plane o ^ 7 Sx + 2y-\-7z = 8. Art. 171] INTERSECTION OF LINE WITH PLANE 22? 4. Show that the two straight lines x ~ 2 =2 y — 6 = 3 z and 4 x — 11 = 4 ?/ — 13 = 3 2 meet in a point. Find the coordinates of this point and show that the two lines lie in the plane 2 x — 6 y + 3 z + 14 = 0. 5. Find the equations of the line passing through (1, — 6, 2) and per- pendicular to the plane 2x — y + 6z=0. 6. Find the equations of the line passing through the point (—2, 3, 2) which is parallel to each of the planes 3 x — y + z — and x — z = 0. 7. Show that the six planes, each containing an edge of a tetrahedron and bisecting the opposite edge, meet in a point. 8. Prove that the six planes, each passing through the middle point of one edge of a tetrahedron and perpendicular to the opposite edge, meet in a point. 9. What is the equation of a plane passing through the point (1, 3, — 2) and perpendicular to the line X — 3 _ y — i _ z c, 2 ~ 5 ~^ ■ 10. Find the equation of the plane determined by the parallel lines x + l ^ y-2 ^z ^j^^^ x-3^ y +'i ^ z-l 3 2 1 3 2 1 11. Find the equations of the line tangent to the sphere x- + y- + z- = 9 at the point (2,-1,-2) and parallel to the plane x -\- 3 y — 5 z — 1 = 0. 12. What are the equations of the line passing through the point (xi, j/i, ^i) and perpendicular to the plane Ax + By + Cz + D = ? 13. What is the equation of the plane passing through the point (xi, yi, Zi) and perpendicular to the line X — Xi _ y — J/2 _ Z — Z2 <> a h c CHAPTER XIV EQUATIONS AND THEIR LOCI 172. Second fundamental problem. The two fundamental probleins of solid analytic geometry were stated in Art. 154. The first of these has been illustrated in the preceding articles by finding the equations of certain well-known loci, such as the sphere, the right circular cone, the plane, the straight line. In this chapter we shall consider the second fundamental problem ; that is, given an equation in the three variables x, y, z, to find the form and properties of the locus. 1 73. Construction of a surface from its equation. The following rules serve as a guide in sketching, or constructing, a surface from its equation. (1) Symmetry. If the equation contains only even powers of one of the variables, the surface is symmetrical with respect to the coordinate plane from which that variable is measured. For example, if the equation contains only even powers of z, and the point (a, h, c) is on the surface, theii the point (a, h, — c) will also be on the surface. The XF-plane is then a plane of symmetry. If the equation contains only even powers of two of the vari- ables, the surface is symmetrical with respect to the coordinate axis along which the third variable is measured. For example, if the equation contains only even powers of y and z, the surface is symmetrical with respect to the XZ-plane and also with respect to the Xy-plane, and hence with respect to their intersection, or the X-axis. The X-axis is then a line of symmetry. If an equation contains only even powers of all three of the variables, the surface is symmetrical with respect to each of the coordinate planes and therefore with respect to their intersection, or the origin. The origin is then a point of symmetry. 228 Arts. 172-174] THE QUADRIC SURFACES 229 (2) IntercejM. The length of the segments from the origin to the points where a surface meets the coordinate axes are called its intercepts. These are found by putting two of the variables equal to zero and solving the resulting equation for the third variable. (3) Traces. The sections of a surface made by the coordinate planes are called the traces of the surface. The equations of the traces are found by putting each variable in turn equal to zero. (4) Plane sections parallel to the coordinate planes. The equa- tion of a surface and the equation z= Jc, a, constant, are together the equations of the curve of intersection of the surface with a plane parallel to the XF-plane. A series of sections parallel to the XF-plane can be found by allowing k to vary. Similarly, sections parallel to the other coordinate planes can be found. To construct a surface, it is customary to plot the traces upon the coordinate planes and a series of sections parallel to at least one of the coordinate planes. 174. The quadric surfaces, or conicoids. The locus of an equa- tion of the second degree in x, y, z is called a quadric surface, or conicoid. It can be shown that any equation of the second degree in X, y, z is reducible by a proper transformation of the coordi- nate axes to one or the other of the two forms Ax^ + Bij^ + Cz^ = D, (1) Aa>^ + Bij^ = 2cz. (2) If the coefficients in (1) are all different from zero, the surface is called a central quadric, the origin being the center. By the preceding article we see that the surface is symmetrical with respect to each of the coordinate planes, with respect to each of the coordinate axes, and with respect to the origin. If the coefficients in (2) are all different from zero, the surface is called a noncentral quadric. The surface is clearly symmetrical with respect to the XZ- and I"Z-planes and with respect to the Z-axis, but it is not symmetrical with respect to the Xl'^plane nor with respect to the X- and Y'-axes. If one or more of the coefficients in either (1) or (2) are zero, the surface is called a degenerate quadric. 230 EQUATIONS AND THEIR LOCI [Chap. XIV. 175. The ellipsoid. If all the coefficients in (1) of the preced- ing article are positive, the equation can be written in the form a" b" c- (1) and the surface is called an ellipsoid (Fig. 132). Fig. 132 The intercepts on the X-, Y-, Z-axes are respectively ± a, ± b, ± c. The numbers a, b, c are the lengths of the semiaxes. The traces on the coordinate planes are all ellipses, represented in the figure by ABCD, BEDF, and AECF. The equations of these traces are respectively t + t = i^ -^^^' = 1, and ^ + 5! = 1. Equation (1) can be written in the form 3,2 y1 + ■ hHi-- = 1, from which we see that any section of the ellipsoid parallel to the XY^plane is an ellipse whose semiaxes are V^-S' and b\\l Arts. 175, 176] HYPERBOLOID OF ONE SHEET 231 Hence, the section will be a real ellipse only vylien z is confined to the range — c<^ z^c, and reduces to a point if z is either — c or + c. Similarly, the sections parallel to the I'Z-plane will be real only when x is confined to the range — a a, the transverse axis of the hyperbola is parallel to the .Z-axis- If X = a, the section of the surface is the two straight lines c Similarly, the sections parallel to the XZ-plane form a system of concentric hyperbolas. The form of the surface is shown in Fig. 133. Art. 177] HYPERBOLOID OF TWO SHEETS 233 EXERCISES 1. Construct the following hyperboloids : (a) 4 a;2 + 9 1/2 - 16 z'^ = 144 ; (6) x^ + t - -^'^ = 25 ; (c) x'^ + 16 f- - z^= 64. 2. Show that ^_^ + 5!=iand-^ + ^ + ^ = l a? b- C' a'- 6'- c^ are the equations of hjrperboloids of one sheet ; the first surrounding the Y-axis, and the second, the A'-axis. 3. ShoWthat (,._iy2 (y-3)2 (z - 2^ ^ ^ 4 9 1 is the equation of an hyperboloid of one sheet whose center is the point (1,3,2). 4. Show that (x - ly (y - m)- _ (z - ny _ j a^ 62 (.2 is the equation of an hyperboloid of one sheet whose center is the point (I, m, n). 5. Show that 3 x2 + 4 ?/2 — ^2 _ g ^ _ is the equation of an hyperboloid of one sheet. Find the coordinates of its center. 6. Construct the surface whose equation is x2 - ?/2 + 2 2-2 - 6 X + 2 y + 4 s + 9 = 0. 7. What are the equations of the planes parallel to the coordinate planes which cut the surface 9 x'^ — ?/" + 9 z- = 36 in pairs of straight lines ? 177. The hyperboloid of two sheets. If two of the coefficients A, B, C in (1), Art. 174, are negative and one is positive, D being positive, the surface is called an hyperboloid of two sheets. Sup- pose B and C are negative, the equation can then be written in the form O i> O a2 &2 c2 ' from which we see that the surface does not meet either the y-axis or the Z-axis and consequently has no trace upon the YZ-^\sLne. The traces upon the other coordinate planes are hyperbolas. 234 EQUATIONS AND THEIR LOCI [Chap. XIV. The sections of the surface parallel to the l"Z-plaue form a system of concentric ellipses given by the equation b^ & a^ from which we see that the section will be a real ellipse only when x is con- fined to the range — a> cc> a. The hyperboloid of two = c. The form of the siir- FiG. 134 sheets is a surface of revolution if h face is shown in Fig. 134. EXERCISES 1. Construct the hyperboloids 4 x^ — 9 2/^ — 16 0'2 = 1 and x"^— 4 2/2 — 4 22 = 1. Which is a surface of revolution ? 2. Show that _:^ + r'_?'^land-^'-^ + ?^=l a^ h'^ c? «2 y2 f^2 are equations of hyperboloids of two sheets ; the first surrounding the F-axis, and the second, the Z-axis. 3. Show that x'^ _ 2 ?/2 _ 4 j;^ - 2 a; — 8 ?/ - 8 = and y"^ - x'^ - 2 s"- + 6x — 2?/ — 40 + 6=0 are equations of hyperboloids of two sheets. Find the coordinates of the center of each. 178. The elliptic paraboloid. If the coefficients A and B \\\ (2), Art. 174, have the same sign, the surface is called an elliptic paraboloid. The equation can be written in the form where c may be either positive or negative. The trace on the XF-plane is a point, namely, the origin ; while the traces on the other coordinate planes are the parabolas />j2 -1/2 — = 2 C2; and •— = 2 cz. Arts. 178, 179] HYPERBOLIC PARABOLOID 235 The sections of the surface parallel to the Xy-plane form a system of concentric ellipses given by the equation -\-^ = ^cz, from Avhich we see that the section will be real only when c and z have the same sign. Hence the surface lies above or below the Xl'-plane according as c is positive or negative. The form of the surface is shown in Fig. 135, where c is supposed to be posi- tive. 179. The hyperbolic parabo- loid. If the coefficients ^.l and B in (2), Art. 174, are opposite in sign, the surface is called a hyperbolic paraboloid. Its equa- tion can be written in the form IT- Of,. 62~ The trace on the Xl'-plane is here a pair of straight lines >JC Fig. 135 intersecting at the origin, and the sections parallel to the XY- plaue form a system of concentric hyperbolas which recede from the trace on the XF-plane as z increases or diminishes. The transverse axis of one of these hyperbolas is parallel to the X-axis if c and z have the same sign, and parallel to the F-axis if c and z have opposite signs. The surface is saddle-shaped. A mountain pass between two solitary peaks resembles roughly a hyperbolic paraboloid (Fig. 136). EXERCISES 1. Construct the following surfaces (a) X- + y'^ — 8 z. (c) a;2 — 4 z^ = 16 y. (ft) 2/2 + -2 = 4 a;. (d) y- ~ x^ = 10 z. 236 EQUATIONS AND THEIR LOCI [Chap. XIV. 2. Reduce each of the equations x^ +2y^ — (Jx + 4:y + Sz + 11 = and z^ — Sy'^ — 4iX + 2z — 6y + l = 0toa, standard form and determine the type of paraboloid of which each is the equation. 3. A point moves so that it is equidistant from two nonintersecting straight lines. Show that its locus is a hyperbolic paraboloid. Fig. 136 4. Discuss the equations z = xy and z = of these equations ? x^ +xy + y'^. What are the loci 180. The quadric cone. If tlie constant term D in (1), Art. 174, is zero and the coefficients A, B, C are not all of the same sign, the locus of the equation is a quadric cone. Suppose that C is negative, and A and B are positive ; the equation can then be written in the form 00" , 2/. a^ b'^ c^ (1) from which we see that the sections parallel to the X Emplane form a system of concentric ellipses which increase in size indefinitely from a point (when 2 = 0) as 2 increases or decreases indefinitely. Again, if P = (x^, y^, zj is any point whose coordinates satisfy (1), we can prove that the line joining the origin to P lies entirely upon the surface. For the coordinates of any point on this line are clearly x = rx^, y = ry^, z. rz. Arts. 180-182 PAIRS OF PLANES 237 where r is any number. But these coordinates satisfy (1) by virtue of the hypothesis that the coordinates of P satisfy (1). Therefore tlie cone may be generated by a line whicli rotates around the origin and intersects an ellipse whose axes are parallel to the X- and y-axes, Fig. 137. 181. Cylinders. If either (1) or (2), Art. 174, contains but two of the variables, the corresponding locus is a cylinder (Art. 157). The cylinders are therefore de- generate quadrics. 182. Pairs of planes. If an equation of the second degree is written with its right member equal to zero, and its left member is then the product of two expressions of the first degree in the variables, the corresponding locus is a pair of planes. For the equation is satisfied by the coordinates of any point which render either fac- tor equal to zero. Thus, Fig. 137 is the equation of the pair of planes ^ + ?^=Oand ?-?/=0. a b a b EXERCISES 1. Construct the cones whose equations are (rt) 9 x2 - 36 if -I- 4 ^2 = 0, and (5) 16 x^ - 4tf - z^ = 0. 2. If in (1), Art. 174, Z> = and A, B, and C are all of the same sign, what is the locus of the equation ? 3. Show that x'^ + iy'^ — z'^ — 2 x + 8 y + 5 = is the equation of a cone whose vertex is the point (1, — 1, 0). 238 EQUATIONS AND THEIR LOCI [Chap. XIV. 4. In general, a^ b c^ is the equation of a cone whose vertex is the point (?, m, n). 5. Construct the cone whose equation is x2 + 2/2 — 2 ^ + 2 ?/ + 4 2 - 1 = 0. 6. Discuss the equations (a) 4 2/2 - 25 = ; (6) 2 2/^ + 5 s^ = ; (c) 2/^ - a;2 - 4 2/ + 6 x - 5 = 0. What is tlie locus of each equation ? 7. Show that the left member of 3-2 _ 2/--2 + 5;2 4- 2 xs — 5 a; — 2/ — 5 2 + 6 = is divisible hj v -\- y + z — 2. What is the locus of the given equation ? 183. Ruled surfaces. If a straight line moves according to a given law, it describes, or generates, a ruled surface. Thus, if a line moves so as to be constantly parallel to one of the coordinate axes, it describes a cylinder parallel to that axis. Again, if a line rotates about a fixed point and intersects a fixed curve, it generates a cone whose vertex is the fixed point. Cones and cylinders are ruled surfaces. If the equations of a straight line contain a parameter k, then when Ti is allowed to vary, the line will move and thus describe a ruled surface. For example, let the equations of the line be hx + y -\- z — k = and x — hy -\- kz -\- 1 — {). From the first of these equations we obtain k^'-l±^. (1) 1-x ^ ^ and from the second, ?. _ 1 + ^ (o\ ~y -z Therefore the coordinates of all the points that lie on the line must satisfy the equation y_±z^l^x^ 1—xy—z or a;2 ^ y^ - z'- = 1, (4) Arts. 183, 184] EQUATION OF GENERATOR 239 "whatever the value of Jc. Conversely, any point whose coordi- nates satisfy (4) must lie on the given line for some value of k. For (4) is equivalent to (3) and the value of k is determined from either (1) or (2). The locus of (4) is an hyperboloid of one sheet. Therefore the given line generates this surface when k varies. EXERCISES 1. Find the equation of the ruled surface generated by the line whose equa- tions are x + y = kz, X - ij = j- fC 2. Find the equation of the ruled surface generated by the line 12/; 2 1 «i k (a) when — = 2. (/>) ^Yhen k + m = 1, (c) when km = 3, and (d) when the m perpendiculars from the origin upon the two planes are in' the ratio 1 : 2. 184. Equation of generator. It frequently happens that the equation of a surface indicates at once that it is a ruled sur- face. For example, the equation (x -\-yy -\-(x-\- )j)z — 3 = can be written , , , ^ r. (x + ij)(x + y + z) = 3. o Hence the straight line x-\-y = -, x-{-i/-\-z = k lies wholly on ft the surface, whatever value is given to k. This line is a generator for any value of k. EXERCISES 1. Show that the hyperboloid of one sheet is a ruled surface. Suggestion. The equation of the surface can be written Hence the system of straight line.s (•^ + ''\ = k(l + -Y ■{l—±\ = l(i-'^ lies wholly on the surface. 2. Show that a second system of straight lines lies wholly on the hyper- boloid of one sheet. 3. Show that the hyperbolic paraboloid is a ruled surface, having two systems of straight lines lying upon it. 240 EQUATIONS AND THEIR LOCI [Chap. XIV. 4. Show that the three other nondegenerate conicoids are not ruled surfaces. 5. Show, by the method of this section, that the cone — + ^ — — = Ois n"^ tfi r'^ a ruled surface. a o c 6. Prove that y'^ — ^ys + 4:z'^ + xy — ^xz^b is a ruled surface. Are there two systems of lines lying upon it ? What is the form of the surface ? How do the generators lie with respect to each other ? 185. Tangent lines and planes. When two of tlie points in which a straight line meets a surface coincide at a point P, the line is called a tangent line to the surface and P is called its point of contact. In general, all the tangent lines at P lie in a plane called the tangent plane. P is the point of contact' of the plane. To find the equation of the tangent plane at a given point on a surface, consider the ellipsoid — \-~-\ — = 1. a^ &^ c^ Let Pi = (xi, III, Z]) and P2 ^ (-''^'2) Ihii ^2) be any two points in space. The equations of the line P1P2 are, in parametric form (Art. 168), x^+rx, y^+ry, z^±rz, .^. r^l '-^ r + 1 ' r + 1 '^ ^ To find the coordinates of the points in which this line pierces the ellipsoid, substitute the values of x, y, and z in the equation of the surface and solve for r. These values of r, when placed in equa- tions (1), give the coordinates sought. The equation for r is readily found to be \^a^ b^ c^ J \ ci^ V' & Now suppose the point Pi = {x^, y^, z^ lies on the ellipsoid. The absolute term in (2) is then zero, and one of the roots is i\ = 0, as it should be. The other root is "V a' b^ c" a^ ¥ c- Art. 185] TANGENT LINES 241 In order that this root should be zero also, and therefore the two points of intersection coincide at Pj, it is necessary and suffi- cient that the numerator should vanish. Hence, when the point P2 lies on the plane a^-_^M-|_?i?_ 1=0, (3) a^ If c^ the line P1P2 will touch the surface at Pi ; and therefore (3) is the equation of the tangent plane at P^. When Pi does not lie on the surface, equation (3) represents a plane called the polar plane of Pj with respect to the ellipsoid. A line perpendicular to the tangent plane at the point of con- tact is called the normal to the surface at this point. EXERCISES 1. Show that the point ('1, 2, 2^^ ^ Ues on the ellipsoid ^ + l^ + €i =1. Find the equation of the tangent plane at this point, and the equations of the normal. 2. Derive the equation of a tangent plane to the hyperboloid of one sheet. 3. Derive the equation of a tangent plane to the hyperbolic paraboloid. 4. Derive the equation of a tangent plane to the quadric cone and show that any tangent plane passes through the vertex. 5. Show, by means of equation (2), Art. 185, that when P2 lies on the polar plane of Pi, the segment P1P2 is divided externally and internally in the same ratio by the points where the line P1P2 meets the surface. 6. Show that the length of a tangent line to a sphere from the point (xi, 2/1, Zi) is equal to the square root of the result of substituting xi, ?/i, zi for X, y, z in the left member of the equation of the sphere, the right member being zero. 7. Show that the locus of points from which tangents of equal length may be drawn to two spheres is a plane. This plane is called the radical plane of the two spheres. 8. Prove that the radical planes of three spheres meet in a line called the radical axis of the three spheres. 9. Prove that the radical axis of three spheres is perpendicular to the plane of their centers. 10. Show by definition that the tangent plane to a ruled quadiic contains the two generators which pass through the point of contact. 242 EQUATIONS AND THEIR LOCI [Chap. XIV. 186. Circular sections. Certain planes cut the conicoids in circles. For example, consider the ellipsoid • t + t + t^i = o. 4 9 1 The coordinates of all points on the curve of intersection of this ellipsoid Avith the sphere x^ + ?/^ + z^ — 4 = will satisfy the equation whatever value is given to k. When k = \, the equation becomes §^ _ ^' = (1) 4 3(3 ^ ^ and therefore represents two planes. Each plane cuts the sphere, and therefore the ellipsoid, in a circle. Any plane parallel to either of the planes (1) cuts the ellipsoid in a circle. For, let 2^ + ^^_;b = 0and2^-V— -m = (2) be the equations of any two planes parallel to the two planes (1). Combining the product of the equations (2) with the equation of the ellipsoid, it is easily seen that all points common to the planes (2) and the ellipsoid must satisfy the equation — \- — -\ hkl z ?/ 4- m z h y — km — 1=0. 444V 2 "^6/ V 2-^67 But this is the equation of a sphere and hence the planes (2) meet the ellipsoid in circles. There are thus two systems of parallel circular sections, each being parallel to one of the planes EXERCISES 1. Find the equations of the planes which cut circles from the ellipsoid 9:fc'^ + 25?/2 + 169 02 ^ i. 2. For what values of k and m will the equations (2) be the equations of tangent planes to the ellipsoid ? Arts. 186, 187] ASYMPTOTIC CONES 243 3. Find the equations of the system of planes which cut the hyperboloid of one sheet — + ^ —■ 9 25 169 1 in circles. 187. Asymptotic cones. Consider the hyperboloid of one sheet y'^ _ ^^ _ -1 CI? ¥ Let Pj = (a*!, y^, z^ be any point in space. The equations of the line joining the origin to Pi are x = rx^, y = ry„ z = rz^, ?' being a parameter; and the coordinates of the points in which this line meets the hyperboloid are found by substituting these values of x, y, and z in the equation of the surface and solving the resulting equation for r. We thus obtain for r the ^ equation. 1 a- +f It follows, from this equation, that as Pi is made to approach the cone ; + y -~=o, (1) the values of r increase in absolute value, becoming in- finite Avhen Pj lies on the cone. Therefore no gener- ator of the cone (1) ever meets the hyperboloid. Moreover, the sections of the cone and the hyper- boloid, parallel to the XT- plane, approach coincidence as the cutting plane recedes from the Xl^plane. The cone (1) is called the asmyptotic cone. Fig. 138 244 EQUATIONS AND THEIR LOCI [Chap. XIV. EXERCISES 1. Find the e(juation of the asymptotic cone of tlie hyperboloid of two sheets. 2. Show that tlie asymptotic cone of the hyperbohc paraboloid consists of two planes. 3. Show that a plane determined by any generator of the hyperboloid of one sheet and the center is tangent to the as}^mptotic cone. 4. Show that neither the ellipsoid nor the elliptic paraboloid has an asymptotic cone. 188. Projecting cylinders of a curve in space. The cylinders whose generators intersect a given space-curve and are perpen- dicular to one of the coordinate planes are called the projecting cylinders of the curve. Arts. 188, 189] PARAMETRIC EQUATIONS 245 To find the equations of the projecting cylinders, eliminate x, y, and z in turn from the equations of the curve. For example, consider the curve whose equations are a-2 -f- 7/2 ^ 8 2, a;2-?/2 = 4 2. Eliminating x, y, and z in turn from these equations, the three projecting cylinders are found to be y^ = 2z, x^ = 6 z, and x"^ — oy- = 0. The first two are parabolic cylinders, shown in the figure. They intersect in the given curve. The third equation decomposes into the two planes x + Vo ^ = and x — Vo y = and shows that the given curve consists of two parabolas lying in these planes. EXERCISES 1. Construct the following curves : (a) x'^ + 2/2 = 2.5, y + ^ = 0. (b) X- + y- — ix = 0, X + y + z =: 3. (c) x^ — if = i, X + y + z - 0. 2. Find the equations of the projecting cylinders of the following curves : (a) x^ + y^-2y = 0, y'^ + z^ = 4. (b) 2y-^ + s^ + ix-iz = 0, y-^ + 3z^ -8x = 12z. (c) x'^ + tf + Z' = 2.5, x2 +4y2 — z^:= 0. The last is a spherical conic. 3. A point moves so as to be constantly 2 units from the Z-axis and 2 units from the point (2, 0, 0). Find the equations of its locus and plot the curve. 189. Parametric equations of curves in space. If the coordi- nates of a point in space are each functions of a parameter, the locus of the point is a line in space, straight or curved. For example, the equations in Art. 168 are the parametric equations of a straight line in space. Again, the equations X = 9-^, y = r^, z = r are the parametric equations of a curve in space. The equations of the projecting cylinders of this curve are found by eliminating r from each pair of equations. Thus, the projecting cylinders are x^ = y^, X = z^, and y = z"^. 246 EQUATIONS AND THEIR LOCI [Chap. XIV. 190. The circular helix. An important curve in mechanics is the circular helix. Its parametric equations are X = a cos 6, y = a sin 6, z = hO, where 9 is the parameter. The equations of the projecting cylinders are x^ -\- y^ a cos -, y ■ z a sm - h Hence the curve lies on the right circular cylinder x^ + y^ = q2_ If & is a positive number, the XZ- cylinder stands on the curve ic = acos- or ABCDE ; and the b yZ-cylinder stands on the curve y = a sm The helix is there- FiG. 140 fore a curve wound around the circular cylinder, the distance be- tween two consecutive turns being 2 67r (Fig. 140). EXERCISES 1. Plot the curves : (a) x = 2r, y = r^, z = — — ■ {b) x ^ecosi?, y=6sm9, g = -^ + ^°^^^ . 4 2. Construct a circular helix when 6 is a negative number. 3. Show that the equations X — a cos d, y = bsind, z = md are the equations of a helix wound on an elliptic cylinder. 4. Construct the curve X = a sec 0, y = b tan 9, z = md. 5. The three equations Jcx + 2y-\- kz-2k = 0, x + y + kz — Jc = 0, kx — y — zk—1 — 0, Art. 190] THE CIRCULAR HELIX 247 are the equations of three coaxial pencils of planes (Art. 152). Express the coordinates of the point of intersection of the three planes, for any value of k, in terms of k and thiis show that the three pencils generate a space-curve. Construct the curve. 6. Shov? that the point (2, 1, 5) lies on the curve X' + if + z'^ = SO, 2/2=^, and find : (a) the equations of the tangent line to this curve at the given point, (&) the equation of the normal plane (perpendicular to the tangent line) at the given point. ANSWERS Art. 3. Page 10. 1. BA, AB, OB; 8, 8, 3 ; No. 2. 73'' on the Fahrenheit scale. 4. 206°, 170^ 5. 310^. 6. 115°. Art. 7. Page 13. 4. P1P2 = o, P2P3 = 3, P1P3 = ^■ 5. P1P2 = V37, P2P3 = 3, P1P3 = i. 6. A sqiiare, one side 4 units, diagonal 4V2, area 16. 7. (2, 0) ; each V2 ; each 2 ; 1. Art. 9. Pages 15-16. 1. (2.598, - 1.5) ; (2, - 3.46-4) ; (- 1.25, 2.73). 2. (\/58, -66°.8); (2V5, 26°.6) ; (- VM, 59°.03). 4. ?/ = ± 3, e = ± 36°.8. 5. (-171,0). 9. ^=(-8.66, 5) or (10, 60^). P= (0, 15) or (15, 0°). C'=(10.39,6) or (12, -60°). Art. 11. Pages 19-20. 1. (a) 5 and 2; slope, |. (b) 1 and - 9 ; slope, — 9. (c) 2 and 8 ; slope, 4. (d) 1 and — 5 ; slope, — 5. 2. 19° 6' ; 40° 54'. 3. (fo) 26° 34'; 75° 58' ; 146° 19'. Art. 12. Page 21. 1. 13.86. 2. VT3. 3. 5.97. 4. 12.73; 2.23; 14.92. 5. (80, -), 144 miles. 4 Art. 15. Page 24. 1. 38°27'. 2. 11.402, 7.616, 7.211 ; 100° 30', 41° 3', 38° 27 . 4. 1.792. 5. 10. 6. 35° 24'. ' 249 250 ANSWERS Art. 17. Pages 25-26. 1. (0, 3.5) ; (1, - 1.5) ; (3, 0). 5. (0, 2) and (3, 3). 6. (1.125, .25). 7. (1, 0) and (0, -2). Art. 19. Page 28. 1. _6. 3. -7.5. 4. 6.897. 6. 2.5. Art. 20. Page 30. 1. 88.5. 3. 22.935. 4. 151. Art. 21. Pages 31-32. 1. 122.5. 2. 25 acres and 1 a sq. rd. 3. 60,294 sq. ft. fx = -94.2 -27.8 188.0 54.1 ■ ^"^ \y= 66.0 157.3 68.3 -166.3 (6) N. 36° E. ; S. 67=^ 36' E. ; S. 29° 46' W. ; N. 26° 48' W. ; N. 1° 12' E. ; N. 50° 45' W. (c) 42,277.06 sq. ft. 5. 150.5. 6. 300, slope, f. Art. 30. Page 39. 1. Line of symmetry, x = 1 ; intercepts on X-axis, — 1 and 3 ; intercept on I^'-axis, —3; turning point (1, —4). 2. Line of symmetry, y — I; intercept on X-axis, | ; intercept on F-axis, 1. 3. (±2, ±2). 6. (_b_ iac-b-^ \ \ 2a ia } 7. Area = 4 x v'25 — x^, where x is one half the length of one side. Turn- ing point for x = — ^. Max. rect. is a square whose side is 5\/2. ^ 432 8. Number of sq. ft. of lumber is — — \- x^, where x is the length of the side x of the base. Height of the box requiring the least amount of lumber is 3 feet. Art. 34. Page 42. 3. (a) y = a. (b) (x"- + y^ + 2 ax)'^ = 4 cfi{x'^ + y'^)- (c) {x^ + y-^ — ax)~ = a2(x2 + y^). 4. ?• = 2 cos d. Art. 41. Page 50. 9. p» = 4. 10. I =— ^ + 1. 50000 11. ,^(93000000)^^ ^• = 1.93, nearly. ANSWERS 251 Art. 45. Page 56. 1. (a) x- + y^-2y-S = 0. (b) x^ + 2/2 + 4 x =: 0. (c) x^ + y^ + 8 X - 6 y + 16 - 0. (d) x'^ + ij- - 2 x - i y - SI = 0. 2. x- + ?/2 - 4 X — 6 2/ = 0. 3. x2 + y- + 4 x -6y -IS = 0. 4. x'-2 + 2/2 - 5 X + i 2/ - I = 0. 5. 10 X - 8 ?/ + 3 = 0. 6. 2(a — c) X + 2(6 - d)y — 6- + d"^ - a- + c^ = 0. 7. x'^ + 2/2 = 16. 8. x^ + 2/2 = 25. 9. x2 + 2/2 _ 6 X + 4 2/ - 12 = 0. Art. 46. Page 57. 1. (a) (3, 0) ; 5. (b) (3, - 2)^ 3 V2". (c) (f , 4) ; J/, (d) (- 1, 2) ; 0. (e) (4,0); 4. (/) (f, J..) ; ii V5. (gr) imag. (h) imag. 2. (19, -Sg^-) ; 19.665, nearly. Art. 48. Page 59. 1. (a)x-y+l = 0. (b) Sx-\-2y + l=0. (c) x + 9 y + IS = 0. (d) 7 X + 4 2/ — 5 = 0. 5. (a)-l;|,|. (6)|;3,_l. (c) 3 ; - |, 2. (d) _ | ; 4, |. Art. 50. Pages 62-63. 1 . X- , 2/2 1 c, 2. ± 4, 0. ^ 25 9 ^ ^ 36 9 ^ '^ 25 25 ^ '^ 100 50 (e) ^%l::^i. ^ ^ 50 25 '161 ■ 1 6 ■ 4. 5. («) a = V2, & = V3, c = iV3. (6) a = \/3, /) = v'2, c = ^\/3. (c) « = V2, 6 = A\/6, c = V|. (r/) (1 = 1, 6 = i\/2, c = ^v^. 6. (a) 2Vi, (6) i|^, (c) fV2, (d) 1. Art. 52. Pages 65-66. J x2_y2^j 2. —--^ = 1. 3. 7.03+ and 1.03+. 9 7 36 28 4. («) 3, 2; 1 Vis. (6) 2, 3; iVlS. (c) 1, 4; VT?. (c7) 2\/to, Vm; iV5. 5. (a) \/6, 2; iViO. (&) 4, 2; V5. (c) Ap, Vn; a/^-^*- ^m ^ TO 6. (4a) 2f. (4&) 9. (4c) 32. (id) Vm. (5a) 6. (56)16. (5c)-—"- 252 ANSWERS Art. 53. Page 67. 1- (1,0), 2. 2. y2 = 8(x-l). 3. 0-2= 4(2/ -1). 4. (a) y^ = 8x. (b) .r2 = 8 2/. 6. (o) 8. (5) 4. (c) 6. (d) 10. Art. 57. Pages 70-71. 1. r^ -6 ?• cos (^- 60^) = 7; VaT. 2. ?• cos (e - (iO°) = 5 ; a; + ?/ VS = 10. 3. >• = 8 sin e ; a;2 j^y2 -gy_ 6. r = ± 2 a cos(6' — 45°) ; x- + ?/-±rt.<:V2 ± a?/ V^ = 0. Art. 59. Pages 72-73. 1. r = "^ . 2. /■ = ^ . 5. r = 4 — 3 cos ^ 3 cos 6 — 2 1 — cos c~ — a'^ , _ a- — c^ 9. r = '^ ~ " — . 10. rr= " -^ . 12. r = _ " . « — c cos ^ rt + cos d ^'2^ COS e — 1 Art. 62. Page 76. 1 /„^ „ 7 .) /T v'S ... Vl05 , \/165 1. (a) a = 6. h = 2, c = V5, e = . in) a = , b = , _ '3 ^ ' 7 11 c = Vf 0, e = V^j. (c) ffl = 10, & = 5, c = 5 V5, e = i V5. (d) a = 2 Vff; 6. = ^^, c = f Vl^, e ^ Vff. (e) rt = 8, & = 5, c = VSQ, e = ^^. ^ _ ,/Qq ^ (/)rt = 8, 6 = 5, c=V89, e=-^-5^. 2. 2 2/ +x = and 2?/-x =0. 3. 5i. 4. r = ^ , 6 =± arc tan 4. 5 cos 6* - 3 ' Art. 72. Page 87. 3. a = 2, T= ^^„-. 3: = 2 cos + Vc- — 4 siii-

= 67°30'; center (- 1, 0). 2(\/2-l) 2(v'2-{-l) (c) ^^ y^ =- 1 ; 61 = 67° 30' ; center (- 1, h). 2(V2-1) 2(\/2-Hl) (d) 3 x2 -h 2/2 = 6 ; 61 = 45° ; center (1, - 2). (e) ^^ ^ ■+ -J^— = 1 ; 61 = 13° 18' ; center (1, 1). .3 + V5 3 - v'5 (/) Imaginary lines. (gr) — ^—3 -f- — y—^ = ; 61 = arc tan 2 ; center (3, 3). 3 + V5 3 - \/5 ANSWERS 257 Art. 115. Page 154. 1. (a) ?/2 = - -|-x ; e = 45° ; vertex (^ —, - -^j nA ,fi - ^^x; e = 4r)° • vertex / ^^^^ 3V2 \ (ft) y _ - ^ i 4 ' 8 j 4V5 . , „_„,.„ l./r .„„/4jv/5 4V5 25 ' 25 (c) .'(•- = — - — - a- d = arc sin - VS ; vertex ( - ^ ' 25 5 V fd) a;2 = — 2^— ; 61 = -arc tan— ; vertex ( — ? — , - ^ - 13VI3 2 5 V 13 Via 507 6?/ „ 1 ^ 12 ^ / 2 2VI8 •^ \ d = - arc tan — ; vertex ' !V13 2 5 4 ?/ .2 VS / (e) X- = ~ ; =arc sm : vertex Art. 116. Page 155. 1. (ffl) X — 2/ — 1 = 0. (&) Imaginary lines. (c) a; + ?/ ± 1 = 0. (d) 3 ..• - 2 2/ + 2 = and 3 a; - 2 y + 3 = 0. Art. 117. Pages 156-157. 1. («) Hyperbola. (')) Real ellipse. (c) Parabola. {d) Hyperbola. 2. (a) Parabola. (ft) Two real lines. (c) Real ellipse. {d) Two real lines. 4. k = \, imaginary. 7. Vertex (|, 4) ; focus (§, ^) ; directrix 4 ,r + 2 ?/ = 7. Art. 118. Page 159. 1. (rt) 3 X + 8 ?/ - 5 = and 3 .x - 8 y + 20 = 0. {h) x -\- y = 4. (c) ?/ + 4 = 0. (d) .'• + 2 y + 3 = 0. (e) 8 x - 37 ?/ + 18 = and 8 X + 13 2/ - 18 = 0. 2. 2/ — 4 X = 8. Art. 119. Page 162. I. (5 ±\/l3)x - 2(7 i 2V13);/ - (13 ± 5VF^) = 0. „ f Axes, X + y = Z and x ~ y = 9. I Asymptotes, (- 15 T 7\/5)x ± 2 2/ Vs + 15 ± 2 VS = 0. 3. 2/-x+l = 0, x + ?/-3 = 0, x + ?/-l = 0. Art. 121. Pages 163-164. 1. 17 X- + 105 xy — 48 2/^ + 210 x + 39 = 0. 2. (rt) 5 X + 5 2/ + 2 = 0. (6) 3 X - 2 2/ = 0. (c) 6 x + 1 = 0. (d) ix-2ay + a^ — c^ + d- = 4. 4. (i -3). 258 ANSWERS Art. 122. Page 165. 1. x^ + 2 xy + y^ — X — li = and x^ — 2 xy + y'^ — x — 7 — 0. 2, x^ + ixy + iy^ — X — 2 = and x^ — i xy + i y"^ — x — 2 = 0. Art. 123. Page 166. 1. (a) x + ^ = ±2{y-i), Sx + l = ±iy-U), x + lb =±(7 y - 10). (b) 7 y + 3(3 = 0, 3V7 ± 12 = 0. (c) (2\/2±V5)x-(\/5tV2)2/=±(4v'5±2V2), x-2y=0, x+y=0. 2. (a) (-5, 0)^ (-4, 3), (3, 4), p^, - |). (&) (±^, -f)" (.) (±1:^0^ ±?^0^, (±2, T2). Art. 124. Page 167. 1. (a) 22 a;2 - 30 xy + 7 2/2 - 22 X + 9 2/ = 0. (&) 24 x2 - 73 xi/ + 29 y^ + 100 x -- 126 y + 40 = 0. Miscellaneous Exercises. Pages 167-168. 2. (2, - 3). 4. x2 + 2/2 - 18 X - 36 ?/ + 81 = and x2 + 2/2 - 2 X - 4 ?/ + 1 = 0. 5. (2, 0) and (5, 0). ^ 6. (W, -ft)- 7. (a)2/2=-|x. (&)g_^=_l. (c) 1+^=1. ,,. ., IOV13 ,. x2 2/2 , (d) 2/ = ^- W — H — 1- 13 _4(5+V5) 4(5 -v'5) (/)f-f=l. (.<7)2/2=^x. W./ = 2|lx. (0 (5 X - 2 2/ + 3) (5 X - 2 2/ - 2) = 0. ( j) x - 3 2/ - 1 = 0. (A;) x2 — 2/2 = -10V2. (?) x2 -2/2 =— 40- (?)i) Imag. lines, (n) 2/-^ = - AVt a;- (0) (X - 2 2/ - 2) (X + 2/ + 1) = 0. Art. 145. Page 200. 2. Each, ^. 3. (7,-4,-3). 4. (1, 0, 11), 7. Art. 147. Page 201. 2. Lengths of the sides, \/83, V2r7^\/54. 4. Terminal point, (i, |, — 3 ± f V23). 6. Direction cosines, f , |, ; f , - l - |. 8. 60° or 120°. Art. 151. Page 204. 1. (a) 90°. (6) arc cos — /p (c) arc cos if. 7. 4^. 8. IJ3. ANSWERS 259 Art. 152. Pages 205-206. 1. (a) (5f, -2f,3i). (6) (3, H,3|). 3. (-lf,2, 4|)and(-i, 0, 2^). 8. -2. J. 1. Art. 155. Pages 208-209. 1. a;2 + ?/2 + ^2 _ 10 a; + 4 1/ - 6 + 37 = 0. x^ + y^ + z^ — ix + 6y + 12 z = 0. x:^ + 7f + s"^ - 2 az = 0. 2. (a) (1, -3, 4), r = 2. (6) (-5, 2, -1), r = 5. (c) (-2, -2, -3), r = 4. (d) (- 3, 0, 0), r = 3. (e) Imaginary. 3. x^ + if + z^ - 2 X - 8 y - 16 z = 0. 4. x^ + ?/2 _|_ ^2 _ 4 ^ _ 217. 5. a;2 + ?/2 + 5;2 _ 4 and x^ + y- + z- = •57(5. Art. 156. Page 210. 1. ^ + ^ + £:=:: 1. 2. 2/2 + 5-2::= 4 px. a2 ft2 52 3. ^'-2^'-^=l and^'-^-^' = -l. a2 &2 ^-2 (jj2 52 ^2 5. 2/2^02_4p;^_ J/-* = 16jy2(x2 + 2;2). Art. 159. Page 212. 1. x2 + y2 = ;s2 tan2 e. 2. -+^-i^^-^ = 0. 4 4 9 4- (rl' 0, II) and (- H^ 0? Vs*)- 5. 6y = as. About the Z-axis. Art. 160. Pages 214-215. 1. (a) xV2 + y + z = 8. (b) x + yV2 - z + 12 = 0- (c) 6x-2y + Sz = 66. (d) 2x + y +■ 2 z + 16 = 0. 2. (a) Sx-2y + 6z = 49. (&) 2 x - 5 ?/ + = 30. (c) Sx + 4:y-2z+29. 3. ra) fx-f2/ + fa; = l. (^b) ^ i^ + ^ y - 1 z = i. ,^v 1 ^^ 2 ^ _ 3 .,. _1_ 2_ §_2;-0 \/21 V2I V2I V21 V14 ' Vli Vli 5. 54. Art. 162. Pages 215-216. 2. 16 X + 6 ?/ - 5; = 14. 4. Area of XZ-proj. = 4. Area of TZ-proj. = 6. Area of XF-proj. = 12. Art. 163. Page 217. 1. ox + 2y +6z = 12. 2. x-3y-2z = 0. 260 ANSWERS Art. 164. Page 218. 1. 4. -3. 2. 1.37, nearly. 3. ^■ + ^-±?:^=l. 4. 58i 3 4 12 ■* 5. 8. 11. 6. x + 6y--iz-l = 0. 7. x-2 + y- + ,s2 — (2/,? + a;.s + x?/) + X + ?/ + .? = -i. Art. 165. Page 219. 1. (4, -4, 2) ; 118°7', 6r53', G0°. 4. 3 x - ?/ - g + 5 = 0. 2. 3 a; + 4 y - 12 ^ - 12 = 0. 5. s x - y + z - 12 = 0. 3. 7 a; + 5 2/ - s - 13 = 0. Art. 167. Pages 220-221. 1. 7 X — ?/ + 2 — 18 = 0. 2. -1- ; impossible ; ^. 3. (a) 11 a; -4 2/ + 2 2; -43 = 0. (&) 8 x + 3 2/ + 5 z - 36 = 0. 4. 2/4-4^-1=0; x+g-5 = 0; 4x-2/- 19 = 0. 5. 6x-5?/-3;s± 6a/15 = 0. 8. 2 x + y + 2 z = 2-s/S. 6. 5 .r - 3 2/ - 7 ,s - 20 = 0. 9. 5 x + 3 2/ + s = 15. 7. x + 1 z = 2. 10. 3 X + 9 2/ + s = 0. Art. 168. Page 222. ^ ^ 3 -1 2 ^ ^ 3 -5 2 ^ ^ ' •" 2. («) (3, -1,0), (0,0, -2), (0,0, -2). (5) (- 4, 8, 0), (I, 0, V), (0, t, f). (c) (2, —3, 0), parallel, parallel. 3. («) 4=, -=^, ^- (6)^, -^1, ^- (c) 0,0,1. V14 VU Vli \/38 V38 V38 4. («) 2(x + 1) = 2(2/ - 2) =(^' 4 3) V'2. (6) -2(x + l) = 2(2/-2) = -(0+3)V2. (c) 2 2/ - x= 5, s + 3 =0. 5. x = y = z. Art. 170. Pages 224-225. 1. X - 2 2/ + 4 = 0, ./■ - 3 ,5 - 2 = 0, 2 2/ - 3 - = ; 3 x - 2 2/ - 6 2:= ; 25 X - 32 2/ - 27 4- 46 = 0. 2. xV5 - 2 2/ = 2\/5 - 10, s = 7 ; (x - 2) V3 = 2: - 7, 2/ = 5 ; 2(2/- 5)V2 = ,s-7, x = 2. 3. 5 27-72^4-4 = 0, 5x-8s-n = 0, 7x-8 2/-19 = 0. 4. ^, 0, -1-. \/2 V2 ■ 5- a -iO), (!, 0, f), (0, -J^,-V)- ANSWERS 261 1 1 a 3 3 3 X — a y — b z m n 1 Vwi- + n- + 1 y/m- + n^ + 1 Vm'-^ + n^ + 1 9. («) -|^, ^,0. (&) 0,1,0. (c)-=-|,o, ^. V29 \/29 Vis Via 10. 51=1^ = ^ = ^^. 11. a; = 2,y = 3. 4 2-5 ' ^ Art. 171. Pages 226-227. 1. (0, 1, -2); (-10, -7, Ifl); (2, -J53, 1). 4. (3, |, i). 5- ■^ = ^ = ^^- 10- 8.7; + 2/-26.s + 6 = 0. 14 1 ■ 11 8 7 ■ 9. 2x + 5w-s = 19. 12 ^•' - ^1 - y - yi = g - '^1 ^ ^ C 13. a{x -xi) + h(y - ?/i) + (•(£• - ,fi) = 0. Art. 183. Page 239. 1. x'^-y^-z- = o. 2. (a) 3x + 2y = 2. {b) 2x^ + 2y-^ + 6yz + 6xz + 5xy-2x-4ij-8s + i = 0. (c) 6 x2 + 6 2/2 - 4 ^-^ + 15 x?/ - 18 X - 18 y + 12 = 0. (fZ) 12 j/2 4. 1.5 5^-2 + 12 a-y - 8 a; - 28 ?/ +12 = 0. Art. 186. Pages 242-243. 1. X- oz=::k, X +Sz = vi. 2. k = 7n = ± \/2. „ 4^ , Vl94 „ , 4 V194 '• 3^+-T3-^ = ''3^--l3-' = '"- INDEX (The numbers refer to the pages.) Abscissa, 11. Addition of directed segments, angles, 9. Adiabatic exi^ansion, 192. Agnesi, Donna Maria, 172. Algebraic functions, 41. Angle which one segment makes with another, 22. which one line makes with another, 105. which a line makes with a plane, 226. which one plane makes with an- other, 218. Area of a triangle, 26, 28. of any polygon, 30. Asymptotes, 81, 133. Asymptotic cones, 243. Axes of coordinates, 11. of ellipse, 62. of hyperbola, 65. Axis of parabola, 123. of pencil of planes, 219. Azimuth, 10. B Bisectors of angles, 106. Boyle's law, 50. Cardioid, 178. Cartesian coordinates, 10. Cassinian ovals, 67. Catenary, 85. Circle, 55. Circular cone, 211. sections, 242. Cissoid, 169. Classification of curves, 96. of quadric surfaces, 229. Clockwise, 9. Cofactors, 150, 217. Colatitude, 196. Common chord, 163. Conchoid, 170. Cone, 236. Conicoids, 229. Conies as sections of a cone, 211. Conic sections, 110. Conjugate axis, 62, 65. diameters, 123. hyperbolas, 134. Construction of a surface, 228. Contour lines, 136. Coordinate axes, 11, 195. planes, 195. Coordinates, cartesian, 11, 195. cylindrical, 197. of point of contact, 115. polar, 13. rectangular, 11. spherical, 196. Cosine curve, 43. Counterclockwise, 9. Cross-sections, 229. Cubic curve, 83. Curves, algebraic, 169. in space, 245. Cusp, 176. Cycloid, 173. Cylinders, 210. 263 264 INDEX D Damped vibrations, 87. Descartes, 10. Determinant, 26. form of equation, 69, 216. Determination of functional corre- spondence, 33. Diameter of conic, 123. Diodes, 170. Directed segments, 8. angles, 8. Direction cosines, 200. Director circle, 114. Directrices of conies, 107. Directrix of a parabola, 66. Discriminant, 150. Discussion of an equation, 77. Distance between two points, 20, 199. of a point from a line, 104. of a point from a plane, 217. Duplication of cube, 170. E Eccentricity, of ellipse, 62. of hyperbola, 6-5. Ellipse, 60. Ellipsoid, 2.30. of revolution, 210. Elliptic paraboloid, 2-34. Empirical equations, 182. Epicycloid, 177. Equations of a line, 58, 221. of first degree, 98. of a plane, 213-216. of second degree, 107, 229. of higher degree, 169. of tangents. 111. Equilateral hyperbola, 135. Exponential curve, 44. Foci of an ellipse, 61. of an hyperbola, 64. of a cassinian oval, 68. Eocus of a parabola, 66. Folium of Descartes, 84. Four-cusped hypocycloid, 177. Function, 33. algebraic, 41. inverse, 45. ti-anscendental, 41. G General equation of second degree, 145. Graph of exponential function, 44. Graphic representation, 34. Graphs, 34. geometric construction of, 34, 42, 44, 47. of inverse functions, 46. of transcendental functions, 41. H Harmonic range, 131. Helix, 246. Hyperbola, 63. Hyperbolic paraboloid, 235. Hyperboloid, 231, 233. Hypocycloid, 175. Intercept form, 58, 215. Intercepts, 38. Intersecting lines, 99. planes, 223. Inverse functions, 45. Involute of a circle, 180. Latus rectum, 63, 66, 67. Law, 53. Lemniscate, 68. Length of a segment, 20, 199. LimaQon, 172. Limiting cases of conies, 143, 155. of quadric surfaces, 236-237. Line, perpendicular to a plane, 226. through a point, 101. through two points, 57. INDEX 265 Linear e(|uations, 98, scale, 7. Lituus, i)l. Locus of a point, 53. Logaritbuiic paper, 189. ■spiral, 9L M Machines, 48. Major axis, 62. Maximum, o5. Midpoint of segment. 24. Minimum, 35. Minor axis. ()2. Monotone function, 35. Multiple-valued functions, 30. N Naperian logarithms, 6. Nicomedes, 17 L Normal form, 102, 213. Normal to a curve, 1 19. Oblate, 210. Oblique axes, 11, 196. Ordinate, 11. Origin, 8. Orthogonal sets of curves, 137. Pairs of planes, 237. Parabola, 66. cartesian equation, 66. polar equation, 71. Paraboloid, elliptic, 231. hyperbolic, 235. of revolution, 210. Parallel segments, 23. Parameter, 73. Parametric equations. 73. Pascal, 172. limagon, 172. Pencil of conies, 163. of lines, 100. Pencil of planes, 219-220. Periodic functions, 43. Perpendicular segments, 23. Plane, 213. through three points, 215. Plotting, 34. Point bisecting a segment, 24. dividing a segment in a given ratio, 25. Polar line, 128. coordinates, 13, 197. equation of circle, 69. equation of ellipse, 71. equation of hyperbola, 71. equation of line, 70. equation of parabola, 71. ! Poles and polars, 127. I Position of a ix)nit in a plane, 10 j Profile, l(i. Projectile, 180. Projecting cylinders, 244. planes, 222. Projections of a segment, 18, 202. Prolate. 210. Quadrant, 12. Quadratic equation, 1 . Quadric surfaces. 229. R Radical axis, 163. Radius vector, 13. Rectangular coordinates, 11, 196. hyi^erbola, 73. Reduction to normal form, 103. Reflection properties, 120. Relation between rectangular coordi- nates and polar coordinates, 14. Removal of term in :nj, 95. Rotation of axes, 92. Ruled surfaces, 238. Rulings on hyperboloids, 239. Semicubical parabola, 173. Simultaneous linear equations, 99. 266 INDEX Sine curve, 42. Single-valued functions, 36. Slope of segment, 18. form of equation of a line, 58. Sphere, 208. Splierical coordinates, 196. Spiral of Archimedes, 40. Steam pressure gauge, 49. Strophoid, 97. Subnormal, 119. Subtangent, 119. Surface of revolution, 209. Symmetry, 37, 77, 228. System of concentric hyperbolas, 135. of circles, 163. of confocal conies, 109. Tangent plane, 240. Tangent to a curve. 111. to a circle, 111. to an ellipse, 112. Tangent to an hyperbola, 112. to a parabola, 112. Temperature a function of time, 50. Three-cusped hypocycloid, 180. Transcendental curves, 173. functions, 41. Transformation of coordinates, 91. from cartesian to polar, 14. Translation of axes, 92. Transverse axis, 62, 65. Trochoid, 180. Trigonometric formulas, 1. Trisection of an angle, 171, 172. Trisectrix of Maclaurin, 173. Vertex of parabola, 66 Vertices of ellipse, 62. of hyperbola, 65. W Witch of Agnesi, 171. Date Due '-*•- " ', i f6B 25 •§» f i \ f. \ f \S'05h3 BOSTON COLLEGE 3 9031 01546435 7 mm mm science ubrary BOSTON COLLEGE LIBRARY UNIVERSITY HEIGHTS CHESTNUT HILL. MASS. 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