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CONSTEUCTIVE GEOMETRY 
 
 OP 
 
 PLANE CURVES. 
 
CONSTRUCTIVE GEOMETRY 
 
 OF 
 
 PLANE CURVES 
 
 WITH NUMEROUS EXAMPLES 
 
 BY 
 
 T. H. EAGLES, M.A. 
 
 INSTRUCTOR IN GEOMETRICAL DRAWING AND LECTURER IN ARCHITECTUrE 
 AT THE ROYAL INDIAN ENGINEERING COLLEGE, COOPERS HILL. 
 
 Honlron : 
 MACMILLAN AND CO. 
 
 1885 
 
 [All Eights reserved.] 
 
A^ 
 
 PRINTED BY C. J. CLAY, M.A. & SON, 
 AT THE UNIVERSITY PRESS. 
 
^ 
 
 PEEFACE. 
 
 The appearance of another text-book on Geometry may 
 perhaps be considered to demand an apology, but I venture 
 to hope that an examination of the following pages will 
 shew them to differ considerably from any existing treatise. 
 The extending use of graphic methods in the solution of 
 many practical engineering problems has appeared to me 
 to demand a corresponding extension in the practice of 
 drawing the curves on which such solutions may frequently 
 depend, and, though the properties of conic sections have 
 been discussed thoroughly both geometrically and analyti- 
 cally, there is so far as I am aware no book treating of the 
 actual delineation of the curves from given data to any- 
 thing like the extent here attempted. Independently how- 
 ever of their applied use, the problems generally will, I 
 think, be found useful merely as drawing exercises in science 
 and other schools. A great deal of attention is devoted to 
 the construction of regular polygons, circles packed into 
 another circle and similar fancy figures, by methods which 
 no practical draughtsman ever uses, while the construction 
 of an ellipse is at the most limited to drawing it from the 
 principal axes or from a pair of conjugate diameters ; and 
 the time spent on these and similar exercises might, I think, 
 
 E. 3578oo ^ 
 
VI PREFACE. 
 
 be more profitably devoted to work bringing out the nature 
 and properties of this and other curves. 
 
 I can say from experience that the practice of sketching 
 a curve freehand through a series of previously found points 
 is a most valuable element in teaching mechanical drawing, 
 while the finding the points furnishes abundant exercise in 
 handling square and compasses, and impresses on the student 
 in a very striking manner the necessity for neatness and 
 accuracy in their use. 
 
 Each problem may of course be drawn on paper without 
 reference to the proof of the principle on which its con- 
 struction depends, but I consider that for the advanced 
 student at any rate it must be much more satisfactory to 
 work with as complete an insight as possible into the 
 methods he is using instead of groping along by mere rule 
 of thumb, so that in nearly all cases notes in proof of the 
 property made use of have been added, although such proofs 
 may be found in numerous published works, and are indeed 
 so completely common property that I have not thought it 
 necessary to give direct references to the pages from which 
 they have been taken. 
 
 I cannot however here omit to notice my indebtedness 
 to Dr Salmon's classical work on Conic Sections, or to 
 Chasles' Gdometrie Sup^rteure for the chapter on Anhar- 
 monic Ratio and the Anharmonic Properties of Conies. 
 Chap. VIII. will, I hope, convince a draughtsman that he 
 can if he likes make use of an engine very little known in 
 England and of enormous power. The methods of Modern 
 Geometry deserve to be brought into much closer relation 
 with the drawing-board than has hitherto been the case. 
 
 The chapter on Plane Sections of the Cone and Cylinder 
 involves some elementary notions of Solid Geometry or 
 Orthographic Projection, but the explanations given will, I 
 hope, enable the average student to work through the chapter 
 
PREFACE. Vll 
 
 without referring to any special treatise on Projection. The 
 ordinary pseudo-perspective diagrams usually given in books 
 on Conies are I think unsatisfactory, and the method of 
 referring the solid to two rectangular planes seems to me 
 in every way preferable. When the mental conception of 
 a plan and elevation is once thoroughly realised the student 
 is well repaid by the exactness with which he is able to 
 lay down on paper any point or line on the surface of the 
 cone. 
 
 The later chapters cannot be read without some know- 
 ledge of trigonometry, but the practice of translating a 
 trigonometrical expression into something which can be 
 represented to the eye is a valuable one, and the hints 
 given in the chapter on the Graphic Solution of Equations 
 will I trust be found useful. 
 
 My warmest thanks are due to my friend and colleague 
 Professor Minchin for much valuable advice and assistance 
 most freely and readily given : without his help the book 
 would have been much less complete than it is, whatever 
 its imperfections may be found to be. 
 
 It would be too much to hope that a work of this 
 character should have been compiled and gone through the 
 press without some errors creeping in. I hope they are not 
 more numerous than from the nature of the case may be 
 considered unavoidable, and I shall be thankful for any such 
 being brought to my notice. 
 
 Coopers Hill, 
 Oct. 1885. 
 
 62 
 
TABLE OF CONTENTS, 
 
 CHAPTER I. 
 
 INTRODUCTORY. 
 PROBLEM PAGE 
 
 1. To draw a line bisecting the angle between two given lines . 3 
 
 2. To find a fourth proportional to three given lines ... 6 
 
 3. To divide a line of given length similarly to a given divided 
 
 line 6 
 
 4. To draw a line through a given point and through the inter- 
 
 section of two given lines ib. 
 
 5. To find the geometric mean between two given lines ... 7 
 
 6. To divide a given line so that the rectangle contained by its 
 
 segments is equal to the square on a given line . . . ib. 
 
 7. To divide a line medially, or in extreme and mean proportion . 8 
 
 8. To find graphically a series of terms in geometrical progression 9 
 
 9. Given two ratios, to determine graphically their product . . 10 
 
 10. To determine graphically the square root of any number . . 11 
 
 11. To find the harmonic mean between two given lines ... 13 
 
 12. To find the third term of an harmonic progression, the first two 
 
 terms being given 14 
 
 Harmonic range and harmonic pencil ib. 
 
 Harmonic properties of a complete quadrilateral ... 16 
 
 13. Two pairs of conjugate points being given, to find the centre and 
 
 foci of the involution 17 
 
 14. Through a given point to draw a line meeting two given lines 
 
 so that the segments between the point and each line are 
 equal 19 
 
X CONTENTS. 
 
 PEOBLEM PAGE 
 
 15. To draw a triangle with its sides passing through three given 
 
 points and its vertices on three concurrent Hues . . 20 
 
 16. To draw a triangle with its vertices on three given lines and its 
 
 sides passing through three given points, one on each line ib. 
 
 17. To determine the locus of the vertex of a triangle on a given 
 
 base and with sides in a given ratio 21 
 
 18. To construct a rectangle equal in area to the sum or difference 
 
 of two given rectangles 22 
 
 19. From a given point P in a given straight line P3I, to draw lines 
 
 making equal angles with PM and cutting a second given 
 
 line C3I at equal distances from C 23 
 
 CHAPTER II. 
 
 THE CIRCLE. 
 
 20. To describe a circle to pass through three given points . . 29 
 
 Pole and Polar 30 
 
 Self -con jugate triangle 32 
 
 21. To describe a circle to pass through two given points and to 
 
 touch a given straight line ib. 
 
 22. To describe a circle to pass through a given point and to touch 
 
 a given line in a given point 33 
 
 23. To describe a circle to touch two given straight lines, one of 
 
 them in a given point ib, 
 
 24. To describe a circle to pass through a given point and to touch 
 
 two given lines 34 
 
 25. To describe a circle to touch three given lines .... 35 
 
 26. To describe a circle to touch a given circle and a given straight 
 
 line in a given point ib. 
 
 27. To describe a circle to touch a given circle and to pass through 
 
 two given points 36 
 
 28. To describe a circle to touch a given circle and two given 
 
 straight lines 37 
 
 29. To describe a circle to touch a given circle, to touch a given line 
 
 and to pass through a given point 39 
 
CONTENTS. XI 
 
 PEOBLEM PAGE 
 
 30. On a given straight line to describe a segment of a circle which 
 
 shall contain a given angle 40 ' 
 
 31. To draw a line touching two given circles 41 
 
 Properties of a system of two or more circles .... 42 
 
 32. To describe a circle to touch two given circles and to pass 
 
 through a given point 46 
 
 33. To describe a circle to touch two given circles and a given 
 
 straight line 47 
 
 34. To describe a circle to touch three given circles .... 49 
 
 35. To draw a circular arc through three given points without using 
 
 the centre 51 
 
 CHAPTER III. 
 
 THE PARABOLA. 
 
 36. To describe a parabola with given focus and directrix . . 57 
 Tangent and normal 59 
 
 37. To draw a parabola with given axis and vertex and to pass 
 
 through a given point 61 
 
 38. To draw a parabola with given focus and axis and to pass 
 
 through a given point 62 
 
 39. To describe approximately by means of circular arcs a parabola 
 
 with given focus and vertex 63 
 
 40. To describe a parabola with given focus and to pass through 
 
 two given points 64 
 
 41. To describe a parabola with given focus, to pass through a given 
 
 point and to touch a given line 65 
 
 42. To describe a parabola with given focus and to touch two given 
 
 lines 66 
 
 43. To describe a parabola with given directrix and to pass through 
 
 two given points 68 
 
 44. To describe a parabola with given directrix, to pass through a 
 
 given point and to touch a given line . . . . ib. 
 
 45. To describe a parabola with given directrix and to touch two 
 
 given lines 69 
 
Xll CONTENTS. 
 
 PROBLEM PAGE 
 
 46. To describe a parabola with given axis and to pass through two 
 
 given points 69 
 
 47. To describe a parabola with given axis, to pass through a given 
 
 point and to touch a given line 71 
 
 48. To describe a parabola with given axis and to touch two given 
 
 lines 73 
 
 49. To describe a parabola to touch two given lines at given points 75 
 
 50. To describe a parabola to touch three given lines, one of them 
 
 at a given point 78 
 
 51. To describe a parabola to pass through three points and the 
 
 axis to be in a given direction 79 
 
 52. To describe a parabola to touch three given lines and the axis 
 
 to be in a given direction 81 
 
 53. To describe a parabola to pass through two given points and to 
 
 touch two given lines 82 
 
 54. To describe a parabola to pass through three given points and to 
 
 touch a given line 84 
 
 55. To describe a parabola to pass through a given point and to 
 
 touch three given lines 85 
 
 56. To describe a parabola to pass through four given points . . 87 
 
 57. To describe a parabola to touch four given lines ... 88 
 
 58. To determine the centre of curvature at any point of a given 
 
 parabola 89 
 
 59. To describe a parabola to touch two given circles, the axis being 
 
 the line joining their centres 91 
 
 CHAPTER lY. 
 
 THE ELLIPSE. 
 
 60. To describe an ellipse with given axes (three methods) . . 99 
 Tangent and normal 102 
 
 61. To describe approximately by means of circular arcs an ellipse 
 
 having given axes (two methods) 103 
 
 Sundry properties of the ellipse 105 
 
 62. To determine the axes of an ellipse from a given pair of con- 
 
 jugate diameters 110 
 
CONTENTS. , Xlll 
 
 PROBLEM PAGE 
 
 63. To describe an ellipse with given conjugate diameters (four 
 
 methods) • . . . Ill 
 
 64. To describe an ellipse with a given axis and to pass through a 
 
 given point 114 
 
 Qo. To describe an ellipse with a given axis and to touch a given line 115 
 
 66. To describe an ellipse, the directions of a pair of conjugate dia- 
 
 meters, a tangent and its point of contact being given . 116 
 
 67. To describe an ellipse, the centre, two points on the curve and 
 
 the directions of a pair of conjugate diameters being given ib. 
 
 68. To describe an ellipse, the centre, the direction of the major 
 
 axis and two tangents being given 118 
 
 69. To describe an ellipse, the centre, the directions of a pair of 
 
 conjugate diameters, a tangent and a point on the curve 
 being given 119 
 
 70. To describe an ellipse, the centre, two tangents and a point on 
 
 the curve being given 121 
 
 71. To describe an ellipse, the centre and three tangents being given 122 
 
 72. To describe an elhpse, the centre, two points on the curve and 
 
 a tangent being given 123 
 
 73. To describe an ellipse, the centre and three points on the curve 
 
 being given 124 
 
 74. To describe an ellipse, the foci and a point on the curve being 
 
 given 125 
 
 75. To describe an ellipse, the foci and a tangent to the curve being 
 
 given ib. 
 
 76. To describe an ellipse, a focus, a tangent with its point of 
 
 contact and a second point on the curve being given . . 126 
 
 77. To describe an ellipse, a focus, a tangent and two points on the 
 
 curve being given 127 
 
 78. To describe an ellipse, a focus, a point on the curve and two 
 
 tangents being given ib. 
 
 79. To describe an ellipse, a focus and three tangents being given . 129 
 
 80. To describe an elHpse, a focus and three points being given . ib. 
 
 81. To describe an ellipse, two tangents with their points of contact 
 
 and a third point being given 131 
 
 82. To describe an elHpse, two tangents and three points being given 133 
 
 83. To describe an ellipse, three tangents and two points being given 134 
 
XIV CONTENTS. 
 
 PEOBLEM PAGE 
 
 84. To describe an ellipse, five tangents being given . . . 136 
 
 85. To describe an ellipse, four tangents and a point being given . 137 
 
 86. To describe an ellipse, five points being given .... 138 
 
 87. To describe an ellipse, four points and a tangent being given . 139 
 
 Pole and Polar 140 
 
 Harmonic Properties 141 
 
 88. To determine the centre of curvature at any point of a given 
 
 ellipse 145 
 
 CHAPTER V. 
 
 THE HYPERBOLA. 
 
 89. To describe an hyperbola, the foci and a vertex, the vertices and 
 
 a focus, or the axes, being given 153 
 
 Tangent and normal 157 
 
 90. To describe an hyperbola, an asymptote, focus and a point on 
 
 the curve being given ib. 
 
 91. To describe an hyperbola, an asymptote, focus and tangent 
 
 being given 158 
 
 92. To describe an hj^perbola, an asymptote, directrix and a point 
 
 being given ih. 
 
 93. To describe an hyperbola, the asymptotes and a point being 
 
 given 159 
 
 94. To describe an hyperbola, the asymptotes and a tangent being 
 
 given 160 
 
 Sundry properties of hyperbola ib. 
 
 95. To describe an hyperbola, transverse axis and a point being 
 
 given 166 
 
 96. To describe an hyperbola, transverse axis and tangent being 
 
 given ib. 
 
 97. To describe an hyperbola, a pair of conjugate diameters being 
 
 given ib. 
 
CONTENTS. XV 
 
 PEOBLEM PAGE 
 
 98. To describe an hyperbola, the centre, directions of a pair of con- 
 
 jugate diameters and two points being given . . . 168 
 
 99. To describe an hyperbola, the centre, directions of a pair of con- 
 
 jugate diameters, a tangent and a point being given . . 170 
 
 100. To describe an hyperbola, the centre, two tangents and a point 
 
 being given 171 
 
 101. To describe an hyperbola, the centre, a tangent and two points 
 
 being given 172 
 
 102. To describe an hyperbola, the centre and three tangents being 
 
 given 17a 
 
 103. To describe an hyperbola, the centre and three points being given ib. 
 
 104. To describe an hyperbola, the foci and a point on the curve being 
 
 given 174 
 
 105. To describe an hyperbola, the foci and a tangent being given . 175 
 
 106. To describe an hyperbola, a focus, tangent with its point of 
 
 contact and a second point on the curve being given . . ib. 
 
 107. To describe an hyperbola, a focus, a tangent and two points 
 
 being given 176 
 
 108. To describe an hyperbola, a focus, two tangents and a point 
 
 being given 178 
 
 109. To describe an hjrperbola, a focus and three tangents being given 179 
 
 110. To describe an hyperbola, a focus and three points on the curve 
 
 being given ib, 
 
 111. To describe an hyperbola, two tangents with their points of 
 
 contact and a third point on the curve being given . . 181 
 
 112. To describe an hyperbola, three tangents and two points on the 
 
 curve being given 182 
 
 113. To describe an hyperbola, two tangents and three points on the 
 
 curve being given 183 
 
 114. To describe an hyperbola, five tangents being given , . . 184 
 
 115. To describe an hyperbola, five points on the curve being given . 185 
 
 116. To describe an hyperbola, four tangents and one point being 
 
 given 186 
 
 117. To describe an hyperbola, four points and one tangent being 
 
 given 187 
 
 118. To determine the centre of curvature at any point of a given 
 
 hyperbola 188- 
 
XVI CONTENTS. 
 
 CHAPTER VI. 
 
 THE RECTANGULAR HYPERBOLA. 
 PROBLEM PAGE 
 
 Hints for solutions from various data .... 191 — 195 
 
 CHAPTER VII. 
 
 RECIPROCAL POLARS. 
 
 119. To find the polar reciprocal of one circle with regard to another 201 
 
 CHAPTER YIII. 
 
 ANHARMONIC RATIO. 
 
 120. Given the anharmonic ratio of four points, and the position of 
 
 three of them, to determine the fourth .... 214 
 
 121. Givenj^any number of points on a straight line, and three points 
 
 on a second line corresponding to a given three on the first, 
 
 to complete the homographic division of the second line . 218 
 
 122. Given a pencil of rays, and three rays of a second pencil 
 
 corresponding to a given three of the first, to complete the 
 second so that the two shall be homographic . . . ib. 
 
 123. Given two homographic ranges in the same straight line, to 
 
 determine the double points 220 
 
 124. Given two pairs of conjugate points and a fifth point of the 
 
 involution, to determine its conjugate .... 226 
 
 125. Given A , a and B,b in a. straight line, to find in the same line a 
 
 point 31 such that 
 
 1MB . 3Ib = ^ ^^ 
 
 126. Given two straight lines Aa, Bb and the points Ay B and JR, to 
 
 draw Bab so that 
 
 m=^ • '^ 
 
CONTENTS. XVll 
 
 PROBLEM PAGE 
 
 127. Given two straight lines Aa, Bh and the points A, B and R, to 
 
 draw Bah so that 
 
 Aa . Bb . = V 231 
 
 128. To draw a triangle having its vertices on three given lines and 
 
 its sides passing through three given points . . . 232 
 
 129. Given two points A and B and a line L ; given also t\yo lines 
 
 Sm, Sn and a point ; to find on L a point Q such that if 
 Om, On be drawn parallel respectively to AQ and BQ the 
 line mn shall 
 
 (1) be parallel to a given direction E, 
 
 (2) pass through a given point P 235 
 
 CHAPTER IX. 
 
 CONE AND CYLINDER. 
 
 130. To determine the section of a cone by any plane 
 
 131. To cut a conic of given eccentricity from a given cone 
 
 132. From a given cone to cut a conic of given eccentricity and 
 
 having a given distance between focus and directrix . 
 
 133. To determine the section of a cyhnder by a given plane . 
 
 134. To determine the sub-contrary section of an oblique cylinder 
 
 135. To determine the section of an oblique cylinder by any plane 
 
 136. To determine the sub-contrary section of an oblique cone . 
 
 137. To determine the section of an oblique cone by any plane 
 
 CHAPTER X. 
 
 CYCLOIDAL CURVES. 
 
 ^^38. To describe a cycloid, the diameter of the circle being given 
 
 Tangent and Normal, and Centre of Curvature 
 
 139. To describe a trochoid, the diameter of the circle and the distance 
 
 of the tracing point from its centre being given 
 
 Tangent, normal and centre of curvature .... 
 
 X.140. To describe an epicycloid, the radii of the rolling and directing 
 
 circles being given 
 
 Tangent, normal and centre of curvature .... 
 
 241 
 
 248 
 
 249 
 250 
 253 
 254 
 257 
 259 
 
 267 
 269 
 
 270 
 271 
 
 272 
 274 
 
XVlll CONTENTS. 
 
 PEOBLEM PAQB 
 
 141. To describe a hypo-cycloid, the radii of the rolling and directing 
 
 circles being given . . . . . . . .275 
 
 142. To describe an epi-trochoid, the rolling and directing circles and 
 
 the position of the tracing point being given . . . ib. 
 
 Tangent, normal and centre of curvature 277 
 
 ,^43. To describe a hypo-trochoid, the rolling and directing circles 
 
 and the position of the tracing point being given . . ib. 
 '*144. To describe the companion to the cycloid, the generating circle 
 
 being given ib. 
 
 Tangent, normal and centre of curvature 278 
 
 CHAPTER XI. 
 
 SPIRALS. 
 
 145. To describe the spiral of Archimedes, the pole and two points 
 
 on the curve being given 282 
 
 Tangent and normal 284 
 
 ^146. To describe a spiral of Archimedes, the pole, initial line, unit 
 
 and constant of curve being given ib. 
 
 147. The Reciprocal Spiral — pole, initial line, unit and constant of 
 
 curve being given 286 
 
 Tangent, normal and centre of curvature 288 
 
 , — ^148. The Lituus — pole, initial line, unit and constant of curve being 
 
 given 289 
 
 Tangent and normal 292 
 
 » 149. Logarithmic or Equiangular Spiral — pole, initial line, unit 
 
 and constant of curve being given 294 
 
 Tangent, normal and centre of curvature 296 
 
 150. Pole and two points on the curve being given .... ib. 
 
 151. To inscribe a Logarithmic Spiral in a given parallelogram . 297 
 
 152. Involute of circle 299 
 
 CHAPTER XII. 
 
 MISCELLANEOUS CURVES. 
 
 153. Harmonic Curve or Curve of Sines 305 
 
 Tangent and normal ib. 
 
CONTENTS. XIX 
 
 PROBLEM PAGE 
 
 154. ''Ovals of Cassini 307 
 
 Tangent and normal 308 
 
 155. ^ Cissoid of Diodes 309 
 
 Tangent and normal 310 
 
 156. ^' Conchoid of Nicomedes 312 
 
 Tangent and normal - . . ih. 
 
 157. ; Witch of Agnesi 313 
 
 Tangent and normal 314 
 
 158. ' Common Catenary — unit being given 316 
 
 Tangent and normal 318 
 
 159. Common Catenary — vertex, axis and point on curve . . ih. 
 
 160. Common Catenary — point of suspension, tangent at that point 
 
 and depth of loop 320 
 
 161. ' Common Catenary — axis, point and tangent at that point . ih. 
 
 162. Tractory or Anti-Friction Curve 322 
 
 163. 'TheLima9on . ^ 324 
 
 164. ■ The Cardioid . > Inverse Curves 328 
 
 165. The Lemniscate . ) 329 
 
 166. To determine the point on a spherical mirror at which an 
 
 incident ray shall be reflected in a given direction . . 331 
 
 167. To determine the point on a spherical mirror at which an 
 
 incident ray shall be reflected to pass through a given 
 
 point . 333 
 
 168. X Magnetic Curves 335 
 
 169. ^, Equi-potential Curves 338 
 
 170. ■ The Cartesian Oval .342 
 
 Tangent and normal 347 
 
 171.- Elastic Curves 348 
 
 172.,/Curves of Pursuit 355 
 
XX CONTENTS. 
 
 CHAPTER XIII. 
 
 SOLUTION OF EQUATIONS. 
 
 PROBLEM PAGE 
 
 173. TosolYex^-2Ax + B^ = 361 
 
 174. Tosolye x^ + 2Ax + B^ = 362 
 
 175. Tosolvex"-2Ax-B'^ = ib. 
 
 176. Tosol\ex^ + 2Ax-B'=0 363 
 
 177. To solve a cos ^ + & sin ^ = c 364 
 
 178. To draw locus represented by sin ^ + sin = a .... 365 
 
 179. To solve r^co8 2e = a^ 
 
 366 
 
 r . sin a- d = b . sin a ' 
 
 180. To draw locus represented by a cot ^ - a + 6 cot - j3 = c . . 368 
 
 181. To solve a cos ^+6 cos = c 
 
 370 
 
 fc cot + Z cot = 7n * 
 
 182. To solve -.~+-^^_., 
 
 sm ^ sm ) 371 
 
 cos 6 = k cos 
 
 J 
 
i- LJF-Oi<iMi/A 
 
 CONSTEUCTIYE TEEATISE ON 
 PLANE CUEYES. 
 
 CHAPTER I. 
 
 INTRODUCTORY. 
 
 The Instruments required for the accurate representation on 
 paper of almost all known curves are few in number and of 
 simple construction. For accurate work however it is essential 
 they should be of good quality, and be kept in good order. A 
 limited number of good instruments is in every way to be pre- 
 ferred to a larger number of inferior articles, and where economy 
 is an object therefore, in preference to the usual large and small 
 single jointed compasses found in cheap boxes of mathematical 
 instruments the author strongly recommends the purchase of one 
 medium size, double jointed pair of compasses with pen and 
 pencil points, which can be used for both large and small circles 
 if care be taken to adjust the legs so that the lower portions of 
 both may be perpendicular to the paper. This is a sine qud non 
 for good work and it is of course impossible with the ordinary 
 single jointed instruments. In addition to the above a pair of 
 dividers, a drawing pen for inking in straight lines, a protractor 
 which should also contain a diagonal scale of half-inches, a couple 
 of set squares (45*^ and 60"), pencil and paper may be considered 
 a complete equipment for the work of the following pages. 
 
 More may be learnt as to the proper way of handling these 
 tools by ten minutes' observation of a practised draughtsman 
 than from pages of explanation, but failing the opportunity of 
 this practical instruction, the following hints may be useful. 
 
 E. 1 
 
2 . ■/ IIS^ ;OT 'INSTRUMENTS. 
 
 Pai^Uel 'j.ili^S;shX)tild; Ije drawn by means of the set squares ; 
 (they are far better than parallel rulers). The edge of one must 
 be adjusted in the required direction and held firmly on the 
 paper, the other should be placed in contact with a second edge 
 of the first and held in that position, and the first may then be 
 made to slide along the second till it comes into the position of 
 the required parallel line. A line perpendicular to another and 
 passing through a given point should be drawn by adjusting an 
 edge containing the right angle of one of the squares to the given 
 line, placing the second square in contact with the hypotenuse of 
 the first and sliding the first along the second until its third side 
 passes through the given point, when the required perpendicular 
 can be drawn. 
 
 If a line is to be drawn through two given points, the point 
 of the pencil should first be placed on one of the points, the 
 square can then be brought up to the pencil and worked against 
 it as a centre till it coincides with the other, when the line can 
 be drawn, and care must be taken that the line passes accurately 
 through both points, as owing to the thickness of the edge of the 
 square it is quite possible to make a slight but quite appreciable 
 error. This is particularly the case if the pencil is cut to a 
 chisel edge instead of to a circular point, and the author would 
 express his decided conviction as to the superiority of the circular 
 point. It is of course quite impossible to draw accurately unless 
 a good sharp point to the pencil be constantly maintained. 
 
 Lines whether straight or circular should be bisected, tri- 
 sected, &c. by trial, mechanical methods however good in theory 
 being unnecessary and indeed objectionable in practice. A veiy 
 little practice in handling a pair of dividers will enable this to 
 be done with great ease and with all attainable accuracy, if the 
 amount by which the first shot exceeds or falls short of the 
 desired result is noted and the legs of the dividers closed up or 
 extended to the necessary estimated fraction of this amount. If 
 the required number of parts admits of division, the line should 
 first be divided in the smaller number of parts necessary, i.e. if 
 it is to be divided into six parts, it should be first bisected, and 
 
INTRODUCTORY. 3 
 
 then each half trisected ; if into nine parts it should 'be first 
 trisected, and so on. Care must be taken by a light handling of 
 the instruments, not to damage the paper, until it is found that 
 it can be marked with the points in the right places, and when 
 a point is being marked on a line with the dividers, special care 
 should be taken to press in the point on the line and not merely 
 somewhere in its neighbourhood. In handling the instruments 
 they should be constantly kept in as nearly vertical planes as 
 possible. A point when found should be marked by a light 
 pencil ring round it and not by a smudge made with a blunt 
 pointed pencil, which entirely obscures the exact position of the 
 point. 
 
 Problem 1. (Figs. 1, 2.) To draw a line bisecting the angle 
 between two given lines. 
 
 It is frequently necessary to do this when the lines are so 
 nearly parallel or are otherwise so situated that their point of 
 intersection does not fall wdthin the limits of the sheet of paper, 
 or drawing-board, and since the method of proceeding in this 
 case includes the ordinary simple case, it is the one chosen as 
 
 Fig.l. 
 
 an example. Let AB, CD (fig. 1) be the given lines. Draw Gil 
 parallel to ^^ at any convenient distance (BE) from it, and draw 
 GK parallel to CD at a distance DF equal to BE from it. This 
 can be done by drawing BE perpendicular to AB from any point 
 B on it, and DF perpendicular to CD from any point D on it 
 and making BE = DF, and then using two set squares in the 
 
 1—2 
 
4 
 
 BISECTION OF AN ANGLE. 
 
 way referred to in the introduction. The distance BE should 
 be so chosen as to bring the point G about as in the figure, 
 i.e. BE should be somewhat greater than half the least distance 
 between the given lines. If the angle EGF be now bisected, 
 its bisector will obviously also by symmetry bisect the angle 
 between AB and CD. Take any equal distances GH^ GK on GE^ 
 GF respectively or, what comes to the same thing, with centre 
 G and any radius describe an arc IIK, and with centres H and K, 
 and with any (the same) radius describe arcs intersecting in L. 
 Then GL will be the required bisector. For the triangle GHL 
 is obviously equal and similar in all respects to the triangle GKL. 
 This method is scarcely satisfactory when the lines are nearly 
 parallel, on account of the smallness of the angle EGF and the 
 difficulty of determining accurately the point of intersection G 
 of two nearly coincident lines, and an alternative method evading 
 this difficulty is shewn in fig. 2. As before, let AB, CD be the 
 
 
 M 
 
 
 / ■ 
 
 \ / 1 N 
 / / \ 
 
 ^ 
 
 
 
 1 
 
 C 
 
 
 /D 
 
 V 
 
 two given lines. At any point B of the one line draw a line as 
 BF, and at any point D of the other construct an angle GDH 
 equal to the angle EBF. The exact size of this angle is im- 
 material but preferably it should not diffipr much from half a 
 right angle. [An angle {GDH) can be constructed equal to a 
 given angle {EBF) by describing arcs EF, GH, with the angular 
 points B, D as centres and with any (equal) radius, and then 
 
INTRODUCTORY. 5 
 
 making the chord GR equal to the chord EF by means of a pair 
 of dividers.] 
 
 Let BF and DH intersect in K. The bisector of the angle 
 BKD will, by symmetry, he 'parallel to the required bisector, i. e. 
 bisecting the angle BKD by the line KL, the direction of the 
 required bisector is known. To find its position, draw any line 
 AC perpendicular to KL meeting the given lines in A and C. 
 The required bisector must evidently pass through M the centre 
 point of AG. It can therefore be drawn through this point 
 parallel to KL. 
 
 Problem 2. (Fig. 3.) To find a fourth proportional to three 
 rjiven lines AB, GD, EF, or to find a line of such length (l) that 
 AB : GD :: EF : I, 
 
 or that the rectangle contained by the two lines AB and I shall be 
 equal in area to the rectangle contained by GD and EF. 
 
 All questions involving proportionals depend on the construc- 
 tion of similar triangles. Draw any two lines OK^ OL meeting 
 
 Fig.3. 
 
 in and containing any angle. From along one line set off 
 OG~AB, the first term of the proportion, and OK = EF, the third 
 term of the proportion. From along the other line set off 
 OH = GD, the second term of the proportion, then through K 
 draw KL parallel to GH meeting OH in L. OL will be the re- 
 quired fourth term. For obviously by the similar triangles OGH, 
 OKL, 
 
 OG : OH :: OK : OL, 
 
 i.e. AB : GD :: EF : OL. 
 
6 
 
 SIMILAR DIVISION OF T\VO LINES. 
 
 A similar construction will obviously give a third proportional 
 to two given lines AB, CD ; i. e. a line of length (I) such that 
 
 AB : CD :: CD : I, 
 or that the rectangle contained hj AB and I shall be equal in area 
 to the square on CD ; the only difference being that in this case 
 the lengths OH and OK will be equal to each other. 
 
 Problem 3. (Fig. 4.) To divide a line of given length (AB) 
 
 similarly/ to a given line CD divided in any manner as at B^F 
 
 (There may he any manner of points of division.) 
 
 Draw any two lines as OG and OH. Make OG==AB, 0H= CD, 
 0K=CE,0L= CF. . . and draw KM, LN. . . parallel to HG, The 
 
 line OG, i.e. AB will be divided in M, N... similarly to CD in 
 F, F.... 
 
 Problem 4. (Fig. 1.) To draw a line through a given point 
 and through the intersection of two given lines. 
 
 It is of course the simplest possible thing to do this when the 
 axjtual point of intersection of the two lines is available. As in 
 Problem 1 however it is frequently necessary to draw a line the 
 direction of which depends on an inaccessible point. Let AB, CD 
 be the two given lines and M the given point. (M may be 
 between the lines as in iig. 1 or on the farther side of either with 
 regard to the other.) Draw any line through M meeting the 
 given lines in N and 0, and at any convenient distance from M 
 
INTRODUCTORY. 
 
 draw a second line parallel to NO meeting the given lines as at 
 A and G, If we divide ACy as in Q, in similar segments to those 
 in which M divides NO (Problem 3) the line QM will be the 
 required line passing through the intersection oi AB and CD. 
 The most convenient method for dividing ^C is probably thus : 
 join CN, draw PM through M parallel to CD and meeting ON in 
 F, and through P draw PQ parallel to AB meeting ^C in Q. AC 
 is obviously divided in Q similarly to NC in P and therefore 
 to NO in M. 
 
 Problem 5. (Fig. 5.) To find the geometric mean between two 
 given lines AB, CD, i.e. to find a line of length (l) such that 
 
 AB : I :: I : CD, 
 or that the square on I shall be equal in area to the rectangle con- 
 tained by AB and CD. 
 
 Draw any straight line EOF and set off on it on opposite 
 sides from 0, OE^AB, OF=CD. On EF describe a semicircle 
 
 Ah 
 
 Fig.5. 
 
 and from draw OG perpendicular to EF meeting the circum- 
 ference in G. OG will be the required mean proportional or 
 geometric mean. For, since the angle in a semicircle is a right- 
 angle (Euclid III. 31), .-. the angles OEG, EGO are together equal 
 to the angles EGO, OGF, said .: the angle 0^6^ = the angle OGF, 
 .'. the right-angled triangles OEG, OGF are similar and 
 
 .-. EO : OG :: OG : OF, 
 
 i.e. AB : OG :: OG : CD. 
 
 Problem 6. To divide a given line so that the rectangle con- 
 tained by its segments is equal to the square on a given line which 
 
8 MEAN AND EXTREME PROPORTION. 
 
 must obviously he not greater than half the line to he divided 
 (fig. 5). 
 
 This is the converse of the last problem. Let EF be the given 
 line, on it describe a semicircle. Draw the radius KL perpendi- 
 cular to EF and on it make KM equal to the side of the required 
 square. Through M draw a parallel to EF meeting the circle in 
 G and from G drop a perpendicular on EF meeting it in 0, will 
 be the required point of division. 
 
 The construction is obvious from the last problem. 
 
 Problem 7. (Fig. 6.) To divide a line medially^ or in extreme 
 and mean proportion, i.e. to find a point (F) in a line AB such 
 that 
 the whole line AB : the greater segment (BF) 
 
 :: BF : the lesser segment (AF), 
 or that the rectangle contained by the whole line and the lesser 
 segment is equal in area to the square on the greater segment. 
 
 Bisect AB in C, from A draw AT> perpendicular to AB and 
 make AD = AC = \AB. Join BD and on it from B cut off 
 
 BE = BA ; from B on BA cut off BF = BE. F will be the required 
 point. This construction is simplified from Euclid ii. 11, the proof 
 may be shewn thus. 
 
 The sq. on BB = sq. on ^^ + sq. on ^Z> (Euclid i. 47). 
 Also „ = sq. on EB + sq. on EB + 2 rect. EB . EB, 
 
 but EB = AB and EB = FB, (Euclid ir. 4), 
 
 .-. sq. on AB = sq. on i^"^ + 2 rect. FB . AB. 
 Again sq. on AB = sq. on FB + sq. on AF+ 2 rect. AF . FB 
 
 (Euclid II. 4), 
 = sq. on FB + rect. AF(AF+ FB) 
 + rect.AF.FB. 
 
INTRODUCTORY. 9 
 
 .-. 2 rect. FB.AD^rect. AF. AB + rect AF. FB, 
 
 i. e. rect. FB {2AD -AF\= rect. AF. AB, 
 but 2AD = AB and AB-AF= FB, 
 
 .'. finally sq. on i^^- rect. AF . AB. 
 
 Problem 8. (Fig. 7.) To find graphically a series of terms in 
 geometrical progression, being given either two successive terms or 
 one term and the common ratio. 
 
 Draw two lines Oe, OF meeting in at any convenient angle. 
 On one mark off the 1st given term as OA, and on the other the 
 
 2nd given term as OB, or if the common ratio be given a length 
 0B= 1st term multiplied by the common ratio. 
 
 [In the figure OA the first term = 2, and OB — 2- A: ; the com- 
 mon ratio therefore is 1*2, the unit being 3'78.] 
 
 With centre and radius OB describe an arc cutting OA in h ; 
 through b draw bC parallel to ^^ cutting OB in C. 00 will be 
 the required third term of the series. Similarly make Oc on OA 
 = 0C and through c draw cD parallel to AB cutting OB in i>, 
 OD will be the required fourth term, and so on in succession. 
 Terms on the other side of OA can also be determined as shown 
 at OB^, OC^,&c. 
 
 The construction evidently depends on the similarity of the 
 triangles OAB, ObC, &c. 
 
 by which OC : OB :: Ob : OA, 
 
 I e. since Ob = OB, OB' = OA . OG, 
 
10 PEODUCT OF TWO RATIOS. 
 
 or each term is a mean proportional between the two on opposite 
 sides of it, in other words the series is in geometrical progression. 
 
 Since OB = r . OA, the above expression for OB^ becomes 
 t\OA = OC, 
 and so also r^ . OA = 01) and so on. 
 
 Very careful drawing is required to ensure accuracy, and the 
 scale should be as large as possible, as otherwise, since errors are 
 cumulative, the lengths obtained for the fourth or fifth and suc- 
 ceeding terms may differ considerably from their true values. 
 
 Problem 9. Given two ratios y and — to determine the ratio 
 
 m 
 
 7 — , or to divide a given line so that the ratio of its segments shall 
 equal the product of two given ratios (Fig. 8). 
 
 Draw any line AB and on it make AD— a, DB = h. With 
 centre B and radius l + m describe an arc, and with centre 
 
 A and radius AC the length of the given line to be divided 
 describe an arc intersecting the former in C. Make BF on BG = I 
 so that FG = m. Draw AF, GD intersecting in and draw BO 
 meeting A G in F. E will be the required point of division : 
 
 AE AD BF_a l^ 
 ^•®* ~EG~'DB'FV~h'm' 
 
 AD sin AOD DB sin BOD 
 
 ^"^ OD~^^^ODA 6 D ~~ sin ODB' 
 
 AD sin AOD 
 DB~ sin BOD* 
 
INTRODUCTORY. 11 
 
 . ., , BF sinBOF sin BOF 
 
 similarly -^ = ^^^ = ^^j^ , 
 
 and 
 
 AE sin AOE sin BOF AD BF 
 
 EG sin EOC sin BOB BD'FC 
 It follows of course that in any triangle if lines be drawn from 
 the vertices ABC meeting the opposite sides in F, E, D and all 
 passing through the same point 0, AD . BF . CE = DB . FC . EA, 
 i.e. that the continued products of the alternate segments taken 
 in order are equal. 
 
 Problem 10. To determine graphically the square root of any 
 number (n), i.e. to determine a line the length of which : length of a 
 line containing n units measured on any scale :: 1 : sjn. 
 
 This is sometimes, though misleadingly, called determining the 
 square root of a given line. The fact is that the expression the 
 square root of a given line has no meaning unless we take the line to 
 represent, by the number of units it contains, a given area; and 
 then the line to be found is the side of a square, the number of 
 square units in which is equal to the number of units contained in 
 the line — the same scale of course being used for each. If a triangle 
 ABG be drawn, right angled at A and having the sides AB^ AG 
 each one inch long, the side BG is the side of a square of two 
 square inches area, and in this sense BG may be said to be the 
 square root of a line two inches long, or of the number 2, the 
 unit being one inch, but if the unit be half-an-inch the same 
 line BG represents the square root of 8, since (Euc. i. 46) 
 BG^ ^AB' + AG' = 2' + 2' = 8. 
 
 If we use a diagonal scale of half-inches, the length BG 
 may be read on it to two places of decimals, and the number 
 so obtained is the square root of 8 to two decimal places. 
 Any question relating to the square root of a number, must 
 therefore always be taken as involving the application of some 
 particular scale. The square root of any proposed number can be 
 found by splitting the number up so as to make it equal to the 
 sum or difference of two or more squares, and then constructing 
 right-angled triangles having sides equal to the sides of these 
 
12 EXTRACTION OF SQUARE ROOT. 
 
 squares. Thus ^3 = ^4 - 1, = J^^ - 1^ so that if a right-angled 
 triangle ABC be drawn, right angled at G and having ^^ = 2 inches, 
 and AC — \ inch, BQ represents ^3, an inch being the unit. 
 (The triangle may be constructed by drawing a semicircle on AB 
 as diameter and making ^C in it = 1 inch.) If the unit is half-an- 
 inch BC represents JAB''-AV\ i.e. Ji'-2' or ^12. 
 
 ^5 = Js^ - 2^, i.e. is the perpendicular of a right-angled 
 triangle the hypotenuse of which is 3 and the base of which is 2, 
 or it may be determined as the hypotenuse of a right-angled 
 triangle one side of which is 2 and the other 1, since ^^5 = J2^ + V. 
 If we halve the unit the same line would represent ,^20. 
 
 J6 = j2'' + j2% i.e. if J 2 be first determined, JQ is the 
 hypotenuse of a rigbt-angled triangle the sides of which are 2 and 
 
 J2, or it may be determined from JQ = v 3^ - ^3^. 
 
 J 7 = J 2^ + J 3^, and can be determined if ^3 is known. 
 ^8 has already been given. 
 
 J 10 = J¥Tv. 
 
 Jll= Ji^ - J 5^, and can be determined if ^ 5 is known. 
 
 J\2 has been given above; and the method is probably 
 sufficiently exemplified by the above, but we will take two 
 examples of larger numbers 
 
 Jil = iJW+Jll^, thus being made to depend on ^IJ. 
 
 JT79 = Jl3'' + JW thus being made to depend on JTO, it 
 
 might also be written =vlP + ^4:7^or could be determined in 
 other ways. No definite instructions can be given as to the best 
 mode of working in any particular case, but as a rule triangles 
 having sides of nearly equal magnitude should be selected, since 
 the intersections of lines cutting at very acute angles cannot be 
 accurately determined. 
 
INTRODUCTORY. 13 
 
 Definition. Three magnitudes are said to be in harmonic 
 progression when the first is to the third as the difference between 
 the first and second is to the difference between the second and 
 third : and the second magnitude is said to be an harmonic mean 
 between the first and third. 
 
 Thus if the magnitudes represented by the lengths of three 
 lines (as AB, AC, AD, fig. 9) are in harmonic progression and the 
 lines be superimposed with a common extremity as in that fig. : — 
 then AB : AJ) :: BC : CD. 
 
 The reciprocals of magnitudes in harmonic progression are in 
 arithmetic progression and conversely: — for, if AB, AC, AD are 
 in harmonic progression then by definition 
 
 AB:AB::BO:CD,or^=%, 
 and if -. „ , — ^ , -— are in arithmetic progression then by 
 definition, -l+_=^^, 
 
 but this may be written 
 
 J 1_^ J 1_ 
 
 AB AC~AC AD' 
 
 AC-AB AD-AC 
 
 AB . AC AC. AD' 
 BC CD 
 
 ab=aD' 
 
 an identical expression with the above. 
 
 Problem 11. (Fig. 9.) To find the harmonic mean between two 
 given lines AB, AD, i.e. to find a line of length I such that 
 AB : AD :: the difference between AB and I 
 
 : the difference between AD and I. 
 
 Set off the given lengths from the same point {A) on any line 
 and in the same direction along it, as AB, AD. Take any point 
 E outside AD and join AE, DE. Through B draw FBG parallel 
 
14 HARMONIC PROGRESSION. 
 
 to BE meeting AE v^ F and make BG-=BF. Join EG cutting 
 AD in C and AG will be the required harmonic mean. For by the 
 
 A? 
 
 Fig.9. 
 
 
 similar triangles ABF, ADE, 
 
 AB : AD :: BE : DE, 
 Also by the similar triangles GBG, CDE, 
 
 BG : GD :: ^G^ : DE, 
 and ^G^ = BE, 
 
 ,\ AB : ^7) :: BG : CA 
 
 :: AG-AB : AD -AG. 
 Problem 12. (Fig. 9.) To find the third term of a harmonic 
 j^rogression, the first two terms being give7i. 
 
 The above construction may be adapted to find the third term 
 of a harmonic progression the first two terms being given. Sup- 
 pose AB and AG given. Superpose them with a common ex- 
 tremity as in the fig. 9. Take any point F outside AG. Join 
 FB and produce it to G making BG = BF. Join AF and GC 
 producing them to meet in E and draw ED through E parallel to 
 FB meeting AG (produced if necessary) in D. AD will be the 
 required third term. 
 
 Dep. When four points in a straight line as ABGD in fig. 9 
 fulfil the condition 
 
 AB : AD :: BG : GD, 
 
 they constitute a Harmonic Range, and if through any point E 
 outside the line the four straight lines EA, EB, EG, ED be drawn 
 these four lines constitute a Harmonic Pencil, which is denoted by 
 E{ABGD}. Any straight line drawn across the pencil is called a 
 
INTRODUCTORY. 15 
 
 Transversal, and every transversal of a harmonic pencil is divided 
 harmonically in the points in which it intersects the lines of the 
 pencil : i. e. the four points of intersection constitute a Harmonic 
 Range. For in fig. 9 draw any transversal as HKLM, and through 
 K drsiw /Kg parallel to FD and therefore to FG, meeting EA, EC 
 in y* and g. Obviously since BF = BG, .'. Kf^ Kg. 
 
 By similar triangles HKf, HME 
 
 HK : HM :: fK : EM, 
 and by similar triangles KLg, MLE 
 
 KL : LM :: gK : EM, 
 but Kf^ Kg, 
 
 .-. HK : HM :: KL : LM, 
 or HKLM constitute a Harmonic Range. 
 
 A particular case of a Harmonic Pencil is furnished by the 
 pencil formed of two straight lines and the bisectors of the angles 
 between them, as shewn in fig. 10, where AD bisects the angle 
 
 BAG and AE is drawn perpendicular to AD, and therefore bisect- 
 ing the exterior angle between AC and BA produced. For draw 
 any transversal as BFGE, and through F draw PFK parallel to 
 AE and meeting AB, AG in P and N. 
 Then PF=^ FN and 
 BF : BE :: PF : AE, by similar triangles BPF, BAE, 
 FG : GE :: FN : AE, by similar triangles FGN, EGA, 
 .'. BF : Be'^:: FG : GE, 
 or the pencil is harmonic. 
 
16 HARMONIC RANGES AND PENCILS. 
 
 A line of given length may obviously be divided harmonically 
 in an infinite number of ways, since a line of length HK = BE 
 can be drawn from any point //on AB to terminate on AE and 
 HL : HK :: LM : MIL 
 
 Harmonic Properties of a complete Quadrilateral, 
 
 If FBeA, FDe^C be harmonic ranges (fig. 11), the straight 
 lines ACy ee^, BD meet in a point, as also AD, BC and ee^. 
 
 \\ 
 
 Fig.ll. 
 
 \ N 
 
 
 \ 
 
 'x 
 
 c) 
 
 y •"' ^^^ 
 
 iy 
 
 nM?'-< 
 
 /..-■■■ 
 
 
 / ,''' 
 
 
 Z^"' 
 
 \ "-^-i. 
 
 For a BD, AG meet in E, draw Ee ; then the pencil 
 E {AeBF) is harmonic and FC is a transversal, so that e^ must lie 
 on^e. 
 
 Similarly SS. AD and BG meet in 0, the pencil 0{AeBF) is 
 harmonic and FC a transversal, so that e^ must lie on Oe. 
 
 If ABGD is any quadrilateral, E the intersection of the sides 
 AG and BD, F of the sides AB and GD, the intersection of the 
 diagonals AD and BG', it follows conversely that EA, EO, EB, EF 
 form a harmonic pencil, as also FE, FG, FO and FA. If EO meet 
 AB in e and GD in e^, AeBF and GefiF are therefore harmonic 
 ranges, and if FO meet ^C in/ and BD in/,, 4/Ui^ and BfDE 
 are both harmonic ranges. 
 
 Further if AD meet /^'^ in a and 5(7 meet it in h, BOGh is a 
 harmonic range since it is a transversal of the pencil F (EGfA)^ 
 therefore AF, Aa, AE and Ab form a harmonic pencil, and there- 
 fore FaEb is a harmonic range, i. e. FE is divided harmonically in a 
 and h. 
 
INTRODUCTOEY. 
 
 17 
 
 Def. a system of pairs of points Aa^ Bb, &c. on a straight 
 line such that XA . Xa = XB . Xb = ,.. = XP^ = XQ^ is called a 
 system in Involution, the point X being called the centre^ P 
 and Q the foci of the system, and any two corresponding points 
 A, a, conjugate points. , 
 
 Problem 13. Two pairs of conjugate points A, a and B, b, 
 being given, to find the centre and foci of the involution. 
 
 The existence of a focus is only possible when both points of 
 a pair are on the same side of the centre, and hence two cases 
 arise, 1st, in which one pair of points lies within the other, and 
 2nd in which each pair lies wholly outside the other. 
 
 Case 1. (Fig, 12.) Let ab be less than AB. Through a the 
 extreme point of the range draw any line ac, and through B the 
 
 Fig.l2. 
 
 more distant from a of the second pair of points draw a parallel 
 line Bd. Make ac = ab, Bd = BA, then dc will intersect ABba in 
 X the required centre — for 
 
 Xa : ac :: XB : Bd, 
 .'. Xa : ab :: XB : BA, 
 .'. Xa+ab : Xa :: XB + BA : XB, 
 i. e. Xb : Xa :: XA : XB, 
 therefore by definition X is the centre of the system. 
 
 Take a mean proportional between either XA and Xa or 
 XB and Xb, which determines the distance XP and XQ from X 
 of the foci. 
 
 E. 2 
 
18 
 
 CENTRE AND FOCI OF POINTS IN INVOLUTION. 
 
 Case 2. (Fig. 13.) Through the extreme points of the system 
 draw any two parallel lines as he, Ad. Make he = ha the distance 
 
 Fig.13. 
 
 Air 
 
 from h of the nearer point of the opposite pair and make Ad = AB 
 the distance from A of the similar point, then cd will cut -46 in 
 X the required centre — for 
 
 XA : Ad :: Xh : he, 
 i.e. XA : AB :: Xh : ah, 
 .-. XA : AB-XA :: Xh : ah-Xh, 
 or XA : XB :: Xh : Xa. 
 
 The foci must be determined as in Case 1, 
 Since XA : XF :: XP : Xa, 
 
 '.'. XA-XP : XA + XP :: XP-Xa : XP + Xa, 
 i.e. AP : AQ :: Pa : aQ, 
 or each pair of conjugate points forms, with the foci of the system, 
 a harmonic range. 
 
 It follows of course that if APaQ be an harmonic range and 
 X the centre point of PQ, 
 
 XA.Xa = XP' = XQ\ 
 
 The following relations between two pairs of conjugate points 
 Aa and Bh, and their centre X and foci P and Q are sometimes 
 useful. 
 

 INTRODUCTORY. 
 
 Since 
 
 XA : Xb :: XB : Xa, 
 
 
 .\ XA : Ab :: XB : aB, 
 
 or 
 
 XA : XB:: Ab : aB 
 
 and since 
 
 Xb : Xa :: XA : XB, 
 
 
 .'. Xb : ab :: XA : AB, 
 
 or 
 
 Xb : XA :: ab : AB ( 
 
 19 
 
 (1), 
 
 (2); 
 
 therefore, multiplying (1) and (2), 
 
 Xb : XB :: Ab.ba : 
 Again, since QbPB is harmonic, 
 
 .-. Qb : QB :: Pb 
 
 or Qb : Pb 
 
 .'. Qb^Pb : Pb 
 
 or 2ZP : Pb 
 
 .'. P¥ : PB'' 
 
 AB.Ba, 
 
 PB, 
 
 PB, 
 
 QB : PB, 
 QB^PB 
 
 2XB : PB, 
 XP' : XB' 
 :: Xb : XB 
 :: Ab.ba : AB.Ba. 
 This determines the ratio in which Bb is divided by P. 
 
 Problem 14. (Fig. 14.) Through a given point P to draw a 
 line meeting two given lines AB and CD in B and D so that 
 PB = PD. 
 
 Through P draw any line meeting one of the given lines as at 
 A. On AP produced make Pa = PA and draw aD parallel to BA 
 
 Fig.l4 
 
 meeting the other given line in D. The line DPB will be the 
 line required, i.e. PB = PD (by the similar and equal triangles 
 APB, aPD). 
 
 2—2 
 
20 TKIANGLE WITH VERTICES ON GIVEN LINES, &C. 
 
 Problem 15. To draw a triangle with its sides passing through 
 three given points A, B, C, and with its vertices on three given con- 
 current lines OD, 0£J, OF (Fig. 15). 
 
 Take any point (as ^) on any one of the given lines and from 
 it draw lines to any two of the given points (as UA, EB) meeting 
 
 the other lines in a and h. Let the lines AB and ah meet in M, 
 Through M draw a line MG passing through the remaining point 
 (C) and meeting the lines Oa and Oh in P and Q. PQ will be one 
 side of the required triangle which can be completed by drawing 
 the lines PA^ QB which will intersect in i? on the third given line. 
 There are generally six solutions as lines can be drawn through 
 each point terminated by either pair of lines. 
 
 Problem 16. To drav) a triangle with its vertices on three 
 given lines AP, BQ, CQP, and with its sides passing through three 
 given points A, B, C one on each line (fig. 16). 
 
 Let two of the given lines (as AP, BQ) meet in ; the third line 
 
INTRODUCTORY. 21 
 
 meets the others in P and Q. Draw the lines AQ and BP inter- 
 secting in I>, and draw OD intersecting FQ in B. Take a mean 
 
 proportional FM between GF and FJE (Problem 5), and a mean 
 proportional QW between CQ and QE. With centres F and Q 
 and radii respectively equal to FM and ^^ describe arcs inter- 
 secting in K. Draw a line bisecting the angle FKQ, intersecting 
 FQ in Z. Z will be one of the vertices of the required triangle 
 which can be completed by drawing BZ intersecting ^P in X and 
 AZ intersecting ^^ in F. X and Y are the other vertices and 
 XY will pass through G. 
 
 Problem 17. To determine the locus'^ of the vertex of a triangle 
 on a given base AB and with sides BP, AF in a given ratio a : b. 
 (Fig. 17.) 
 
 On the given base AB describe any one triangle with sides 
 BP : AF :: a : h. 
 
 Bisect the angle AFB by FD meeting AB in D and di*aw PC 
 perpendicular to FD meeting AB in G. 
 
 On Z) (7 as diameter describe a circle, which will be the required 
 locus of the vertex. 
 
 * For definition of locus, see p. 29 post. 
 
22 
 
 SUM OR DIFFERENCE OF TWO RECTANGLES. 
 
 I'roof, Take any point Q on the circle, and draw QA^ QD, 
 QB, QU. Since PB bisects the angle AP£ 
 
 .'. BD : AD .: a '. h (Euc. vi. 3), 
 
 Fig.17. 
 
 and since DPC is a right angle and PD bisects the angle APB 
 .'. P{ADBG) is a harmonic pencil (p. 15), 
 
 .-. also Q{ADBC) is a harmonic pencil, and consequently since 
 DQC is a right angle, QD bisects the angle AQB, 
 
 .-. BQ : AQ :: BD : AD :: a : b. (Euc. vi. 3.) 
 
 Problem 18. To construct a rectangle equal in area to the sum 
 or difference of two given rectangles A BCD, DEFG (fig. 18). 
 
 Apply the smaller rectangle to the side of the larger as in the 
 figure. Complete the rectangle ABHE. Draw DH cutting FG 
 
 Fig. 18. 
 
 --- - ^ 
 
 — m 
 
 — C 
 
 in K. Through K draw LM parallel to AB and the rectangle 
 ABML will be equal in area to the sum of the two given rect- 
 angles. (Euc. I. 43.) 
 
 The dotted lines and the small letters in the fig. shew the con- 
 struction for the difference of two rectangles. 
 
INTRODUCTORY. 23 
 
 Problem 19. From a given point P in a given straight line 
 PM to draw lines making equal angles with PM and cutting a 
 second given line CM at equal distances OD, CE from a given 
 point C (fig. 19). 
 
 From P and C draw PJ^, CF perpendicular to CM. Make 
 the angle MPF equal to the angle MPN and let PF meet CF in 
 
 F. With centre F and radius FP describe a circle cutting CM in 
 D and E which will be the required points. 
 
 Proof. CD = CE since CF is perpendicular to DE. 
 
 The angle DFP is double the angle DEP. (Euc. iii. 20.) 
 
 Half the angle DFP together with the angle FPD = a. right 
 angle. 
 
 The angle DEP together with the angle EPN= a right angle. 
 .-. the angle i^PZ) = the angle ^PiV, 
 and .-. the angle MPD = i\iQ angle MPE. 
 
 The point C must evidently lie on the opposite side of M to N. 
 
 This is also a solution of the problem to construct a triangle, 
 given the vertex, the bisector of the vertical angle, and the difier- 
 ence of the segments of the base made by that bisector: for 
 DM-ME=WM. 
 
24 . EXAMPLES. 
 
 Examples on Chapter L 
 
 1. Draw a circle of radius 2*87. In it place a chord AB of 
 length 4-8, and draw BC making 60° with AB. If (7 is on the 
 circle shew that the side ^C of the triangle is approximately 4-96. 
 
 Shew that the geometrical mean between 3*76 and 2*43 is 
 3-02 approximately. 
 
 2. Inscribe a square in a given triangle ABC. 
 
 (Through A draw a parallel A J) to BC ; make AD equal in 
 length to the perpendicular from A to BC, and join i> to the end 
 of the base BC that will enable it to cut one of the sides AB 
 or AC in JE. E is one of the angular points of the required 
 square, the base of which will coincide in direction with BC.) 
 
 3. Bisect a given triangle ABC by a straight line drawn 
 through a given point DmAG. AD<DC. 
 
 (Bisect BC in E and through A draw AF parallel to DE 
 meeting BC in F. DF will be the required line.) 
 
 4. Given the middle points P, ^, R of the sides of a triangle, 
 construct the triangle. 
 
 (The side through P is parallel to QR, and so for the others. 
 TakeP^ = 2, <}i?= 1-8, i?P= 1-3.) 
 
 5. Construct a triangle having given the base AB, the verti- 
 cal angle C, and the difference of the sides AC, CB. 
 
 C 
 (Construct a triangle ABB having the angle ADB= 90** + — , 
 
 J9^ = the given difference and AB the given base. Produce AD 
 to C, and make the angle DBC = the angle BDC.) 
 
 6. Construct a triangle, being given the base AB the difference 
 of the base angles, and the difference of the sides AG, and BC. 
 
 (Make a triangle DBA, with angle DBA = J the given differ- 
 
EXAMPLES. 25 
 
 ence. £A = the given base and AD the given difference of sides : 
 produce AJ) to G and make the angle BBC = angle BDC.) 
 
 7. Construct a triangle, being given the base AB^ the vertical 
 
 angle G and the sum of the sides AG and BG, 
 
 G 
 (Make an angle ADB = - , make DA^AC + BG, and AB = the 
 
 rj 
 given base; make the angle DBG = ^ , and so that BG cuts AD in 
 
 G between A and D.) 
 
 8. Let ^^(7 be any triangle, GD a perpendicular from G on 
 -4^ and E a point on ^^ such that DE-DB. AE is the differ- 
 ence of the segments of base made by the perpendicular, then given 
 AE and any one of the following pairs of data, construct the 
 triangle. 
 
 a. Sum of sides (J(7 + BG) and difference of base angles. 
 
 (We are given in the triangle AGE, AG + GE, AE and vertical 
 angle AGE, i. e. base, vertical angle and sum of sides. The triangle 
 can therefore be constructed (last example) and from it the required 
 triangle ABG.) 
 
 p. Difference of sides and difference of base angles. 
 
 (Make an angle ADE containing 90" + ^ where a is given 
 
 difference. Make DA ==the given difference of sides, and ^jE' the 
 given difference of segments ; produce AD to G and make DEG 
 = EDG ; produce AE to B and make GB=GD = GE. ABG will 
 be the required triangle.) 
 
 y. Sum of sides and vertical angle. 
 
 (Construct a triangle AEG on the given difference of segments 
 ^^ as base, with AG + GE — ^Yen sum of sides and the given 
 vertical angle as difference of base angles (a above), produce AE 
 to B and make GB = GE). 
 
 8. Difference of sides and vertical angle. 
 
 (Make an angle ^^^=half the given angle, make ^i^'^the 
 given difference of sides, produce AF to G and make the angle 
 FEG = EFG, produce AE to B and make GB=GE = GF.) 
 
2d examples. 
 
 : 9. Given the lengths AD^ BE^ CF of the bisectors of the 
 sides of a triangle ABC, to construct the triangle. 
 
 (Construct a triangle FOG making FG = ^AJD, GO = ^BE, 
 FO = ^FG ; produce FO to C and make 0C = 2.0Fso that FC is 
 the given length ; produce GO both ways to B and F and make BG 
 = OF = GO so that BF is the given length. Join BC and draw 
 BFj CF, producing them to meet in A. ABC will be the 
 required triangle.) 
 
 10. Given the lengths AD, BF, CF of the perpendiculars on 
 the sides from the opposite angles of a triangle ABC, to construct 
 the triangle. 
 
 (Determine a length Mb such that Ci^ : BE :: AD : J/6,and 
 on it construct a triangle Mbc, making be = BE and 3Ig = AD. 
 From M drop a perpendicular on be and on it make Md=AD. 
 Through d draw BdC parallel to b, meeting Mb, Mc in B and C. 
 MBC will be the required triangle.) 
 
 11. Given three points D, E, F, to construct a triangle of 
 which these points shall be the feet of the perpendiculars on the 
 sides from the opposite angles. 
 
 (The sides are perpendicular to the bisectors of the angles of 
 the triangle DEF.) 
 
 12. Divide a given straight line AB into two parts AC, CB, 
 such that the difference of the squares on the parts may be equal 
 to the square on a given line DF < AB. 
 
 (Take a third proportional FG to AB and DE. FG will be 
 the difference between the required parts, and ^-6 is their sum, so 
 that AC, and CB are known.) 
 
 13. Divide a given line AB into two parts AC, CB, such that 
 the square on AC may be double the square on CB. 
 
 (TBke AC : CB :: ^/2 : 1.) 
 
 14. Divide a given straight line AB into two parts AC, CB, 
 such that the sum of their squares shall be equal to the square 
 on a given line DE. 
 
EXAMPLES. 27 
 
 4 7? 
 
 (Construct a rectangle equal in area to 2DEy - AB^ (Prob. 
 18). Take a mean proportional between its sides which will be 
 the difference between AG and GB; the sum and difference of the 
 parts being known, the parts are known.) 
 
 15. Divide a given straight line AB into two parts AG^ GB, 
 such that AB' + GB' = 2.AG\ 
 
 (Take JC: GB :: 1 + VS : 1.) 
 
 16. Draw any triangle ABG, bisect AB in I), join GB, and 
 through G draw GE parallel to ^^; shew by drawing a trans- 
 versal, that the rays GA, GB, GB, GB form a harmonic pencil. 
 
 17. Given the directions of one pair of opposite sides of a 
 quadrilateral AB and GB, and the point (F) of intersection of 
 the other pair, shew that the locus of the intersection of the 
 diagonals is a straight line. 
 
 (If AB, GB intersect in F, and G is the intersection of the 
 diagonals, the pencil B (AGGF) is harmonic.) 
 
 18. Find the geometric mean (BB) between two given lines 
 (AB and BG) and shew by construction that the harmonic mean 
 between AB + BB, and BG + BB is 2BB. 
 
 19. A line AB is divided harmonically in G and B, and a 
 part GB of the line which contains two terms GB and BB is 
 bisected in F. Shev.^ that FG is the geometric mean of FA 
 and FB. 
 
 20. Divide a given straight line AB medially in the point G, 
 and produce the line so that the part produced is equal to AG 
 the smaller segment; shew by construction that the rectangle 
 contained hj AG and the whole line thus produced, together with 
 the square on ^^S is equal to four times the square on GB. 
 
CHAPTER II. 
 
 THE CIRCLE. 
 
 Euclid's well known definition is " A circle is a plane figure 
 contained by one line, which is called the circumference, and is 
 such, that all straight lines drawn from a certain point within the 
 figure to the circumference are equal to one another : and this 
 point is called the centre of the circle". A radius of a circle is a 
 straight line drawn from the centre to the circumference, and 
 therefore by the above definition all radii of a circle are equal. 
 
 Hence a circle is completely determined if we know its centre 
 and the length of its radius, and it might seem at first sight that 
 two geometrical conditions would be sufficient to determine it. 
 The position of the centre however must be counted as two con- 
 ditions, and a circle can generally be drawn to satisfy three 
 geometrical conditions, and three are in general necessary and suffi- 
 cient for its determination. Thus an infinite number of circles 
 can be drawn to pass through two points, or to touch two lines, 
 and some other condition, such as the position of a third point 
 through which it must pass or of a line which it must touch in the 
 first case, or of a third line which it must touch or of a point 
 through which it must pass in the second, or such as the length of 
 the radius in either, must be given to make the exact solution of 
 the problem possible. 
 
 The above limitation " in general" is necessary because it is 
 possible to give certain special positions to the lines and points 
 which would render the problem impossible : thus e. g. in the first 
 case a circle cannot be drawn through three points in the same 
 straight line, or at least no circle of finite radius, or if the given 
 conditions are "to pass through two given points and touch a 
 
THE CIKCLE. 29 
 
 given line" the line must obviously lie outside the points, i.e. it 
 must not pass between them, and similarly if the conditions are 
 "to touch three given lines" one at least of the lines must not be 
 parallel to the other two, but notwithstanding these special cases 
 it is generally true that a circle can be drawn to satisfy any three 
 geometrical conditions. 
 
 Definition. When a point is restricted by conditions of any 
 kind, to occupy any of a particular series of positions, that series 
 of positions is called the locus of the point. 
 
 Problem 20. (Fig. 20.) To describe a circle through three 
 given points A, B, C, not in the same straight line. 
 
 If the line joining ^, ^ is bisected in D and DO is drawn per- 
 pendicular to AB, DO will obviously be the locus of the centres of 
 
 all circles passing through A and jB, i. e. any circle through A and B 
 must have its centre on DO, since in the equal right-angled tri- 
 angles ADO, BDO, AO is equal to BO. Similarly, bisecting BC 
 in E and drawing EO perpendicular to BC, EO is the locus of 
 centres of circles passing through B and C. Hence the centre of 
 the circle passing through A, B and G must lie simultaneously on 
 both these loci, i. e. must be at their intersection, and the distance 
 from this point to either A, B or G will be the radius of the 
 required circle. 
 
 Euclid in definition 2 of Book iii. defines a tangent to a circle 
 in these words. " A straight line is said to touch a circle when it 
 
30 TANGENTS TO A GIVEN CIRCLE. 
 
 meets the circle, and being produced does not cut it," and shews 
 in Corollary to Prop. 16, Bk. in. that the line drawn perpendicu- 
 lar to a radius at its extremity fulfils the condition of this 
 definition. This is the most convenient way in which to draw 
 the tangent at any point on the circumference, and the tangent so 
 drawn can easily be shewn to agree with the general definition of 
 a tangent usually given as applicable to all curves, which is as 
 follows : — 
 
 Definition. If two points be taken on a curve and a chord 
 drawn through them ; then, if the first point remains fixed while 
 the second, moving along the curve, approaches indefinitely near 
 to the first, the chord in its limiting position is called the tangent 
 to the curve at the first point. 
 
 To shew that such chord in its limiting position will in the 
 ■circle be perpendicular to the radius at the point, take two points 
 P, Pj, (fig. 20) on the curve, and draw PP^, then since OP=OP^ 
 the angles OPP^ and OP^P are equal and will remain equal how- 
 ever close Pj may be taken to P. But when P^ coincides with P 
 €ach of these angles becomes a right angle, i. e. the tangent at P 
 will be perpendicular to OP. 
 
 To draw a tangent to the given circle from an external point 
 Q. Join OQ and on it as diameter describe a circle cutting the 
 given circle in M and M^. (It will necessarily do so in two points 
 on opposite sides of its diameter.) 
 
 Then QM^ QM^ will be tangents to the circle since QMO is a 
 right angle being in a semicircle. (Euclid, Prop. 31, Bk. in.) 
 It is always possible to draw two tangents to a circle from any 
 external point. 
 
 Pole and Polar. (Fig. 20.) 
 The line MM^ is evidently perpendicular to OQ for the tri- 
 angles QOM, QOM^ are equal in all respects, i.e. the angle MOQ 
 = the angle MfiQ; then if MM^ meets OQ in J^ we have in the 
 two triangles NOM, NOM^^ 0M= OM^, ON common and the 
 angle, NOM = t}iQ angle NOM^^ .-. the angle OiVi/=the angle 
 ONM^^ and .•. each is a right angle. 
 
THE CIRCLE. 31 
 
 The triangle MON is . •. similar to the triangle QOM and 
 ,',0N '. OM :: OM ; OQ, 
 or ON. OQ = r\ 
 
 where r is the radius of the circle. 
 
 Now whether the point Q be taken inside or outside the circle, 
 it is always possible to find on the line 0^ a point N fulfilling the 
 above condition, and a line MNM^ drawn perpendicular to OQ 
 through the point N so determined is called the 'polar of Q with 
 respect to the circle, while the point Q is called the -pole of MK^ 
 with respect to the circle. 
 
 To draw \h% polar of any point Q with respect to a given circle. 
 
 If the given point be without the circle the polar is, by the 
 previous definition, the chord of contact of the tangents drawn from 
 Q to the given circle. If the given point be within the circle, 
 draw OQ and produce it, and through Q draw MQM^ perpendicu- 
 lar to OQ, and meeting the circle in M and M^ and at either if or 
 M^ draw MN or MxN a tangent to the circle, meeting OQ pro- 
 duced in N, then a line through iV perpendicular to OQ will be the 
 required polar. 
 
 Cor. 1. If the given point be on the circle its polar is the 
 tangent at the point, i. e. the polar passes through the pole. 
 
 Cor. 2. If a point A lie on the polar of Q then Q lies on the 
 polar of A. For draw OA and on it drop a perpendicular Qq from 
 Q meeting it in q and the circle in m and w^ : then the triangles 
 OQq, OAN are similar and 
 
 .-. Oq : OQ :: ON : OA, i.e. Oq.OA^OQ. 0N=7^, 
 
 by definition, r being the radius of the circle, i.e. mQm^ is the 
 polar of A which consequently passes through Q. 
 
 Cor. 3. The pairs of tangents drawn at the extremities of any 
 chord through Q intersect in the straight line AB the polar of Q. 
 Hence the polar may be defined as the locus of the points of inter- 
 section of tangents at the extremities of chords through a fixed 
 point. 
 
32 
 
 SELF-CONJUGATE TRIANGLE. 
 
 Given a circle and a triangle ABC, if we take the polar s with 
 respect to the circle, of A, B, C, we form a new triangle A'B'C 
 called the conjugate triangle, A' being the pole of BG^ B' of GAy 
 and G' of AB. In the particular case where the polars of -4, j5, G 
 respectively are BG , GA, AB, the second triangle coincides with 
 the first, and the triangle is called a self- conjugate triangle. 
 
 Problem 21. (Fig. 21.) To describe a circle to pass through two 
 given points and touch a given straight line, lying outside the points. 
 
 Let A and B be the given points and DD^ the given straight 
 line. It will be observed that the point of contact of the line is 
 
 Fig.2l. 
 
 not given — this would be a fourth geometrical condition and 
 therefore if a circle is required to touch a given line at a given 
 point, it can only in general fulfil one other condition as e.g. pass 
 through on^ point outside the line. See next problem. 
 
 Join AB and produce it to cut DD^ in G and indefinitely be- 
 yond as to a. It is a known proposition (Euclid 36, Book iii.), 
 that "if from any point without a circle two straight lines be 
 drawn, one of which cuts the circle, and the other touches it ; 
 the rectangle contained by the whole line which cuts the circle 
 and the part of it without the circle, shall be equal to the square 
 on the line which touches it." 
 
 If therefore a mean proportional cd be taken between GA and 
 GB, and its length be set off from G along I>£>^ as GD, then 
 obviously a perpendicular to BD^ through D will be the locus of 
 the centres of circles touching the given line in D. It AB be 
 
THE CIRCLE. 
 
 33 
 
 bisected in E, and EO be drawn perpendicular to AB, EO will 
 be the locus of centres of circles through A and B. The required 
 centre will therefore be at 0, the intersection of these loci and 
 the distance to ^, ^, or i> will be the required radius. Since 
 the length CD may be set off on either side of C there are 
 obviously two solutions as shewn. 
 
 If the line joining A and B be parallel to the given line, this 
 solution fails, but the point of contact can be at once determined, 
 since by symmetry it is obviously where EO cuts the given line, 
 and a third point through which the circle must pass being thus 
 obtained the solution can be completed by Problem 20. 
 
 Problem ^2. (Fig. 21.) To describe a circle to pass through a 
 given point A mid to touch a given straight line BD^ in a given 
 point D. 
 
 The straight line 1)0 through I) perpendicular to J)D^ is 
 obviously the locus of centres of all circles touching the straight 
 line in E, and the straight line FO through F the centre point 
 of AD, perpendicular to AD is the locus of centres of all circles 
 through A and D. The centre of required circle is therefore at 
 0, the intersection of these loci, and the distance from to A 
 or D will be the required radius. 
 
 Problem 23. (Fig. 22.) To describe a circle to touch two straight 
 lines AB, CD, one of them in a given point A. 
 
 E. 
 
M CIRCLE TO TOUCH TWO LINES AND PASS THROUGH A POINT. 
 
 A locus of the centre is of course the line AO perpendicular 
 to AB. A second locus will obviously be the line BO bisecting 
 the angle ABC, and the required centre will therefore be at 0. 
 If at a perpendicular OC be drawn to BC, the triangles OB A 
 and OBC are equal in all respects and therefore OA =^ OC = the 
 required radius. The given lines make with each other the angle 
 ABB as well as the angle ABC, and therefore bisecting the 
 angle ABB, the centre 0^ of a second circle is obtained touching 
 the other side of AB. 
 
 Problem 24. (Fig. 23.) To describe a circle to j^ctss through a 
 given point G and to touch tioo given lines AC, BF. 
 
 The centre must obviously lie on the line KH bisecting the 
 angle between the given lines in which the given point lieh. 
 
 Fig.23. 
 
 (See Problem 1.) Draw also the line GL passing through G and 
 the intersection of the given lines (Problem 6). Take any point 
 K on HK as centre, and describe a circle touching AC and BF, arid 
 cutting GL in M and L. This is always possible, since KH is the 
 bisector of the angle between these lines. Draw GO parallel tt> 
 MK and GH parallel to KL, and // will be centres of circles 
 fulfilling the required conditions : for if A is the point of contact 
 of the trial circle, KA will be perpendicular to AC, and if OB, 
 HC be drawn perpendicular to AC, KA, OB and HC will all be 
 parallel, and therefore the triangle GOB will be similar to MKA 
 and GlIC to LKA, but the triangles MKA, LKA are isosceles, 
 therefore also OG must be equal to OB and UG to HC, ^ 
 
THE CIRCLE. 
 
 oo 
 
 Problem 25. (Fig. 24.) To describe a circle to touch three 
 given lines AB, BG, CA, not more than two of vnhich are parallel. 
 
 The line AO bisecting the angle BAG will be the locus of 
 centres of circles touching BA and GA^ the line GO bisecting the 
 
 Fig. 24. 
 
 angle BGA will be the corresponding locus for BG and GA. 
 Hence will be a centre for a circle touching all three lines. 
 Since BA makes with BG and GA not only the angles ABG, BAG 
 respectively, but also the angles ABE, BAD, a second solution is 
 obviously obtained by bisecting the exterior angle BAD as shewn 
 by AO^, and similarly for the remaining sides. Hence four circles 
 can be drawn touching three straight lines. The exterior circles 
 are said to be escribed to the triangle ABO. 
 
 Problem 26. (Fig. 25.) To describe a circle to touch a given 
 circle {centre (7, radius GD) and a given straight line AB in a given 
 point A. 
 
 The line AO drawn through A perpendicular to AB i^ a locus 
 of the required centre. Draw a diameter DD^ of the circle 
 parallel to AO. Join AD cutting the given circle in E, and join 
 
 8—2 
 
36 CIRCLE TO PASS THROUGH TWO POINTS AND TOUCH A CIRCLE. 
 
 CE producing it to cut AO in 0, will be the centre of a circle 
 fulfilling the given condition. A second solution is possible, since 
 
 Fig.25. 
 
 A may be joined to either extremity of DD^. 0, is the centre of 
 a second circle. 
 
 Proof. The angle 0^^= angle CDE, (Euc. i. 29.) 
 
 „ CED^ „ CJDE, (Euc. I. 5.) 
 
 CEB = „ OEA, (Euc. I. 15.) 
 
 .-. „ OAE= „ OEA, 
 
 and .-. OE^OA. (Euc. i. 6.) 
 
 Hence a circle through A from centre will pass through E and 
 
 will there touch the given circle, since they will have a common 
 
 tangent perpendicular to CO. 
 
 Problem 27. (Eig. 26.) To describe a circle to touch a given 
 circle (centre A, radius AD) and pass through two given points B, 
 C, which must be either both inside, or both outside the circle. 
 
 Draw a line through EC; bisect EC in E and draw EO per- 
 pendicular to EC. EO is the locus of centres of circles through 
 B and C. Take any point such that a circle described witli 
 centre and radius OB, or OC will cut the given circle as in 
 MN'. Draw a line through JfiV cutting BC in T, and from 2' 
 draw tangents TD, TD^ to the given circle (Prob. 20). Lines 
 joining AD, AD^ will cut EO in points 0,, 0^ which will be the 
 centres of circles fulfilling the required conditions. 
 
THE CIRCLE. 
 
 Two circles can generally be drawn. If the line joining the 
 given points lie wholly without the given circle, one circle will 
 
 ng.26. 
 
 o„^-. 
 
 touch the given circle externally and one internally (as in the fig.); 
 if the line joining the points cut the given circle, and both points 
 lie on the same side of the circle, both circles will touch the 
 given circle externally, and if the points lie on opposite sides of 
 the circle both will touch it internally. If the line joining the 
 given points touch the given circle one circle only can be drawn. 
 
 Proof. The rectangle TM . TN ^x^Q,t. TB , TC (Euc. in. 36, 
 
 Cor.), 
 „ „ = sq. on TD. (Euc. iii. 36). 
 
 .-. sq. on TD = rect. TB . TC 
 .'. TB is a tangent to the circle going through B, D, C. 
 
 Problem 28. (Fig. 27.) 2b describe a circle to touch a given 
 circle (centre A, radius AM) and two given straight lines BG, DE. 
 
 There are several solutions depending on the relative positions 
 of the lines and circle. If the lines are parallel the problem is 
 impossible unless some part of the circle lies between the lines. 
 In this case the line drawn midway between the lines parallel to 
 either of them is evidently a locus of the required centre ; a 
 second locus will be the circle described with centre A and radius 
 equal to the sum oi AR and half the distance between the lines, 
 and since these loci intersect in two points, either may be taken as 
 
38 CIRCLE TO TOUCH A CIRCLE AND TWO LINES. 
 
 the centre of the required circle. If the given lines are not parallel 
 and the given circle cuts one of them, as in the fig., then by 
 
 Fig.27. 
 
 
 drawing lines parallel to the given lines at a distance from them 
 equal to the radius of the given circle, the problem may be reduced 
 to describing a circle to touch these lines (in pairs) and to pass 
 through the centre of the given circle, i.e. may be reduced to 
 Problem 24. Let the given lines intersect in F and consider first 
 the circles which can be drawn in the angle EFC. Draw (?//, LK 
 parallel to FE at a distance from it equal to ^^and similarly HK 
 and GL parallel to FC. FK will bisect the angle OFF and will 
 be the locus of the required centres. Take any point on it as 
 centre and describe a circle to touch GH and GL cutting GA in J/, 
 J/i. Then AO^ drawn parallel to Mfi to cut FK in 0^ determines 
 Oj a required centre and AO^ parallel to MO determines 0^ a 
 second required centre. Similarly for the circles lying in the 
 angle DFG. Any point 0^ on FL being taken as centre and a 
 circle described to touch GH and HK cutting HA in N and N^ ; 
 AO^ parallel to J\^fi4 determines O3, the centre of a third circle 
 fulfilling the required conditions and a line through A parallel to 
 iVC?4 would determine a fourth centre. It is of course accidental 
 that in the figure 0^ falls nearly on GA. 
 
 If the given circle did not cut either of the given lines, it 
 
THE CIRCLE. 
 
 89 
 
 would still be possible to draw four circles toucliing the lines and 
 the circle, but two of them would have internal contact with the 
 given circle, instead of all touching it externally as in the figure. 
 
 If the given circle cut both lines there would be six possible 
 solutions. 
 
 Problem 29. (Fig. 28.) To describe a circle to touch a given 
 circle {centre A, radius AF) to touch a given line BC and to pass 
 through a given point D. 
 
 If the given point be within the circle, the given line must 
 not be wholly outside the circle. 
 
 From A draw AC perpendicular to the given line and meeting 
 the circle in E and F. First join ED and on it determine a point 
 
 Fig. 28 
 
 G such that the rect. ED .EG=- rect. EC . EF, i. e. take EG a fourth 
 proportional to ED, EG, EF. [Making Ef (on ED) = EF, draw 
 fg parallel to DC meeting EG in g and make EG = Eg.'\ Then a 
 circle through D and G and touching the given line will also touch 
 the given circle and the problem is reduced to Problem 21. If 
 ED, EC intersect in T, a mean proportional (TE) must be taken 
 between TG and TD so determining the point of contact. 2'B 
 may be set off along EG on either side of T and hence there are 
 two solutions giving external contact. Second. — Join FD and on 
 it determine a point G^ such that rect. FD . FG^ = rect. on FG, FE. 
 i. e. take a fourth proportional to FD, FG, FE. G^ must be taken 
 on the opposite side of F to D because C and E are on opposite 
 sides of F. Then circles through D and G^ touching the given line 
 
40 SEGMENT OF A CIRCLE TO CONTAIN A GIVEN ANGLE. 
 
 will also touch the given circle, and this case also reduces to 
 Problem 21. There are again two possible circles because if 
 JJF and BC intersect in T^ the mean , proportional (T^H) between 
 T^D and Tfi^ may be set off on either side of T^. 
 
 Proof. Join B the point of contact of circle tlirough D and G 
 to E meeting the given circle in K and join FK. Then the tri- 
 angles EKF and ECB are similar 
 
 .-. EC : EB :: EK : EF, 
 or rect. EO . EF= rect. EB . EK, 
 
 but „ = rect. ^/>.^(? (const.). 
 
 .♦. rect. ED .EG = rect. EB . EK. 
 .'. K must be on circumference of circle through BDG. (Euc. 
 III. 36 Cor.) 
 
 Join OK, then angle 0^7t = angle 0KB (Euc. i. 5), 
 „ AKE= „ AEK{ „ ), 
 „ AEK= „ 07rj5 (Euc. I. 29), 
 .-. „ AKE= „ OBK, 
 and therefore OKA is a straight line, i.e. the two circles will 
 touch at K. 
 
 Problem 30. On a given straight line AB to describe a segment 
 of a circle which shall contain a given angle (Fig. 29). 
 
 Bisect AB in C and through C draw CO perpendicular to AB. 
 {CO is of course a locus of the centre.) Make the angle OCD 
 
 Flg.29. 
 
 equal to the given angle (p. 4) and through A draw AO parallel 
 to CD meeting CO in 0. will be the centre of the required 
 circle. (Euc. iii. 20.) 
 
THE CIRCLE. 
 
 41 
 
 Pkoblem 31. (Fig. 30.) To draw a line toucJdng two given 
 circles^ neither of which lies wholly inside the other. 
 
 A and AB are the centre and radius of the larger circle and C 
 and CD those of the smaller circle. 
 
 Join AC, cutting larger circle in B. 
 
 From B on AC make BM^BN=CD, and with A as centre 
 describe circle MM^M^. From C draw tangents CM^^ CM^ to 
 
 ss,^:- 
 
 .,Sc 
 
 iw- 
 
 touch this circle (Prob. 20). Produce A3f^, ^^2 *^ n^^^t the circle 
 in E and G, and lines ED, GF through E and G parallel to CM^ , 
 CJ/g will be taiigents to both circles. These tangents meet in (Oj) 
 a point lying on ^C produced, and are the only pair that can be 
 drawn if the given circles intersect. If the smaller circle lies 
 wholly outside the larger, as in fig. 30, a second pair can be drawn 
 by describing a circle through N with A as centre, drawing tangents 
 CiYj, CN^ to it, and drawing HJ, KL parallel to these lines re- 
 
42 SYSTEM OF TWO OR MORE CIRCLES. 
 
 spectively, which will intersect in (0) a point on AC between the 
 given circles. The construction is obvious, since UM^ = JBM= CD. 
 Common tangents to two circles may be drawn practically with 
 all attainable accuracy by adjusting a set square to touch the 
 circles, and drawing a line by its edge ; but the points of contact 
 should always be determined by drawing the radii perpendicular 
 to the tangent. 
 
 Properties of a system of Two or more circles. 
 
 The points 0, 0^ in which common tangents to two circles 
 intersect are called the centres of similitude of the two circles. 
 As is easily seen, they are the points where the line joining the 
 centres is cut externally and internally in the ratio of the radii : 
 and in this sense both exist when the circles cut each other, in 
 which case of course only one pair of common tangents can be 
 drawn, and even when one circle lies wholly inside the other, so 
 that it is impossible to draw any common tangent. 
 
 If through a centre of similitude we draw any two lines meeting 
 the first circle in the points R, It^, S, S^, and the second in the 
 points p, /3j , (T, (T^f then the chords R!S, pa- will be parallel, as also 
 the chords R^S^ and p^<r^ ; and the chords RS and pjO-j, R^S^ and 
 pa- will intersect respectively in points P and Q on a line perpen- 
 dicular to the line joining the centres of the circles. 
 
 This line is called the radical axis of the two circles. 
 
 The rectangle OR . OR^ is constant, since it equals the square 
 on OH the tangent from (Euc. iii. 36), i. e. 
 
 0R.0R^ = 0S.0S^ 
 and Op . Op^ = Oa . Oa-^ . 
 
 Proof. In the triangles OAR^ 00 p^ the angle AOR -the 
 
 angle COp and OA : 00 v. AR : Op, 
 
 .'. also OR : Op :: AR : Op (Euc. vi. 7), 
 
 O 7? 
 i.e. the ratio yr- is constant and equal to the ratio of the radii 
 Op 
 
 of the circles wherever the line ORp be drawn, 
 .-. OR : OS :: Op : Oa- 
 
THE CIRCLE. 43 
 
 and the angle BOS = the angle pOa ; 
 
 .'. the triangles BOS, pOcr are similar in all respects, so that the 
 
 angle ORS = the angle Opa- and po- is parallel to RS. 
 
 Similarly R^S^ is parallel to p^a-^ , which proves the first part of 
 the proposition^. 
 
 Again, since SRR^S^ is inscribed in a circle, the angle P^O = the 
 angle SS^R^ - the angle Sa-^P. The triangles RRpi and Fa-^S are 
 therefore similar, since the angle RRp^ is common to both. 
 
 .-. FR : Pp, :: Po-^ : PS, 
 i.e. PR.PS = Pp^.Pa;, 
 
 but PR . PS = square of tangent from P to circle A, 
 
 and Pp,.Pa;= „ „ „ C; 
 
 .-. the tangents from P to the two circles are equal, and 
 
 .-. ^'-A£\' = PC\'-aDl'; 
 similarly tangents from Q to the two circles are equal. 
 
 But the locus of the intersection of equal tangents to two 
 circles is a straight line perpendicular to the line joining their 
 centres, and dividing the distance between them so that the differ- 
 ence of the squares of the parts is equal to the difference of the 
 squares of the radii : for if X be such a point and PX perpendicular 
 to AC, at every point on it we shall have 
 
 PA' - AX' = PX' ^PC^'- CX\ 
 .'. Pa\'-JC^ = AX'--CX'; 
 and as above PA' - PC = AB' - CD' -^ AX' - CX'. 
 
 Hence the line PQ in the figure must be such locus which proves 
 the second part of the proposition. 
 
 Definition. A line drawn perpendicular to AC, the line 
 joining the centres of two given circles, through a point X on it, 
 such that the difference of the squares oi AX and CX is equal to 
 the difference ofvthe squares of the radii of the two circles is called 
 the radical axis of the two circles. 
 
 As already shewn, it is the locus of the intersection of equal 
 tangents to the two circles. 
 
44 
 
 EADICAL AXIS AND AXES OF SIMILITUDE. 
 
 It may be constructed as in the last proposition or immediately 
 from the definition Ly bisecting AC in. a (fig. 30), and making aX 
 towards C, the centre of the smaller circle, a fourth proportional 
 to 2AC, AB + CD, and AB- CD, 
 
 i. e. by making aX : li-r : : B + r : 2 AC, 
 where B and r are the radii of the circle, and drawing a line 
 through X perpendicular to AC j for in this case 
 
 AC X 2aX= B' - r' hnt AC ^ AX+ XC and 2aX = AX- CX 
 .'. (AX + CX) {AX- CX) = B'-r' = AX' - CX\ 
 
 The radical axis bisects the distance between the polars with 
 respect to the two circles, of either centre of similitude, which fur- 
 nishes another method of constructing it. 
 
 Given three circles (centres C, (7,, C^, radii r, r^, r^] the line 
 joining a centre of similitude of C and C^ to a centre of simili- 
 tude of C and C^ will pass through a centre of similitude of C^ 
 and C^. Let S^ and S^ (fig. 31) be the centres of similitude of 
 
 Fig.31. 
 
 C and Cj, and S^ a centre of similitude of C and C^, and let S^S^, 
 Cfi^ meet in S, S will be a centre of similitude of C^ and C^. 
 
 For since CSJ : C,SJ :: r : r^ :: CS^ : C,S^, 
 .'. CSJ : CS^ :: C^S^ : C^S^, 
 or CS^C^S^ is a harmonic range ; therefore aS' (CSJC^S^) is a har- 
 monic pencil, and therefore if CC^ cuts SS^' in S^\ 
 
 CS^'C^S^ is a harmonic range, and since S^ is a centre of 
 similitude of C and (7,, S^ must be the other. 
 
THE CIRCLE. 45 
 
 Tlirougli C„ draw C^L parallel to CS^ and meeting aS'^S'^ in L, 
 Then by similar triangles 
 
 C,S : C^S :: C,L : C\S„ 
 
 and C„L 
 
 CA ■■■■ 
 
 CS, ■■ 
 
 CS„ or 
 
 C,L-. 
 
 CA ■ ('''% 
 
 cs. 
 
 C\S : 
 
 C,S :: 
 
 ,CA,-' 
 
 • CS 
 
 OS, _ 
 
 c,s,, 
 
 CA 
 
 CA 
 
 CS, 
 
 cs, 
 'OA 
 
 
 
 or aS' is a centre of similitude of C, and C^ . 
 
 Since for each pair of circles there are two centres of similitude, 
 there will be in all six for the three circles, and these will be dis- 
 tributed along/oi^r axes of similitude, as represented in the figure. 
 
 Corollary. If a circle (centre A) touch two others (centres C 
 and Cj) the line joining the points of contact will pass through 
 a centre of similitude of G and C^. For when two circles touch, 
 one of their centres of similitude will coincide with the point of 
 contact. If A touch G and C^ either both externally or both 
 internally, the line joining the points of contact will pass through 
 the external centre of similitude of C and G^. If J. touch one 
 externally and the other internally, the line joining the points of 
 contact will pass through the internal centre of similitude*. 
 
 Given any three circles, if we take the radical axis of each pair 
 of circles, these three lines will meet in a point, which is called 
 the radical centre of the three circles. 
 
 For let the radical axes of A and C and of B and C intersect 
 in R (fig. 34), then the tangents from R to A and G are equal, as 
 also the tangents from R to B and C; therefore the tangent from 
 R to A must be equal to the tangent from R to B, i.e. R must be 
 a point on the radical of A and B, which proves the proposition. 
 
 If two circles have a common radical axis, and points L and Z, 
 be taken on the line joining their centres at a distance from its 
 * Salmon's Conic Sections. 
 
46 LIMITING POINTS OF A SYSTEM OF CIRCLES. 
 
 intersection (X) with the radical axis equal to the tangent which 
 can be drawn from A'' to either circle, these points are called the 
 limiting points of the entire system of circles which have the 
 same (common) radical axis. They "have many remarkable 
 properties in the theory of these circles, and are such that the 
 polar of either of them, with regard to any of -the circles, is a line 
 drawn through the other perpendicular to the line of centres. 
 These points are real when the circles of the system have common 
 two imaginary points, and are imaginary when they have real 
 points common"^." 
 
 When they are real it is evidently impossible for the centre of 
 any circle of the system to lie between them, and the more nearly 
 the centre approaches to either of them the smaller must the cor- 
 responding radius be. The limiting points themselves may there- 
 fore be considered as circles of the system of infinitely small radius. 
 
 If a system of circles have a common radical axis, and from any 
 point on it tangents be drawn to all the circles, the locus of the 
 points of contact must be a circle, since all these tangents are 
 equal ; and it is evident that this circle cuts any of the given 
 system at right angles, since its radii are tangents to the given 
 system. It is the circle passing through the limiting points of the 
 system. 
 
 Com^ersely all circles which cut the given system at right 
 angles pass through the limiting points of the system. 
 
 Problem 32. (Fig. 32.) To describe a circle to touch two given 
 circles {centres A and B, radii AD, BE respectively) and to pass 
 through a given point C. 
 
 Take S a centre of similitude (p. 42) of ^ and B ; draw CS 
 and find the poles P and P^ of this line with respect to each circle, 
 (i.e. draw AP, BP^ perpendicular to CaS' and intersecting the 
 chords of contact of tangents from aS' in P and P^. Draw Xll 
 the radical axis of the given circles (p. 44) : draw AC, bisect it 
 in m and make mM on it towards C of length such that 
 AC : AD :: AD : ImM ; 
 * Salmon's Conic Sections. 
 
THE cmcLE. 47 
 
 draw MR perpendicular to AC meeting the radical axis in It 
 The lines UP, RP^ will cut the circles in the points of contact 
 a h- a b of the required circles and their centres can be at once 
 found by producing Aa^ Ba^ &c. to meet in and 0^. 
 
 In the figure the circles touch one of the given circles in- 
 ternally and one externally because S is the internal centre of 
 similitude. If the external one be taken two more circles can be 
 drawn, one touching both externally, the other both internally. 
 
 Problem 33. To describe a circle to touch two given circles 
 {centres A and B, radii AC, BD respectively) and a given straight 
 line EF (fig. 33). 
 
 Draw the radical axis of the given circles, meeting EF in R, p. 44. 
 From A and B drop perpendiculars on the given line meeting it 
 in E and F and the circles in C, C^, D and D^ respectively. 
 Join CjZ^j cutting AB in S a centre of similitude of A and B. 
 Find P and P^ the poles of this line with respect to the circles 
 (p. 31). Draw RP, RP^ cutting the circles in ab, afi^. Then 
 aa^, bb^ are the points of contact of circles fulfilling the required 
 conditions, and the intersections of Aa, Ba^ and of Ab^ Bb^ givt- 
 the corresponding centres. The above circles each touch both 
 of the given circles externally or both internally since 8 is the 
 external centre of similitude of .4 and B (p. 45). If C, be joined 
 
48 
 
 CIRCLE TO TOUCH TWO CIRCLES AND A LINE. 
 
 to Z> or C to D^ cutting AB in the internal centre of similitude, 
 the poles of these lines give the points of contact of circles touch- 
 ing one of the given circles internally and the other externally — 
 
 and if C be joined to J) the poles of this line give another pair of 
 circles touching both externally or both internally. One of these 
 latter is shewn in the fig. There are altogether 8 solutions. 
 
 Second Solution. This problem may also be solved by 
 dropping perpendiculars from A and B on the given line as AE, 
 BF^ bisecting the parts lying between the circles and the 
 lines as CE^ DF, in G and H and describing parabolas having 
 A and B as foci and G and H as vertices respectively (Prob. 36). 
 The first will necessarily be the locus of the centres of circles 
 touching the line and the circle A externally, and the second will 
 be the locus of the centres of circles touching the given line and 
 the circle B externally, and hence their intersection (0) will de- 
 termine the centre of a circle touching both circles externally 
 and the given line. Similarly if C^E be bisected in G^ and DJ^^ 
 in //, and parabolas be described having A and B as foci and 
 G^, ZT, as vertices respectively, each of these curves will be the 
 
THE CIRCLE. 
 
 49 
 
 locus of centres of circles touching the line and the corresponding 
 given circle internally. Hence the points of intersection of 
 these four parabolas determine the centres of circles fulfilling 
 the conditions of the problem. 
 
 Oj gives internal contact with both circles, 
 0^ gives internal with A external with B, 
 
 O3 gives external or internal „ „ 
 
 and so on. 
 
 The proof of the construction is obvious from the definition of 
 a parabola subsequently given. 
 
 Problem 34. To describe a circle to touch three given circles 
 {centres ABC, radii AD^ BE, CG respectively) (Fig. 34). 
 
 If the circle be required to touch the three either all externally 
 or all internally draw the external axis of similitude SS^ p. 45, 
 
 +— ^. 
 
50 CIRCLE TOUCHING THREE CIRCLES. 
 
 and take the poles PPJ^^ of this line with respect to each circle, 
 p. 31. 
 
 Find the radical centre R of the three circles (p. 45). Then 
 the lines RP, RP^j RPq cut the circles in the points ab, a^b^, ajb^j 
 in which the required circles must touch them : and the centre of 
 the circle touching all three externally is given by the intersection 
 of Aa, Ba^, Ca^, which three lines will meet in a point, and the 
 centre of the circle touching all three internally is given by the 
 intersection of Ab, £b^, Cb^. 
 
 A similar construction with the remaining three axes of 
 similitude, will determine the circles touching one internally and 
 the remaining two externally and vice versa. 
 
 There are altogether eight solutions. 
 
 Second solution. Join AP cutting the circles in B, D^, E and 
 
 jE',. Bisect DE in K and 7>,^, in if . BK will necessarily be 
 
 1 111 J 
 
 equal to AK^. With B and A as foci, and K, K^ as vertices de- 
 scribe an hyperbola (Prob. 89), the branch of which through K 
 will be the locus of the centres of circles touching circles A and B 
 externally, and the branch of which through K^ will be the locus 
 of centres of circles touching these circles internally. Similarly, 
 join BC cutting the corresponding circles in F^, F, G, G^. Bisect 
 FG in X, and Ffi^ in L^ and with C and B as foci, and L, L^ as 
 vertices, describe an hyperbola, the two branches of which will be 
 the loci of centres of circles touching circles B and G externally 
 and internally. The intersection of corresponding branches of the 
 two hyperbolas will therefore determine Oj, 0^, the centres of 
 circles touching the three given circles all externally or all in- 
 ternally. 
 
 AgaiQ bisecting DE^ in M and B^E in M^ and taking B, A 
 as foci and if, M^ as vertices, an hyperbola can be described the 
 branches of which will be the loci of centres of circles touching 
 circles A and B, the one internally and the other externally, and 
 the intersections of this hyperbola with that through L and L^ in 
 Og, 0^ will give centres of two more circles fulfilling the given con- 
 ditions. The hyperbola through N and N^ , points corresponding to 
 
THE CIRCLE. 
 
 5] 
 
 M, J/j, will determine 0^ and 0^, two additional centres corre- 
 sponding to O3, 0^ and lastly, by its intersection with the two 
 branches through i¥^ and M^ will determine 0^ and 0^. 
 
 The construction is obvious from the definition of the hyper- 
 bola subsequently given. 
 
 Problem 35. (Fig. 35.) To draw a circular arc through three 
 given points A, B, C without using the centre. 
 
 Let AB be greater than either ^C or BC. With centre A 
 and radius AB describe an arc BD meeting AC in D, and with 
 
 centre B and the same radius describe an arc AE meeting BC in 
 F. From D on each side of it set off on the arc any equal 
 distances 7)1, and set off the same distances from U on the arc 
 BA, similarly make 1)2 = E2, and so on. The line joining A to 
 any point above D will intersect the line joining B to the corre- 
 sponding point below E and vice versa in points (as F, G) on the 
 required arc. 
 
 Proof. It is easily seen that the angle AFB = angle AGB =r the 
 angle AGB, &c., and therefore AFCGB all lie on a circular arc. 
 
 4-2 
 
52 EXAMPLES. 
 
 Examples on Chapter II. 
 
 1. Describe a circle to pass through two given points, P and 
 Pi, and to bisect the circumference of a given circle (centre 6', 
 radius GA). 
 
 (Draw PC and produce it to D so that PC .CD = AG\ The 
 circle through P, D, P^ fulfils required condition.) 
 
 2. Draw two circles cutting orthogonally, and shew by con- 
 struction that any line through the centre of either cutting both 
 circles is divided harmonically at the points of intersection. 
 
 3. Given the base AB oi 2^ triangle and the sum of the squares 
 of the sides AC^ + BC^^ draw the locus of the vertex. 
 
 (A circle, centre at E the middle point of AB^ and radius 
 
 4. Draw two circles (centres A and B) cutting orthogonally, 
 and draw their common chord meeting AB in C. Draw BE a 
 chord of the first circle passing through B^ and shew that a circle 
 can be described through ADEC. 
 
 5. The centre J. of a circle lies on another circle which cuts 
 the former in j5, C; ^i> is a chord of the latter circle meeting BC 
 in E, shew that the polar of D with respect to the first circle passes 
 through E. 
 
 o. At two fixed points A, B are drawn AC^ BD at right angles 
 to AB and on the same side of it, and of such magnitude that the 
 rectangle AC, BD is equal to the square on ^^ : prove that the 
 circles whose diameters are AC, BD will touch each other, and that 
 their point of contact will lie on a fixed circle. 
 
 (The circle on AB as diameter.) 
 
 7. With three given points A, B, G not lying in one straight 
 line as centres describe three circles which shall have three common 
 tangents. 
 
 (Bisect the angle BAG by AD meeting BC in D, 
 „ CBA by BE „ GA in E, 
 „ „ AGBhyCF „ AB in F, 
 
 then ED, DF, FE will be the required conmion tangents.; 
 
EXAMPLES. 53 
 
 The question is obviously, given the centres of the escribed 
 circles of a triangle, to draw the triangle. 
 
 8. A and B are two given points on the same side of a given 
 straight line CD, which AB meets in C. Determine the points on 
 OB on each side of C at which AB subtends a greater angle than 
 at any other point on the same side. 
 
 (The points of contact of circles through A and B, and touching 
 CB. Prob. 21.) 
 
 9. A and B are two given points within a circle; and ^^ is 
 drawn and produced both ways so as to divide the whole circum- 
 ference into two arcs. Determine the point in each of these arcs 
 at which AB subtends the greatest angle. 
 
 (The points of contact of circles through A and B touching the 
 given circle. Prob. 27.) 
 
 10. Shew by construction that the circle which passes through 
 the middle points of the sides of any triangle ABC will pass through 
 the feet of the perpendiculars from A, B, on the opposite sides, 
 and if be the intersection of these perpendiculars, will also pass 
 through the middle points of OA, OB, 00. Shew also that it will 
 touch the inscribed and escribed circles of the triangle, and that its 
 radius is half that of the circumscribing circle. 
 
 (The circle is called the nine point circle.) 
 
 11. Given four points ABCD in a straight line taken in order. 
 Shew that the locus of the point P moving so that the angle 
 APB = the angle OPD, is a circle which may be constructed in 
 the following manner. Let AB be less than CD, and take h be- 
 tween and D so that hD = AB. The centre is on the given 
 straight line at a distance from A, such that 
 
 AO : AO :: AB : Oh, 
 and the radius (?•) is such that 
 
 r-=OB, 00=0A, OB. 
 
 1 2. Find the locus of a point such that the area of the triangle 
 whose angular points are the feet of the perpendiculars from it on 
 the three sides of a given triangle, has a constant area. 
 
 /^ 
 
54 EXAMPLES. 
 
 [It is a circle of radius p, concentric with the circle circum- 
 scribing the given triangle; and p is determined from the equation 
 
 where E is the radius of circumscribing circle, k is the given 
 constant area and A is the area of the given triangle. If 4:k < A, 
 p is given by the equation 
 
 (Salmon's Conic Sections, Chap, ix.)] 
 
 As a numerical example, draw any triangle ABO, and take 
 p : 7? :: Vf : 1, 
 
 shew that in this case k = - . 
 
 1 3. Given on a straight line four points in the order F,A,B,Q; 
 describe a circle passing through A and B such that tangents 
 drawn to it from F and Q may be parallel. 
 
 [With centres F and Q and radii JFA, FB, JQA, QB 
 respectively describe two circles. A circle passing through ^ and 
 B, and through the points of contact of a common tangent to these 
 circles will be the one required.] 
 
 14. Given a fixed circle and an external point 0. Draw the 
 tangent at any point F of the circle and complete the rectangle 
 which has OF for side and the tangent for diagonal. Shew that 
 the angular point opposite will lie on the polar of 0. 
 
 15. From the obtuse angle ^ of a triangle ABO draw a line 
 meeting the base in D so that AD shall be a mean proportional 
 between the segments of the base. 
 
 [Find the centre of the circle circumscribing ABO. On AG 
 as diameter describe a circle cutting the base in D, the required 
 point.] 
 
 16. Find on a given line AB a point A such that its polar 
 with respect to a given circle shall pass through a given point 0. 
 
 [Find F the pole of AB, then the pole of OF will lie on AB 
 i.e. will be the required point A.'\ 
 
EXAMPLES. 55 
 
 17. Given a point A, a line through it AB, and a circle centre 
 C; draw a triangle APB which shall be self-conjugate with 
 respect to the circle (p. 32). 
 
 Take P the pole of the given line and from C draw CB perpen- 
 dicular to AF meeting AB in B, APB will be the required 
 triangle ; for since B is on the polar of P the polar of ^ will pass 
 through P, and is perpendicular to GB, i.e. is the line AP. 
 
 18. Given a triangle APB obtuse-angled at P, to draw the 
 circle with respect to which the triangle shall be self-conjugate. 
 
 The centre (C) of the circle must evidently be the intersection 
 of the perpendiculars from the angular points on the opposite 
 sides. Let the perpendicular from P on AB meet it in i>. The 
 radius of the circle will be a mean proportional between CP 
 and CD. 
 
 19. Given a circle, describe a triangle which shall be self- 
 conjugate with respect thereto, and with its sides parallel to those 
 of a given triangle ahp^ obtuse-angled at p. 
 
 Through C the centre of the given circle draw CA perpendicu- 
 lar to hp, GB perpendicular to ap and CM perpendicular to ah. 
 The vertices of the required triangle will liej one on each of these 
 lines. Through any point m on CM draw dme perpendicular to 
 CM meeting CA in d and CB in e, and through d draw df perpen- 
 dicular to CB and Bf perpendicular to CA ; f will necessarily lie 
 on CM. 
 
 If D is the point on Cm through which the side of the required 
 triangle perpendicular to Cm passes : — 
 
 where r is the radius of the given circle, i. e. CD is a mean pro- 
 portional between r and a length I determined by taking a fourth 
 proportional to Gf, Cm, and r ; for if 
 
 Gf : Cm '.'. T : I, 
 
 l = r^, and .'. CD\'^lr. 
 
CHAPTER III. 
 
 THE PARABOLA. 
 
 If a line be drawn through the centre of a given circle perpen- 
 dicular to the plane of the circle, the surface generated by a 
 straight line which passes through a fixed point on the first line 
 and moves round the circumference of the circle is called a right 
 circular cone. It will be shewn in Chap. ix. that the intersection 
 of this surface with any plane must be one or other of the follow- 
 ing: — a point, a pair of straight lines, a circle, a parabola, an 
 ellipse or an hyperbola. The construction of these last three 
 curves from their definition as the sections of a cone seems ct 
 priori to be the natural way of treating the subject; but the fact 
 is they are more easily constructed from some of their known 
 plane properties, and therefore, deferring the consideration of 
 them as lying on the surface of a solid, each will at first be defined 
 as the locus of a point moving in a plane so that its distance from 
 a fixed point is always in a constant ratio to its distance from a 
 fixed line, both point and line being in the plane of motion. 
 
 The fixed point is called the focus, and the fixed line the 
 directrix. 
 
 In the parabola the ratio is one of equality, i.e. the distance 
 from the fixed point is always equal to the distance from the 
 fixed line. 
 
 In the ellipse the ratio is one of less inequality, i. e. the dis- 
 tance from the fixed point is always less than the distance from 
 the fixed line. 
 
 In the hyperbola the ratio is one of greater inequality, i. e. the 
 
THE PARABOLA. 57 
 
 distance from the fixed point is always greater than the distance 
 from the fixed line. 
 
 The eccentricity of a conic is the numerical value of this ratio. 
 
 A parabola can generally be drawn to satisfy four geometrical 
 conditions, and four conditions are in general necessary and suffi- 
 cient to determine the curve. Thus an infinite number of para- 
 bolas can be drawn to pass through three given points or to touch 
 three given lines, or to pass through two points and touch a given 
 line, or to fulfil any three similar conditions, and in each case a 
 fourth condition must be added to make the exact solution possi- 
 ble. At the same time four conditions may sometimes lead to 
 more than one solution, just as, more circles than one can fre- 
 quently be drawn satisfying three given conditions; and occasion- 
 ally some limitation as to the position of the points or lines given 
 as data of the problem, is necessary to enable a real curve to be 
 drawn. 
 
 If the focus is given in any particular problem, this is equiva- 
 lent to two geometrical conditions, and therefore in general only 
 two others can be fulfilled, i.e. given the focus and two points 
 through which the curve is to pass, the problem is completely 
 determinate and a parabola cannot be drawn to have a given point 
 as focus and to pass through any three random points. The direc- 
 trix being given is also equivalent to two geometrical conditions, 
 and therefore along with it, only two others can be fulfilled, such 
 as, e. g. to pass through a given point and touch a given line, or to 
 touch two given lines, or to touch a given line at a given point, 
 &c. 
 
 Problem 36. (Fig. 36.) To draw a parabola the focus F and 
 the directrix MX being given. 
 
 Draw FX from F perpendicular to MX. Bisect FX in A 
 and A will be a point in the curve. With F as centre and any 
 radius greater than FA (as F 3) draw a circular arc D^D^ set off 
 from A towards i^ a distance J 3' equal to AZ, and at 3' erect a 
 perpendicular to FX meeting the circular arc in DD^. These will 
 be points in the curve, and similarly drawing any number of arcs 
 
58 
 
 GIVEN FOCUS AND DIRECTRIX. 
 
 with F as centre and setting off from A towards F^ distances 
 equal to the distances of the arcs beyond -4, and erecting per- 
 
 Fig.36 
 
 pendiculars to FX at these points meeting the corresponding 
 arcs, any number of points on the curve may be determined and 
 the curve drawn through the points thus obtained. 
 
 The construction is obvious : at any point as P draw FN per- 
 pendicular to AX meeting it in N\ then the distance FP from the 
 focus is to be equal to FM the perpendicular distance from the 
 directrix. But FF= FA + the distance of the arc beyond A \ and 
 PM-XN = XA + the same distance. 
 
 .\FP = PM since FA^AX. 
 
 (See also the next problem.) 
 
 Def. From the construction the curve is evidently symmetri- 
 cal about FX which is called ilie axis. The point A where the 
 curve cuts the axis is called the vertex', and any line parallel to the 
 axis is called a diameter of the curve. 
 
THE PARABOLA. 5^ 
 
 The parabola consists of one infinite branch. Like the focus 
 and the directrix, the vertex and axis are each equivalent to two 
 conditious in the construction, but it should be noticed that certain 
 pairs of these lines and points given together are equivalent not to 
 four but only to three conditions. This apparent anomaly may 
 be thus explained. Suppose directrix and axis are given, these 
 are two lines at right angles to each other and hence the direction 
 of either is implicitly involved in that of the other, and thus in- 
 stead of the two conditions of position and direction being given 
 independently along with the second line, one only, namely posi- 
 tion, is really given, and the two lines together are therefore 
 equivalent to three conditions only. Similarly focus and axis, or 
 vertex and axis make only three conditions since the position of 
 the axis is partly involved in that of the focus or of the vertex. 
 
 To draw a tangent at any point. 
 
 P and D (fig. 36) being any two points on the curve, if the 
 line through PD meet the directrix in R and DK is parallel to 
 
 PM, then 
 
 FP : FD :: PM : DK 
 
 :: PR : DR 
 
 by similar triangles, and .-. FR bisects the exterior angle between 
 FP and FD (Euc. vi. Prop. A). Hence if the point D move up 
 to and coincide with P so that the chord PD becomes the tangent 
 at P (Def. p. 30), in which case FD of course coincides with 
 FP^ the line FS drawn from the focus to the point in which the 
 tangent at P meets the directrix, must be perpendicular to FP. 
 The triangle SFP is therefore equal and similar to the triangle 
 SMP. Hence the tangent at any point P of a parabola bisects the 
 angle between the focal distance FP and the perpendicular PMfrom 
 P on the directrix. It can therefore be drawn either by bisecting 
 the angle FPM or by making FT on the axis equal to FP, and 
 joining PT ; for in this case the angle jPPI^= angle FTP^ which 
 is equal to the alternate angle TPM, PM, FT being parallel. 
 
 Def. The perpendicular PN from P on the axis is called the 
 ordinate of P. The double ordinate through the focus is called 
 
60 TANGENTS FROM AN EXTERNAL POINT. 
 
 the latus-rectum of the curve, and its length is always equal to 
 iAF. It is sometimes called the principal parameter of the 
 curve. Since 
 
 FP = PM=XN^FT and FA^AX, 
 XN-AX^FT-FA, 
 i.e. AT^AN or iy^=2, AN. 
 Def. The line NT is called the sub-tangent at the point P. 
 
 The line PG perpendicular to the tangent at P is called the 
 normal at P. 
 
 It has been shewn that the tangent bisects the angle FPM, 
 . ' . PG bisects the angle FPL where Z is a point on MP produced, 
 i. e. the angle FPG - angle LPG = angle PGF, 
 and .-. FG=FP=PM=XN, 
 .'. FG-FN=XN-FN, 
 i.e. A'G = FX = 2AF. 
 Def. The line NG is called the sub-normal of the point P. 
 
 The tangent at the vertex is perpendicular to the axis, as is 
 obvious from the symmetry of the curve, and a perpendicular 
 from the focus oti any tangent intersects it and the tangent at the 
 vertex in the same point. 
 
 The focus and directrix being given, tangents to the curve can 
 be drawn from an external point Q thus (fig. 36). With centre F 
 and radius equal to the distance of Q from the directrix describe 
 a circle ; draw tangents to it from Q, and join F to the points of 
 contact a, a^, producing the lines to meet the curve in VV^. QV, 
 QV^ will be tangents, for, if VM^ be the perpendicular on the 
 directrix, and the diameter at Q meet the directrix in X^ and VQ 
 meet it in S^, 
 
 YM^ : QX, :: VS, : Q.S,, 
 
 or FY : Fa :: YS^ : Q,S^, 
 .-. FS^ is parallel to aQ, but aQ is perpendicular to FY, .'. FS^ is 
 perpendicular to FY, and .-. YS^ or YQ is a tangent through the 
 point Q ; similarly for F, Q. 
 
THE PARABOLA. 
 
 61 
 
 A tangent to a parabola parallel to a given line may be drawn 
 by constructing the angle GFP = twice the angle which the line 
 makes with the axis, so determining the point of contact F. 
 
 Problem 37. (Fig. 37.) To draw a parabola, the vertex A, the 
 axis AJ}^ and a point P on the curve being given. 
 
 This might be solved by first finding the focus and proceeding 
 as in the last problem. It can however be solved independently 
 
 Fig.37. 
 
 without using circular arcs, and the method is evidently applicable 
 to the last problem after any one point on the curve has been found. 
 Draw the tangent at the vertex and a diameter through P 
 meeting it in M. Divide 2fP into any number of equal parts 
 (say four), and AM into the same number. Then diameters 
 through the several points on AM will meet lines joining A to the 
 corresponding points on MP (counting from A in the first case 
 and from M in the second) in points of the curve as £, C, D. As 
 the curve recedes from the axis the points found get more and 
 more distant from each other (compare C to D and D to P), but, 
 if desirable, points can be interpolated between any two points 
 already found by subdividing the corresponding spaces on MP and 
 
62 GIVEN FOCUS, AXIS AND POINT. 
 
 AM. In the figure points are thus interpolated between C and D 
 and between D and P. The curve can be carried beyond F by 
 carrying on the divisions on the two lines as in the figure. 
 
 The other half of the curve can be put in by symmetry. 
 
 The tangent at any point I) can be drawn by drawing the 
 
 ordinate DN^, and making AT on the axis equal to A^^, on the 
 
 other side of the vertex; DT will be the tangent at B, as has 
 already been shewn. 
 
 The focus F is found by drawing the normal at any point B, 
 bisecting the sub-normal NG and setting off AF=^NG. 
 
 The construction for the curve depends on the fact that if a 
 diameter be drawn through the centre point of any chord, the 
 tangents at the extremities of the chord intersect on the diameter, 
 and the curve cuts the diameter at the centre point between the 
 chord and the intersection of the tangents. Thus ^P is a chord, 
 the diameter through 2 (on AM) will intersect it in its centre 
 point F, Al is the tangent at A and therefore the tangent at F 
 will also pass through 2, and C, which bisects V2 since 
 
 6'2 : CV :: J/2 : F2 
 will be a point on the curve. 
 
 Similarly B may be shewn to be on the curve, since it bisects 
 the diameter between 1 and the centre point of the chord AO, and 
 D may be shewn to be on the curve as bisecting the diameter 
 between (73 the tangent at C, and the centre point of the chord 
 OF. 
 
 Problem 38. (Fig. 37.) To draw a parabola the focus F, 
 the axis FN^ and a point F on the curve being given. 
 
 The directrix and consequently the vertex can at once be de- 
 termined by drawing FM parallel to the given axis, measuring 
 along it a length FL equal to FF and from L dropping a 
 perpendicular on the axis intersecting it in X. This per- 
 pendicular is, of course, the directrix, and the vertex bisects 
 FX. The curve can then be drawn by either of the preceding 
 methods. 
 
THE PARABOLA. 
 
 63 
 
 Problem 39. (Fig. 38.) To draw a curve formed of cir- 
 cular arcs approximating to a parabola the focus F, and vertex A 
 being given. 
 
 The following method depends on the fact that in the 
 parabola the sub-normal is constant and equal to twice AF. 
 
 Fig.38. 
 
 Draw the axis AFN and on it take ^1 equal to FA, and draw 
 any ordinates as BB^^ GC\, J^D^, &c. With 1 as centre de- 
 scribe an arc through A, extending as far as the centre or- 
 dinate between A and BB^, from L the foot of ordinate BB^ 
 make Z2 equal to twice AF, and with centre 2 and radius to 
 the point where arc through A meets the centre ordinate be- 
 tween A and B describe an arc extending to half-way between 
 B and C ', from M the foot of ordinate CG^ make J/3 equal to 
 twice AF and with centre 3 and radius to the point where 
 arc through B has been stopped describe an arc extending to 
 half way between C and D. Similarly from the foot of the 
 ordinate DD^ measure a distance on the axis equal to twice AF 
 so determining the centre (4) for an arc through D, and con- 
 tinue the process for any number of successive ordinates. It 
 will be seen that the centres are determined by measuring a 
 constant distance from the foot of the successive ordinates 
 
64 
 
 GIVEN FOCUS AND TWO POINTS. 
 
 equal to the known constant length of the sub-normal in the 
 parabola (p. 60), but that the radius of each arc depends 
 entirely on the arc previously drawn, so that the curve must 
 be commenced from the vertex. Each successive arc extends 
 some distance on each side of the ordinate from which its 
 centre is determined. It is convenient, though not essential, 
 to commence with ordinates dividing AF into equal parts, and 
 tolerably close together, and as the curve recedes from the 
 vertex and cuts the ordinates more nearly at right angles the 
 distance between them may be increased. Carefully drawn, the 
 method gives a remarkably close approximation to the real 
 form of the curve, as may be seen by comparing the dis- 
 tance of the point P in the figure from F with the distance 
 NX^ its perpendicular distance from the directrix. The half dis- 
 tance between the ordinates to which each successive arc has to 
 extend, and which furnishes the starting point for the next arc 
 can generally be estimated with quite sufficient accuracy by the eye. 
 Problem 40. (Fig. 39.) To draw a parabola, the focus F, and 
 two points A and B on the curve being given. 
 
THE PARABOLA. 
 
 65 
 
 With centre A and radius AF describe a circle CFD, and 
 with centre B and radius BF describe a circle G^FD^. Draw 
 common tangents CC^ and DD^ to the two circles. (Prob. 31.) 
 These will be the directrices of two parabolas fulfilling the 
 given conditions, and the curves may be drawn by any of the 
 preceding methods. 
 
 The construction is obvious. 
 
 Problem 41. (Fig. 40.) 7'o draw a parabola, the focus F, a 
 point A on the curve, and a tangent YT being given. 
 
 The point of contact of the tangent is not given, as this 
 would be a fifth condition. With centre A and radius AF 
 
 Fig.40. 
 
 describe a circle FM, and on FA as diameter describe a circle 
 EFE^, the centre being G. From F drop a perpendicular FY 
 on the given tangent, and from Y draw tangents YE, YE^ to 
 the given circle. Join FE, FE^ and produce them to meet the 
 larger circle in M, M^ , then MX, M^X^ drawn parallel to YE, 
 YE^ respectively will be the directrices of two parabolas fulfilling 
 the given conditions. 
 
 Proof It is known that the perpendicular from the focus 
 on a tangent passes through the point of intersection of that 
 E. 5 
 
66 GIVEN FOCUS, POINT AND TANGENT. 
 
 tangent aud the tangent at the vertex, hence F is a point on the 
 tangent at the vertex. 
 
 The directrix must evidently touch the circle MFM and 
 must meet the perpendicular on it from the focus at a point 
 double the distance from F that it is from the tangent at the 
 vertex. 
 
 In the triangles AEF, AEM^ AF = AM, AE is common 
 and the angles AEF, AEM are equal, each being a right 
 angle ; 
 
 .-. FE^EM and .-. AM is parallel to EC, since FC=GA\ 
 but EG is perpendicular to YE, and therefore MX which is 
 parallel to YE is perpendicular to MA, and therefore touches 
 the circle MFM^. 
 
 Draw FX perpendicular to MX and let YE meet it in F, 
 
 then FV : FX :: FE : FM; 
 
 .-. FX=2FV, since FM=2FE. 
 
 Hence two parallel lines have been found, one of which touches 
 the circle MFM^, while the other passes through F and bisects 
 the distance between F and the first. 
 
 Problem 42. (Fig. 41.) To draw a parabola, the focus F and 
 two tangents BT, RT^ being given. 
 
 (The problem is impossible if the given lines are parallel, i.e. 
 they must always intersect in some point B; and F must not lie 
 on either of them.) 
 
 Join EF, and at F on each side of RF construct an angle 
 RET, RFT^ equal to 1\RS, the angle between the given lines 
 alternate wifch that in which F is situated. T and T^ will be the 
 points of contact of the given tangents and the problem is reduced 
 to Problem 40. As in that problem two lines can be drawn touch- 
 ing circles with centres T, 1\ and radii TF, ^'^i^" respectively, which 
 will be the directrices of parabolas having F as focus and passing 
 through Tand '1\, but only one of these will in addition touch the 
 given lines at those points. 
 
THE PAKABOLA. 
 
 67 
 
 Proof. The construction depends on the well-known property 
 of the parabola, that the exterior angle between any two tangents is 
 
 equal to the angle subtended at the focus, by the segment of either 
 between the 2)oint of intersection and the 2^oint of contact. For if 
 be any point in ^i^ produced, the angle TFO = twice angle FTE, 
 since (Prob. 36) the angle FTE =- angle which TR makes with the 
 axis. Similarly angle ^ji^6> = twice angle FT^S^; 
 
 .: angle T^FO - angle ^i^0 = twice (angle FT^S^ - angle FTR), 
 
 i.e. TFT^ = twice angle UBT= twice angle T,RS. 
 
 Through F draw FJD parallel to the directrix meeting TS in 2), 
 then FD = FS, since TS bisects the angle between FS and the 
 directrix. Let FE meet the directrix in K. 
 
 By similar triangles 
 
 KS : FD :: KE : FE, 
 KS : KE :: FS : FE\ 
 
68 GIVEN DIRECTKIX AND TWO OTHER CONDITIONS. 
 
 and similarly, if S^ denote the intersection of BT^ with directrix, 
 KS^ : KB :: FS^ : FR, 
 .', KS : KS, :: FS : FS^. 
 Hence the angles KFS, KFS^ are either equal or supplemen- 
 tary. In the figure they are supplementary, i.e. angle KFS= angle 
 RFS^. 
 
 But angle RFT is the complement of angle KFS 
 and „ EFT^ „ „ „ XFS^, 
 
 .-. angle BFT =3iiiglG RFT^, 
 and .-. each of them = J angle ^^i^I", - angle T^RS, 
 which proves the property above referred to. 
 Second Solution. 
 The problem may also be solved by dropping perpendiculars 
 from the focus on the given tangents, their points of intersection 
 determining the tangent at the vertex. 
 
 Problem 43. (Fig. 39.) To draw a pa/rabola, the directrix 
 CC^ and two points A and B on the curve being given. 
 
 This is merely the converse of Prob. 40. With the given 
 points as centres and with radii equal to the distance of each 
 from the given directrix describe arcs intersecting in F and F^y 
 either of which may be taken as the focus. 
 
 Problem 44. (Pig. 40.) To draw a parabola, the directrix 
 MX, a point A on the curve and a tangent YT being given. 
 
 With centre A describe a circle MFM^ touching MX. This 
 will of course be a locus of the focus. At S, the point of inter- 
 section of the given tangent and directrix, construct an angle TSF 
 equal to the angle between MS and TS produced, i.e. = the angle 
 MSK. SF will be another locus, i.e. the focus will be at F, the 
 intersection of the line and circle. The line SF will evidently 
 meet the circle again beyond A and this point of intersection will 
 be the focus of a second parabola fulfilling the given conditions, 
 the point of contact of tangent and the point A being on the same 
 side of the axis. 
 
THE PAKABOLA. 
 
 69 
 
 Proof. That the circle is a locus of the focus needs no demon- 
 stration: that the line is a locus of the focus is proved, since it has 
 been shewn (Prob. 36) that FS is always perpendicular to the line 
 joining F to the point of contact of the tangent through S, and 
 that therefore the two triangles FST, LST, where TL is perpendi- 
 cular to MS, are equal and similar in all respects; and that therefore 
 angle 7^^^^= angle LST wangle MSK. 
 
 Problem 45. (Fig. 41.) To draw a parabola, the directrix 
 KX and tivo tangents RT, RT^ being given. 
 
 At S, the point of intersection of RT with KX, construct an 
 angle TSF equal to the angle TSK. As in the last problem SF 
 
 FIg.4Ia. 
 
 •- — — d, 
 
 will be a locus of the focus. Similarly, if RT^ meet the directrix 
 in aS'j, construct an angle T^S^F equal to the angle T^S^X, and S^F 
 will be a second locus, therefore the intersection of these Hues deter- 
 mines F, the focus. In the figure the directrix and the tangent 
 RT^ do not intersect within any reasonable distance, but the line 
 through their intersection making the same angle with the tangent 
 as the tangent does with the directrix can easily be drawn, as shewn 
 in fig. 41a. Let ab, cd be any two converging lines; from any two 
 points (a, b) on the one, drop perpendiculars ac, bd on the other 
 and produce them: make ce= ca, df^ db, then obviously a6 and ef 
 will pass through the same point on cd and will be equally inclined 
 thereto. 
 
 Problem 46. (Fig. 42.) To draw a parabola, the axis AN and 
 two points P, Q on the curve being given. 
 
 [The two points must not be at equal distances from the axis 
 
70 
 
 GIVEN AXIS AND TWO POINTS. 
 
 whether on the same or on opposite sides of it, nor must they be 
 on the same perpendicular to the axis.] 
 
 Draw the ordinates PiY, QN^, of which let PN be the greater; 
 the vertex will then obviously lie on the same side of N as N^ 
 
 Fig.42. 
 
 and beyond it. On NP produced make Pn equal to QN^, and on PN 
 make Pm also equal to QNy Then Nm is evidently equal toPN—QN^. 
 On the axis make No=Nm and on the same side of iVmake Np^-NP. 
 Through o draw ox parallel to p?i meeting PN in a?, and through 
 P draw PA parallel to N^x meeting the axis in A. A will be the 
 vertex of the required parabola and the problem is reduced to 
 Prob. 37. 
 
 Proof. It is a well-known property in the parabola that 
 PN\'^ = iAF . AN where F is the focus, PN an ordinate and A the 
 vertex. __ 
 
 .-. PN\' : QNy' :: AN : AN^, 
 
 PN\'-QNi' : P^\' :: AN -AN, : AN, 
 (PN+QN,){PN-QN;} : PN\' :: NN, : AN, 
 AN PN 
 
 NN, " {pn+qn;){pn-qn;) 
 
 or 
 
 i.e 
 
 or 
 
 PN 
 
THE PARABOLA. 71 
 
 If a fourth proportional be taken to 
 
 PN, PN+QN,, and PN-QN^, 
 i. e. if a length I be determined such that 
 
 PN : PN ^ QN^ :: PN- QN^ : I, 
 the above equation may be written 
 
 AN PF 
 ~NN, ~ I ' 
 
 i.e. ^iV is a fourth proportional to such length I, NN^ and PN. 
 But this is really what has been done, for 
 
 Np : Nn :: No : Nx, 
 i.e. PN : PN + QN^ :: PN-QN^ : Nx, 
 
 i. e. Nx is the required length I, 
 
 and Nx : NN^ :: PN : AN. 
 
 That Pi^|'= 4^i^. AN may be shewn thus: Join PA and let 
 it meet the directrix in E. Join EF (F being the focus) and pro- 
 duce it to meet the diameter through P in L, while the diameter 
 meets the directrix in M. Then since FA = AX, PL = PM= PF, 
 for ML is parallel to FX, therefore the circle on ML as diameter 
 goes through F, and therefore the angles MFL, MFE are both 
 right angles and 
 
 EX. XM^ FX\^=^AF\% 
 also AN '. AX :: PN : EX by similar triangles, 
 
 PN\^ : EX. MX 
 PiV'h : 4:AF\\ 
 .: PNl'^AAF.AN, since AF-^AX. 
 
 Problem 47. (Fig. 43.) To draw a parabola, the axis AN, a 
 point P on the curve and a tangent OT being given. 
 
 [The tangent must not be parallel to the axis, and the point 
 must lie within the angle formed by the tangent and a sym- 
 metrical line on the other side of the axis.] 
 
 Draw the ordinate PN and let it meet the given tangent in 0. 
 Make NP^ on the other side of the axis equal to PN, and P^ will 
 
72 
 
 GIVEN AXIS, POINT AND TANGENT. 
 
 by symmetry be a point on the curve. Find a mean proportional 
 between OP and OP^ (Prob. 5) and set off its length OE on OP 
 
 Fig.43. 
 
 from towards the axis. Draw through E a parallel to the axis 
 meeting the tangent in ^. ^ is the point of contact of such 
 tangent. Draw QN^, the ordinate of Q^ and the vertex, A^ will 
 bisect N^T, the subtangent of Q (Prob. 36). The problem is there- 
 fore again reduced to Prob. 37. 
 
 Proof. That the diameter through Q, the point of contact of 
 the given tangent, meets OP in E such that OE^ = OP . OP^, may 
 be shewn thus. Let PAP^ be a parabola and OQ a tangent at Q. 
 Take any point a on the given tangent, and draw any two chords 
 as a&c, ah'P^ and let q and q^ be the vertices of the corresponding 
 diameters, and let the diameter through q meet hem. v. through 
 a draw ad parallel to qv meeting the parabola in d, and draw du 
 parallel to he meeting its diameter \xiu. 
 
THE PARABOLA. 73 
 
 Then ah .ac^cw^" -hv\^ (Euc. ii. 6) 
 
 = 4:Fq . [qu — qv) (p. 71) if i^ is the focus, 
 = iFq . ad, 
 and similarly ah' . aP = 4:Fq^ . ad, 
 
 .'. ah.ac : ah' .aP :: Fq : Fq^, 
 i. e. the ratio of the rectangles depends only on the positions of q 
 and 5 J , and is independent of the position of the point a. 
 
 If the lines ahc, ah'P move parallel to themselves until they 
 become the tangents at q and q^, we shall then obtain, if these 
 tangents intersect in ^^ 
 
 t^\' : i^q^l' • Fq : Fq^, 
 and .*. ah.ac : ah' .aP :: t^q^ : t^q^^, 
 
 but the tangent aQ may be regarded as a chord cutting the para- 
 bola in two coincident points, and therefore if the tangent at q 
 meet aQ in t and vq meet it in m 
 
 ah.ac : aQ^" :: 'qt\' : tQ\^ 
 
 :: ~qi\^ : 'tm\\ 
 Also if Qk is the diameter at Q meeting ac in k, by similar 
 triangles 
 
 qt : tm :: ak : a^, 
 
 .-. ah.ac : a^|^ :: "^1^ : ^\\ 
 or ah . ac = ak\ ^, 
 
 which justifies the construction. 
 
 Problem 48. (Fig. 44.) To draw a parabola, the axis UN and 
 two ta/ngents PT, QT heing given. 
 
 [The point T must not be on the axis.] 
 
 If from the point U in which either of the tangents (as QT) 
 cuts the axis, a line UR be drawn making the same angle with 
 the axis as QT but on the opposite side of it, this will, by sym- 
 metry, be a third tangent to the curve. Let it meet the other 
 tangent {PT) in V. Describe a circle through the three points 
 T, U, V (Prob. 20), cutting the axis in F. F will be the focus of 
 
74 
 
 GIVEN AXIS AND TWO TANGENTS. 
 
 the required parabola, and FU will be the distance from F of 
 the point of contact of either of the tangents QU, RU. With 
 
 
 \ Fig.44. 
 
 
 
 ^"-X 
 
 
 
 \ ^jlt>. 
 
 >-.. 
 
 
 - 
 
 / \ y^Q. ^-^^ 
 
 \^r """i><? 
 
 > 
 
 centre F and radius FU describe an arc cutting UR in R, with 
 centre R and the same radius describe a circle, and the directrix 
 will touch this circle and is of course perpendicular to the axis. 
 The problem is therefore reduced to Prob. 36. 
 
 Proof. The fact that the circle through the points of inter- 
 section of three tangents is a locus of the focus is generally true, 
 and is not confined to the case of two tangents meeting on the 
 axis. For draw any tangent pah meeting the parabola in p, the 
 two given tangents in a and h and the axis in c, and let Th meet 
 the axis in t. It has been shewn (Prob. 42) that the angle aht 
 is equal to either of the angles pFb, PFh^ also the angle Fpc = the 
 angle Fcp = the angle bet, 
 
 .'. the remaining angle Fba of the triangle Fpb, 
 
 i. e. if two tangents intersect in b tJm angle which either makes with 
 Fb is equal to the angle which the other makes with the axis. 
 
THE PAEABOLA. 
 
 75 
 
 Similarly, since QT^ FT intersect in T, the angle FT a is eqvial 
 to the angle Ftb^ i.e. htc, 
 
 . '. angle Fha - angle FTa^ 
 or a circle goes round aFTh. (Euc. iii. 27.) 
 
 Problem 49. (Fig. 45.) To draw a parabola, two tangents A7\ 
 BT, and their points of contact A and B being given. 
 
 First method. Divide AT, BT into any (the same) number 
 of equal parts ; the lines joining opposite points on the two tan- 
 
 Fig.4-5. 
 
 gents, (i.e. supposing each divided into 8 parts, the lines joining 
 1 on ^^ to 7' on BT, 2 on ^7" to 6' on BT, and so on,) will be 
 tangents to the curve, which can easily be drawn to touch them 
 all. Or points on the curve may be found successively thus. 
 Bisect AT, BT in the points 4, 4'. The line joining these points 
 is a tangent to the curve at its centre point, i. e. bisect 4, 4' in i* 
 and P is a point on the curve. Similarly the line joining the 
 point of bisection (6) of 4^4 and the point of bisection of (m) 4P 
 
76 GIVEN TWO TANGENTS AND THEIR POINTS OF CONTACT. 
 
 will be a tangent to the curve at its centre point, and the line 
 joining the point of bisection (6) of i'B and the point of bisection 
 (n) of 4'P will be a tangent to the curve at its centre point, and 
 the method of bisecting the tangents successively may be con- 
 tinued. It is obvious that the point m found by bisecting 4P 
 is identical with the point of intersection of the line 44' and 
 the line joining 6 on -4 to 2' on B. The focus may be found 
 as the intersection of the circle circumscribing the triangle formed 
 by any three tangents w4th that circumscribing the triangle 
 formed by any other three, as e.g. the triangles 47^4' and 5:7^3', 
 and the directrix may then be determined by Prob. 40. 
 
 Second method. The focus may be determined independently, 
 without drawing additional tangents, thus. Join AJ3, bisect it 
 in V and join VT. VT will be a diameter of the curve, and 
 the curve will pass through P the centre point of YT. Bisect 
 YT in P. Find a third proportional to YT, A Y (Prob. 5), the 
 length of which will be equal to 2FP if F is the focus. [This 
 may conveniently be done by making Tv on TY equal to AY 
 and drawing a line through v parallel to J. F to meet AT. The 
 length (l) of this line will be the required third proportional, 
 since TF: YA :: Tv or YA : L] 
 
 Describe a circle with centre F and radius equal to ^l which 
 will be a locus of the focus, and the directrix will be a tangent 
 MX to this circle perpendicular to the diameter TY. Then F 
 may be determined as the intersection of a circle, with centre A 
 and radius AM, the distance of A from the directrix, and the 
 previously drawn circle. 
 
 Third method. It has been shewn (Prob. 41) that the exterior 
 angle between any two tangents is equal to the angle which either 
 subtends at the focus. Therefore if on -4 2^ as chord a segment of 
 a circle be described on the side towards B, containing an angle 
 AFT equal to the angle ATK (Prob. 30), where K is on BT 
 produced, this segment will be a locus of the focus. Similarly if 
 a segment containing the same angle be described on BT towards 
 A, it will be a second locus and the focus will be at the inter- 
 
THE PARABOLA. 77 
 
 section of the two, and the directrix may be determined by 
 Prob. 40. 
 
 Proof. That the line joining the intersection of tangents to 
 a parabola to the point of bisection of the chord joining their 
 points of contact', is a diameter may be shewn thus. Let AB be 
 two points on a parabola, AT^ BT tangents at the points, F the 
 focus and AN^ BN^ perpendiculars on the directrix meeting NTN^ 
 parallel to the directrix in N and N^. Join FA, FB and draw 
 Ta perpendicular to FA and Th perpendicular to FB. Then the 
 angle ^^a = angle TAN, 
 
 .'. TN=Ta, and similarly TN^ = Tb. 
 
 But Ta ~ Th, since it has been shewn that angle TFA »= angle 
 TFB. (Prob. 42.) 
 
 .-. TN^TF^. 
 
 If TV be drawn parallel to AN or BN^, i.e. to the axis, meeting 
 AB in F, it will make AY-. VB v. TN i TN^, i.e. AV^VB,, 
 or the diameter through T bisects AB. Since TN= TN^ it follows 
 that any straight line through T terminated by the diameters 
 A and B is bisected in T and more generally that every line 
 through the point of intersection of two tangents terminated by 
 diameters through the extremities of the corresponding chord of 
 contact, is bisected by such point of intersection. 
 
 That P, the point in which the curve meets TV, bisects TV 
 and that the tangent at P is parallel to AB maybe shewn thus: — 
 Since AN, TV and BN^ are parallel lines, it follows that every 
 line meeting the three is bisected by :Z F"; and therefore if the 
 tangent at the point P be drawn meeting AN yd. G and BN in 
 G^, PG = PG^ ; but if it meets AT, BT in 4 and 4', it follows as 
 above that P4 = 4(y, P^' = 4:'G^, and therefore Gi, 4P, Pi' and 
 4'(xj are all equal, which is only possible if GPG^ bisects TV and 
 is parallel to AB. 
 
 Hence Ti - 4^, Ti' = i'B and 44' - ^AB. 
 
 To shew that il F is a mean proportional between VT (or 2PV) 
 and 2FP, draw FU parallel to ^^ or to 44', meeting PV in U, 
 
78 GIVEN THREE TANGENTS AND POINT OF CONTACT OF ONE. 
 
 then the angle FUT = angle 4Pf7, 
 
 = angle i^4^, (Prob. 48), 
 and therefore the circle which it is known can be drawn (Prob. 
 48) through i^4^4', will pass through U. 
 
 Hence, AV being twice P4, 
 
 ~AV\' - 4P4| ^ = 4.PU. PT. (Euc. iii. 35.) 
 But the angle roT= angle FP^\ since FU i^ parallel to P4', 
 
 = angle i'PT = angle FUP, 
 and therefore FP = PU; also PT=PV, 
 
 therefore AV\' = ^FP . PV. 
 
 Definition. A chord through the focus parallel to the tan- 
 gent at P is called the parameter of the diameter through P, and 
 it follows from the above that its length is always equal to 4:FP. 
 (See definition of latus-rectum, p. 60.) 
 
 Problem 50. (Fig. 46.) To draw a 2yctrabola, three tangents, 
 TU, TV, UV and the point of contact P of one of them TTJ being 
 
 Describe a circle through TUV (Prob. 20), then F the focus 
 
 Fig.46. 
 
 1 
 
THE PARABOLA. 79 
 
 lies on this circle (Prob. 48). On FT describe a segment of a 
 circle containing an angle equal to the exterior angle between the 
 tangents meeting in T, i.e. the angle VTY. (Prob. 30.) This 
 segment will be a second locus of the focus (Prob. 42), which will 
 therefore be at the intersection of the segment with the previously 
 drawn circle. 
 
 If P (as in the figure) lies beyond TJ the segment must be 
 described on the side of TP towards F : but if P lies between T 
 and ?7, the segment must be described on the other side of TP^ 
 since the focus can never lie inside the triangle TJJY and the 
 angle it contains must be the angle TJTY, since that would then 
 be the exterior angle between the tangents. 
 
 [The centre for the segment may conveniently be found by 
 drawing TO perpendicular to YT to meet the perpendicular 
 bisector of PT in (7.] 
 
 Construct the angle UFQ equal to the angle PFU. Q will 
 be the point of contact of UY, and the direction of the axis is 
 determined since it is parallel to the diameter joining U to the 
 centre point of PQ. (Prob. 49.) It can then of course be drawn 
 through F. 
 
 Lastly, the vertex may be found since it is the centre point 
 between N the foot of the ordinate from P and the point in which 
 PT cuts the axis (p. 60.) 
 
 The point of contact R, of TY, may of course be determined 
 without drawing the curve by making the angle ^i^jR= angle TFP. 
 
 The construction is evident from preceding problems. 
 
 Problem 51. (Fig. 47.) To draw a parabola, three points 
 A, B, C, on the curve, and the direction of the axis, as BD, being 
 given. 
 
 Draw lines through AB, BC, CA, and let BD parallel to the 
 given direction of axis meet AC in B. Bisect AG in F and draw 
 FL parallel to BB to meet BC in L. Draw LG parallel to AG 
 to meet BB in G. Join A G and it will cut FL in R, the vertex 
 of the diameter through F. 
 
80 
 
 GIVEN THREE POINTS AND DIRECTION OF AXIS. 
 
 If HK~HE, AK will be the tangent at A and the focus may 
 be found by taking HU such that A^^^4.HE .HU, i.e. taking 
 
 Fig.47. 
 
 KxT. „._.4..___ Hi'-_. \-4— 
 
 2HU2i third proportional to 'iHE, AE; drawing UF parallel to AG 
 and making HF=HU. [If we take a third proportional to EKy 
 AE it will be 2HU. This may conveniently be done by making 
 Ea = EA and drawing au parallel to AK. Eu will be the 
 required third proportional. The problem reduces to Prob. -40.] 
 
 Proof. To shew that H is the vertex of the diameter through 
 E. Draw BN parallel to ^ C meeting EH in N. BN = ED, 
 
 and DA.DC = AE\ -' - ED^ = AE'- BJV' (Euc. ii. 5) ; 
 but in any parabola 
 
 AE' = 4..FH.HE, 
 and BN'^4..FH.HN, 
 
 . '. AE" - BN' = ^FR. EN = 4:FE. BI), 
 .'. DA.DC '.AE' :: BD -, HE -, 
 but in the figure 
 
 BD -EL :: DC : CE or EL 
 
 BD.AE 
 
 DG 
 
and 
 
 THE PAKABOLA. 
 
 AE :EH y.AD . DG, 
 iiAD'.EL, 
 
 81 
 
 DC ' 
 i.e. DA.DC : AE' :: BD : HE, 
 
 or HE has been determined of the proper length. 
 
 Problem 52. (Fig. 48.) To draw a parabola, three tangents 
 UT, TV, VU and the direction of the axis, as AN, being given. 
 
 Through T draw MTM^ perpendicular to the given direction 
 of the axis. It is a known property of the parabola that if the 
 
 Fig.48 
 
 portion of any tangent UV intercepted between two others UT, 
 TV\)G projected on any line parallel to the directrix as on MM^ 
 by lines Um, Vm^ perpendicular to MM^ , then any other tangent 
 to the curve between the points of contact of TU, TV will have 
 fche same projected length mm^ on the axis. If therefore TM, 
 TM^ be each made equal to mm^ lines through M and M^ perpen- 
 dicular to MM^ will intersect TU, TV respectively in Q and Q^ , 
 
 E. 6 
 
82 GIVEN THREE TANGENTS AND DIRECTION OF AXIS. 
 
 the points of contact of TU and TV. The problem is therefore 
 reduced to Prob. 49, or it may be completed by utilising other 
 known properties of the curve already demonstrated, e.g. — making 
 the angle TQF equal to angle TQM, QF is a locus of the focus ; 
 similarly Q^F (the angle TQ^F being made equal to angle TQ^M^ 
 is a second locus, and F, the focus, is therefore the intersection of 
 QF, Q,F. 
 
 Again, the circle round UT, TV, VU is known to be a locus 
 of the focus (Prob. 48), and the angle UFQ is known to be equal 
 to the angle TUV. Prob. 42. Therefore, if on UQ a segment of 
 a circle be described containing an angle equal to the angle TUV 
 (Prob. 30), the intersection of this segment with the above circle 
 will determine F. Any number of tangents to the curve between 
 Q and Q^ can be at once drawn without previously determining 
 the focus by measuring the length mm^ anywhere on MM^ between 
 M and M^ and from the extremities drawing perpendiculars to 
 MM^ to meet TQ, TQ^. Any pair of such points being joined 
 will of course give a tangent to the curve. 
 
 Proof. That the projected length on MM^ of the portion of 
 any tangent intercepted between TQ, TQ^ is constant may be 
 shewn thus. Let R be the point of contact of UV and let the 
 diameter through R meet MM^ in t. Draw Qn, Q^n^ parallel to 
 UV meeting tR in n and n^. Then UR = ^Qn (Prob. 49) and 
 therefore tm = ^ tM. Similarly tm^ = J tM^ . 
 
 Therefore mm^ = J il/J/^ = constant, since MM^ is the projection 
 of the chord of contact of two fixed tangents. 
 
 Problem 53. (Fig. 49.) To draw a parabola, two points A, B 
 on the curve and two tangents TL, TM being given. 
 
 [The tangents must not be parallel and the points must not be 
 on opposite sides of either tangent.] 
 
 Draw a line through A and B meeting the given tangents in L 
 and M. Take LG on LM a mean proportional between LA and 
 LB (Prob. 5), and MD on ML a mean proportional between MB 
 and MA. Bisect CD in E. TE will be the direction of the axis 
 
THE PARABOLA. 
 
 83 
 
 of a parabola fulfilling the required conditions and CQ, DQ^ drawn 
 parallel to TE to meet the given tangents will determine Q and Q^ , 
 
 their points of contact. The problem therefore reduces to Prob. 49. 
 or may be completed similarly to the preceding. Since LC and 
 MD may be set off on either side of L and Jf, b& LC^, MD^ in the 
 figure, the point of bisection E^ of G^D^ determines TE^ the 
 direction of the axis of a second parabola fulfilling the required 
 conditions. Further, either C^D or CD^ may also be taken as the 
 segment to be bisected, and there are consequently ybwr solutions. 
 
 The proof depends entirely on the property of the parabola 
 already referred to in Prob. 47. 
 
 6—2 
 
84 GIVEN THREE POINTS AND A TANGENT. 
 
 Problem 54. (Fig. 50.) To draw a'parahola^ three points A^ B, 
 
 G, and a tangent LM being given. 
 
 [The points must all be on the same side of the tangent.] 
 Join two pairs of the given points as AB, BC and let the joining 
 
 lines cut the given tangent in L and M. On LB take LD a mean 
 
 Flg.50. 
 
 proportional between LA and LB (Prob. 5), and on MB take ME 
 a mean proportional between MC and MB. Then by the property 
 of the parabola already referred to (Prob. 47) a line through B 
 parallel to the axis of a parabola through A and B and touching 
 LM, will pass through the point of contact of LM with such 
 parabola ; and a line through E parallel to the axis of a parabola 
 through B and C and touching LM will pass through the point of 
 contact of LM with such parabola. Hence the line joining DE 
 will be parallel to the axis of a parabola which can be described 
 through AB and G to touch the given line, and its intersection 
 with LM will determine the point of contact of such parabola. 
 
 Since LD, ME can be set oflf on either side of L and M (as 
 LD^, ME^, similarly the line joining D^ and E^ will be parallel to 
 the axis of a second parabola fulfilling the conditions of the 
 problem ; its point of contact being P : and similarly DE^ and 
 D^E will determine the direction of the axes of two more such 
 parabolas. The line DE^ determines P^ as the point of contact. 
 
THE PAKABOLA. 85 
 
 Hence there are four solutions, and the problem in either case 
 is reduced to Prob. 51. In the fig. two of the four parabolas are 
 drawn, viz. those whose axes are parallel to D^E^ and DE^ respec- 
 tively; the necessary construction in each case being indicated. 
 
 It might be considered at first sight that if a mean proportional 
 were taken between the segments NA, NC of the line joining AG, 
 the third pair of the given points, cutting the given tangent in N, 
 two additional points would be obtained which, being joined to 
 either D, D^, E or E^, would give the directions of axes of ad- 
 ditional parabolas. This however is not so, since it will be found 
 that the points thus obtained coincide with the intersections of 
 ED, E^D^j and of BE^y ED^ respectively, and therefore no more 
 solutions than the four already mentioned are obtainable. 
 
 Problem 55. (Fig. 51.) To draw a parabola, a point A on 
 the curve and three tangents BG, GD, DB being given. 
 
 [No two of the tangents must be parallel, and the given point 
 must not lie within the triangle formed by the tangents, nor so 
 that any one tangent lies between it and either of the remaining 
 tangents.] 
 
 Let G be the vertex of the triangle formed by the tangents, 
 which cannot be reached from the given point without crossing 
 BD. Through B draw BE parallel to GB and through D draw DE 
 parallel to GB, meeting BE in E. Through G draw GK parallel 
 to BD and join EA meeting GK in K, GB in Z, and BD in M. 
 
 First let A lie between E and K ; complete the harmonic 
 range KAEA^, i.e. find a point A^ beyond E on KL such that 
 
 KA : KA^ :: AE : EA^ (Prob. 12.) 
 
 [Through A, K draw Aa, Ka any two lines intersecting in a, 
 produce a A to a^ making Aa^ = Aa. Join a^E and produce it to 
 meet Ka in b. Draw bA^ parallel to a A and it will intersect KA 
 in the required point.] 
 
 Then A^ will be a point on the curve and the problem reduces 
 to Prob. 53. 
 
 Second, let the given point lie beyond E as A^, then, com- 
 pleting the harmonic range A^EAK (Prob. 11), A will be a 
 
86 
 
 GIVEN THREE TANGENTS AND A POINT. 
 
 second point on the curve and the problem again reduces to 
 Prob. 53. 
 
 Fig.5l. 
 
 ^ 
 
 In completing the figure, one of the tangents employed should 
 be the one situated as BD in the figure, because it is necessary to 
 take a mean proportional between the segments of the chord AA^ 
 included between the tangent and the curve, i.e. to take a mean 
 proportional between MA and MA^: but it will be found that 
 MEy MK are each equal to such mean proportional, and therefore 
 E and K can be at once used without any further construction. 
 If GB is the second tangent made use of, a mean proportional LG 
 or LG^ must be determined between LA^ LA^ (Prob. 5), and 
 two of the four parabolas which can be constructed by means of pairs 
 of the points Ky E, G, G^ to pass through A and A^ and to touch 
 BLj BD will also touch CD. There is an ambiguity as to which 
 particular pairs of points must be selected, but this can easily 
 be settled by trial in any given case. In the fig. it will be found 
 that the pairs E, G and -£', G^ are those required, and that 
 
THE PARABOLA. 87 
 
 the pairs K, G and K, G^ give parabolas which while touching 
 BG, DB, do not touch CD. 
 
 There are in general two solutions. 
 
 Proof. It i? shewn at the end of Chap, iv. among the har- 
 monic properties of conies, that the three diagonals of a complete 
 quadrilateral circumscribing a conic form a self-conjugate triangle. 
 It is easily proved analytically that every parabola touches the line 
 at infinity, i.e. has one tangent situated at an altogether infinite 
 distance. Now BE and DE meet CD, CB respectively in in- 
 finitely distant points, pass, that is, through the points in which 
 this infinitely distant tangent meets CB and CD, they are there- 
 fore diagonals of the circumscribing quadrilateral formed by the 
 three given and the infinitely distant tangent, and its third 
 diagonal must be the line CK since this meets BD in infinitely 
 distant points. E is therefore the pole of the line CK, and con- 
 versely the polar of K passes through E. 
 
 But a straight line drawn through any point is divided har- 
 monically by the point, the curve and the polar of the point 
 (see end of Chap, iv.), therefore A^ must be a point on the 
 curve. 
 
 Problem 56. (Fig. 52.) To draw a parabola to pass through 
 four given points A, B, (7, D. 
 
 [The points must not lie at the angles of a parallelogram, and 
 must be so situated, that being joined in' pairs, the two points 
 of each pair are both on the same side, or on opposite sides of the 
 point of intersection of the joining line*.] 
 
 Join BC, AD to meet in E. Through C draw CK parallel to 
 AB meeting AD in K. Take a mean proportional EG between 
 ED and EK (Prob. 5) and CG will be the direction of the axis 
 of the required parabola. The Problem is therefore reduced to 
 Prob. 51. 
 
 Since the distance EG may be set ofi" on either side of E as 
 EG^, the line CG^ will be the direction of the axis of a second 
 parabola fulfilling the given conditions. 
 
 t* Puckle's Conic Sections. Fourth Edition, Art. 313, Ex. 1. 
 
 ¥ 
 
88 GIVEN FOUR POINTS. 
 
 Froof. From the construction 
 
 EB \EA :: EC : EK, 
 
 \ 
 
 ng.52. 
 
 r^ .-fi- 4^- 
 
 and EK : EG :: EG -.ED; 
 
 .-, EB :EA ::EC :^, 
 
 or EG' : EC :: ED.EA : EB, 
 
 which may be written 
 
 EG' : EC :: ED.EA : EC .EB, 
 
 a relation 'which is known to hold in the parabola. (Besant's 
 Geom. Conies, 3rd Ed., Art. 213.) 
 
 Problem 57. (Fig. 63.) To draw a parabola to touch four 
 given lines AB, BC, CD, DA, no two of which are parallel. 
 
 Let CD meet ^jB in -^ and AD meet BC in G. 
 
 The circle circumscribing the triangle formed by any three 
 of the lines will be a locus of the focus (Prob. 48), which may 
 therefore be determined as the intersection of the circles circum- 
 scribing any two of such triangles. In the figure, the circles 
 circumscribing BCE and ABG are drawn. They intersect in F, 
 
THE PARABOLA. 
 
 89 
 
 the focus. The tangent at the vertex can be at once determined, 
 by dropping perpendiculars from F on any two of the given 
 
 Fig.63, 
 
 tangents as FT, FY^ perpendiculars on AB, BG ; Y and Y^ are 
 points on the tangent at the vertex. (Notes to Problem 36.) 
 
 Problem 58. (Fig. 54.) To determine the centre of curvature 
 at any point P of a given parabola. 
 
 [A circle can be drawn through any three points of a curve, 
 but cannot in general be drawn through a greater number taken 
 arbitrarily. If a circle be drawn through three points of a curve 
 and the outside points be conceived to gradually move up to the 
 centre one, the circle in the limiting position it assumes when 
 the points approach indefinitely near to each other so as ultimately 
 to coincide, is called the circle of curvature at the point, and its 
 centre is called the centre of curvature. The circle is said to 
 pass through three consecutive points of the curve, and obviously 
 has closer contact with it at the point than any other circle can 
 
 L 
 
90 CIRCLE OF CURVATURE. 
 
 have, since it is not possible to draw a circle through fo%ir con- 
 secutive points. The centre of curvature will necessarily lie on 
 the normal at the given point, and any circle having its centre 
 on the normal and passing through the point really passes through 
 two consecutive points of the curve, since curve and circle have a 
 common tangent.] 
 
 F is the focus, PT the tangent, and PG the normal at the 
 point P of the given parabola. 
 
 Join PF and produce it to K^ making FK equal to FP. Draw 
 KO perpendicular to PK to intersect the normal at P in 0. 
 will be the centre of curvature at P. 
 
 Fig.54. 
 
 If the circle of curvature cuts the parabola again in Q^ it will 
 be found that PQ, the common chord, makes the same angle with 
 the axis as PT, the tangent, does, and that 
 
 pq = ^PT. 
 
 The focal chord FR of the circle of curvature is known to be 
 in length equal to 4i^P, and it is on this known value of the 
 focal chord that the construction depends. 
 
 The chord iPY) of the circle of curvature through P parallel 
 
THE PARABOLA. 91 
 
 to the axis is also equal to 4#P, since this chord and PR are 
 equally inclined to the tangent at P. 
 
 The length PO of the radius of curvature may also be deter- 
 mined by taking a fourth proportional to FY, FP and 2FPy where 
 FY is the perpendicular from F on the tangent at P. 
 
 The locus of the centre of curvature of any curve is called the 
 Evolute of that curve ; and the original curve, when considered 
 with respect to its evolute, is called an Involute. The chain- 
 dotted curve in Fig. 54 is the evolute of the portion of the parabola 
 lying above the axis. 
 
 Normals to the curve are tangents to the evolute ; and since 
 the focal radius of curvature at the vertex = 2 . AF^ the evolute 
 must touch the axis at a point = 2 . ^F from A. 
 
 If the ordinate of the point of intersection of the curve and 
 evolute be drawn meeting the axis in iV, it will be found that 
 
 AN —d) . AF = twice the latus rectum. 
 
 The evolute of the parabola is a curve known as the semi- 
 cubical parabola. 
 
 Problem 59. To draw a parabola to touch two given circlesy 
 the axis hei^g the line joining the centres. 
 
 Let G be the centre of the larger circle, c that of the smaller, 
 R and r their radii. Determine a fourth proportional to 2 (7c, 
 R ^r, and R-r. From G towards c set off on Gc a length GN 
 equal to this fourth proportional, i.e. a length such that 
 
 GN :R-r ::R + r : 2Gg. 
 
 Draw JVP perpendicular to Cc meeting the circle in P, and P 
 will be the required point of contact of the curve. The problem 
 therefore reduces to Prob. 47, the given point being also the point 
 of contact of the given tangent. 
 
92 EXAMPLES. 
 
 Examples on Chaptek III. 
 
 1. Draw a parabola, the focus F, the position of the axis 
 {FT) and a tangent {PT) being given. 
 
 (From F draw FY perpendicular to PT meeting it in Y, and 
 from Y draw YA perpendicular to FT meeting it in ^. A will 
 be the vertex of the required parabola.) 
 
 2. Draw a parabola, the focus F, a tangent PT and the 
 length of the latus rectum being given. 
 
 (With centre F and radius equal to one-fourth of the given 
 latus rectum, describe a circle ; from F draw FY perpendicular to 
 the given tangent meeting it in F, and from Y draw tangents to 
 the circle. Either point of contact will be the vertex of the re- 
 quired parabola (two solutions). The given tangent must not cut 
 the circle.) 
 
 3. Draw a parabola, two points (P, Q), the tangent at one of 
 them (PT), and the direction of the axis being given. 
 
 (Bisect PQ in V, draw VT parallel to given direction of axis 
 meeting the given tangent in T ; QT is the tangent at Q, and 
 problem reduces to Prob. 49.) 
 
 4. Draw a parabola, the vertex (P) of a diameter, and a cor- 
 responding double ordinate QQ^ being given. 
 
 (Bisect QQ^ in V. PFwill be a diameter; on VP produced 
 make PT=PV. TQ and TQ^ are the tangents at Q and Q^y and 
 problem reduces to Prob. 49.) 
 
 5. Draw the locus of the foci of the parabolas which have a 
 common vertex (A) and a common tangent PT. 
 
 (The parabola which has A for vertex, the perpendicular on 
 P2^ as axis, and the distance of PT from A as latus rectum.) 
 
 6. Inscribe in a given parabola a triangle having its sides 
 parallel to three given straight lines AB, BC, CA. 
 
 (Draw BD parallel to the axis of the parabola meeting AC in 
 
EXAMPLES. 93 
 
 B and GE parallel to the axis meeting AB in E. Draw a tangent 
 to the parabola parallel to BE (p. 61) and from P its point of 
 contact draw FQ, PR parallel to AB, AC meeting the parabola 
 again in Q, R. PQR will be the required triangle.) 
 
 7. Draw a parabola with a given focus, and to touch a given 
 circle at a given point. 
 
 [Let F be the focus, F the point on the circle, draw FT the 
 tangent, and construct an angle TFM —\hQ angle FFT. The axis 
 of the required parabola will be parallel to PJ/.] 
 
 8. Shew that if tangents be drawn to a parabola from any 
 point 0, and a circle be described with the focus as centre, passing 
 through and cutting the tangents in F and Q, FQ will be per- 
 pendicular to the axis, and its distance from is twice its distance 
 from the vertex. 
 
 9. Draw a circle to touch a parabola in P, and to pass through 
 the focus. Let it meet the parabola again in Q and Q^ : draw a 
 focal chord parallel to the tangent at P, and shew that the circle 
 on this chord as diameter will pass through Q, Q^ , and that the 
 focal chord and QQ^ will intersect on the directrix. 
 
 10. Draw any right-angled triangle BEF (E being the right 
 angle). Describe a parabola with focus F and to touch EB at Z), 
 and shew "^at if any circle be described to pass through B and F 
 and cutting EB produced in F, the tangent to it at F will also be 
 a tangent to the parabola. 
 
 11. Given two lines PR, QR, and a point F on one of them, 
 shew that any point on the circumference of the circle passing 
 through P and R and touching QR may be taken as the focus of 
 a parabola passing through F and to which the given lines shall 
 be tangents. 
 
 12. AJB IB the diameter of a circle; with A as focus and any 
 point on the semi-circumference of which A is the centre as foot of 
 directrix describe a parabola, and shew that it will touch the 
 diameter perpendicular to AB. 
 
94 EXAMPLES. 
 
 13. If APC be a sector, of a circle of which the radius CA is 
 fixed, and a circle be described touching the radii CA, CP and the 
 arc AP, shew that the locus of the centre of this circle is a parabola 
 and describe it. 
 
 14. Given a segment of a circle, describe the parabola which 
 is the locus of the centres of the circles inscribed in it. 
 
 15. If from a point P of a circle PC be drawn to the centre, 
 and R be the middle point of the chord PQ drawn pai*allel to a 
 fixed diameter ACB, describe the locus of the intersection of CP, 
 AR, and shew that it is a parabola. 
 
 16. Describe a parabola with latus rectum = 2*7 units, and in 
 it draw a series of parallel chords inclined at 60° to the axis. 
 Shew that the locus of the point which divides each chord into 
 segments containing a constant rectangle = 4 sq. units in area, is 
 a parabola, the axis of which coincides with the axis of the original 
 parabola and with the latus rectum = 2*1 units. 
 
 17. Draw a parabola to touch the three sides of a given 
 triangle, one of them at its middle point ; and shew that the per- 
 pendiculars drawn from the angles of the triangle upon any- 
 tangent to the parabola are in harmonical progression. 
 
 18. Given two unequal circles (centres (x and ^, radii R and r) 
 touching each other externally, from G the centre of the larger 
 
 R — T 
 
 circle make GR on Gg towards g = . Draw NP perpen- 
 
 2 
 
 dicular to Gg meeting the circle in P and describe a pai-abola 
 
 with Gg as axis and to touch the circle in P (Prob. 47), and 
 
 shew that it will also touch the smaller circle. 
 
 19. Given a point F and two straight lines intersecting in ; 
 describe a parabola with F as focus and to touch the given lines 
 (Prob. 42) ; and shew that if any circle be described passing 
 through and F and meeting the lines in P and Q, PQ will be a 
 tangent to the parabola. 
 
EXAMPLES. 
 
 95 
 
 20. Draw the parabola which is the locus of the centre of a 
 circle passing through a given point and cutting off a constant 
 intercept on a given straight line. 
 
 (The point is the focus and a perpendicular to the line the 
 axis.) 
 
 21. Given four tangents to a parabola, shew that the directrix 
 [is the radical axis of the system of circles described on the diagonals 
 [of the quadrilateral as diameters. 
 
 22. Given the focus F^ a point P on the curve and a point 
 on the directrix, describe the parabola. 
 
 [Tangents from L to the circle described with centre P and 
 radius PF are the directrices of two parabolas fulfilling required 
 [conditions.] 
 
 23. Given a focus F^ a tangent PT^ and a point L on the 
 directrix, describe the parabola. 
 
 [From F draw a perpendicular FY to PT meeting it in F; 
 )roduce FY to / and make Yf = FY : / is a second point on the 
 pdirectrix.] 
 
 24. Given three tangents to a parabola and a point on the 
 [directrix, draw the curve. 
 
 [The ortho-centre of the triangle formed by the tangents is a 
 md ^oint on the directrix.] 
 
CHAPTER IV. 
 
 THE ELLIPSE. 
 
 The ellipse has already been defined (p. 56) as the locus of a 
 point which moves in a plane so that its distance from a fixed 
 point in the plane is always in a constant ratio, less than unity, 
 to its distance from a fixed line in the plane. The corresponding 
 definition in the case of the parabola furnishes immediately the 
 best condition for the geometrical construction of that curve, but 
 this is not so with the ellipse. The ellipse can be more easily 
 constructed geometrically from a property which will be shewn 
 immediately to be involved in the above definition, and in virtue of 
 which the curve may be defined as follows : — 
 
 Dep. The ellipse is the locus of a fixed point on a line of 
 constant length moving so that its extremities are always on two 
 fixed straight lines perpendicular to each other. 
 
 In Fig. 55 let AC A ^, BCB^ be two straight lines intersecting 
 each other at right angles in C. If a length (as ah) be marked ofi" 
 on the smooth edge of a slip of paper, and the slip be moved 
 round so that the point a is always on the line BCB^ and the 
 point h on AGA^, then any point as P on the edge of the paper 
 will trace out an ellipse. When the edge of the slip coincides 
 with AGA^ the tracing point will evidently be at a distance GA 
 from G equal to aP, and when it coincides with BGB^ the tracing 
 point will be at a distance GB from G equal to hP. By this 
 method of construction the curve is evidently symmetrical about 
 both the lines AGA^ and BGB^^ i.e. if C^^ be made equal to 
 C-4, A^ will be a point on the curve, and if GB^ be made eqttal 
 
THE ELLIPSE. 97 
 
 to CB, B^ will be a point on the curve. It is moreover obvious 
 that ACA^ is the longest and BCB^ the shortest line which can 
 be drawn through G and terminated by the curve. 
 
 Def. The line ACA^ is called the major axis, the line BCB^ 
 the 'minor axis, the point C the centre, and the points A, A^ vertices 
 of the curve. 
 
 From B, the extremity of the minor axis, as centre with 
 radius CA (the semi-major axis), describe arcs cutting the major 
 axis in F and F^; through B draw B3I parallel to CA, from F 
 draw FM perpendicular to BF meeting B3f in M, and draw 3fX 
 perpendicular to CA meeting it in X. 
 
 F will be the focus and 3IX the directrix (see definition, page 
 56). 
 
 From the similar triangles FB3f, CFB, 
 
 
 FB : BM :: CF : CA :: 
 
 CA : CX, 
 
 since 
 
 FB = CA and BM- 
 
 -CX; 
 
 
 .-. CF : CA-CF :: CA 
 
 : CX-CA^ 
 
 i.e. 
 
 CF : FA :: CA 
 
 : AX, 
 
 or 
 
 FA : AX :: CF : CA 
 
 :: FB : BM; 
 
 also since 
 
 CF : CA :: CA : 
 
 CX, 
 
 .*. 
 
 CA-CF : CF+CA :: CX- 
 
 ■CA : CA + CX, 
 
 or 
 
 FA : FA, :: AX 
 
 :A,X, 
 
 i.e. i 
 
 FA : AX :: FA, 
 
 : A,X; 
 
 therefore A, B and A, are points satisfying the original definition. 
 Def. a circle described on the major axis as diameter is 
 called the auxiliary circle. 
 
 Through any point F on the ellipse draw the ordinate PX 
 (perpendicular to major axis) meeting the axis in N and the 
 auxiliary circle in Q. Since QX is parallel to BC and CQ = aP, 
 .'. aP is parallel to CQ, 
 .-. PX : QN :: Ph : QC 
 :: BC : AC, 
 or PX\' : QX\' :: BC\' : AC^'; 
 
98 AUXILIARY CIRCLE, 
 
 but it is known that in the circle 
 
 .'. Wf : AN.NA, :: BC^ : AC\\ 
 This is a very important property of the ellipse and will now be 
 shewn to result from assuming the ratio FF : NX to be con- 
 stant. 
 
 Through P draw PA, PA^ meeting the directrix in E and //. 
 Join FH and draw PLK perpeudicular to the directrix meeting 
 FH in L and the directrix in K. 
 Since PK is parallel to A^X, 
 
 .'. PL : PK :: FA, : A,X 
 :: FA : AX. 
 But by supposition FP : PK :: FA : AX, 
 therefore FP = PL, and the angle LFP :^ FLP = the alternate 
 angle LFX ; 
 i.e. FL bisects the angle PFX; 
 
 similarly FF bisects the angle between FX and PF produced, 
 therefore the angle EFH is a right angle, since it is made up of 
 the two angles EFX and HFX. 
 
 By the similar triangles PAX, A EX, 
 PN : AN :: EX : AX, 
 also PX : A.X :: BX : A,X, 
 
 .-. PX\' : AX.XA, :: EX. EX : AX.A.X 
 :: FX^ : AX. A,X, 
 since EFH is a right angle ; 
 
 Le. PN^ is to AN . XA, in a constant ratio. 
 
 Hence taking PN coincident with BC, in which case 
 AN=NA, = AC, 
 BCf : IC,^ :: FX(' : AX.A.X, 
 and .♦. PN' : AN.NA, :: EC : AC\ 
 
 This of course shews that the point P is the same whether deter- 
 mined as the locus of a fixed point on a line of constant length 
 
THE ELLIPSE. 99 
 
 sliding between two fixed rectangular axes or as tlie locus of 
 a point which moves so that its distance from a fixed point (^F) is 
 in a constant ratio to its distance from a fixed line [MX), i.e. the 
 two definitions of the ellipse already given are really identical. 
 From the symmetry of the curve it is evident that F^ is a second 
 focus and M^X^ a second directrix. 
 
 Five geometrical conditions are generally necessary to deter- 
 mine an ellipse, and the ellipse shares with the hyperbola the 
 property of satisfying five geometrical conditions. One or other 
 of these curves can generally be drawn to pass through five given 
 points or to touch five given straight lines, or to pass through two 
 given points and touch three given lines, or to fulfil any five 
 similar conditions. Which curve will satisfy the given conditions 
 depends of course upon the relative positions of the given points 
 and lines, and the necessary limitations will be noticed in discuss- 
 ing the separate problems. As in the case of the parabola the 
 giving of certain points and lines is really equivalent in each case 
 to the giving of two geometrical conditions; of these may be 
 mentioned the centre, the foci, and the axes. 
 
 The eccentricity of the ellipse is (p. 57) the numerical value of 
 the above fixed ratio ; it is generally denoted by e and calling 
 
 CA=a 
 and CB = h, 
 
 its value is eJ-^H, 
 
 a 
 
 as is evident from the similar triangles FBM, FOB. 
 
 Problem 60. (Fig. 55.) To describe an ellijyse having given 
 axes AA^, BB^. 
 
 First Method. Draw two lines perpendicular to each other 
 intersecting in C. Set off CA^ CA^ each equal to ^AA^, and CB^ 
 CB^ each equal to -| BB^. Take a smooth edged slip of paper and 
 mark off on it Pa = CA and Fh = CB (a and h may be on the 
 same or on opposite sides of F). Keep the point a on the minor 
 axis and the point b on the major axis and (as already demon- 
 
 7—2 
 
100 
 
 GIVEN THE AXES. 
 
 strated) the point P will be on the curve. Any number of points 
 may thus be determined. In the lower portion of the figure the 
 
 Fig.55. 
 
 lengths CA, CB are shewn set off on opposite sides of P, and this 
 arrangement is the better when the lengths AA^, J55j are nearly 
 equal, as in that case, when set off" on the same side of P, the 
 distance ah is too short to determine the direction of Pa with 
 accuracy. 
 
 Second Method (fig. 56). Arrange the axes as above, and on 
 each as diameter describe a circle. Draw any number of radii as 
 CI, C2, &c. From the extremities of the radii of the circle on the 
 major axis draw lines parallel to the minor axis, and from the ends 
 of the radii of the circle on the minor axis draw lines parallel to the 
 major axis. The lines drawn from corresponding points (as 7P, 
 TP) will intersect on the required ellipse, which can therefore 
 be drawn through the points thus determined. 
 
 The proof is at once obvious by drawing through any point P 
 on the curve a line parallel to the coiTespouding radius (77, 
 cutting the axes in h and a. Then 
 
 Ca P7 is a parallelogram, and .*. Pa-C7 = CA, 
 Cb PT is a parallelogram, and .-. P6 = C7' = CB, 
 
THE ELLIPSE. 
 
 101 
 
 so that the points found by this conHruc'ti6ii ' are identical with 
 those found by the first. 
 
 Third Method (fig. 56). Determine the foci; — i.e. from the end 
 of the minor axis {B^ as centre describe an arc with radius = CA 
 
 cutting AA^m. F and F^. Stick a pin firmly through the paper 
 at each of the three points B^^ F^ F^^ and tie a fine thread or piece 
 of silk lightly round these pins, keeping it down in contact with 
 the paper while doing so. Take out the pin at By and keeping 
 the string stretched with the point of a pencil, the curve may be 
 drawn by moving the pencil round the circuit. This method is 
 theoretically perfect, but it fails in practice to give a very exact 
 result chiefly owing to the extensibility of the string and the 
 impossibility of keeping it at a constant tension. It is difficult 
 moreover to tie up the loop of the string to exactly the proper 
 length and to keep the string continually in contact with the 
 paper. Its use therefore cannot be recommended, but it illustrates 
 a very important property of the ellipse, viz. That the sum of the 
 focal distances of any 'point on the ellipse is constant and equal to 
 the major axis, which may be proved thus : 
 
102 TANGENT AT ANY POINT. 
 
 In fig. 55, P is atiy point on the ellipse, 
 
 FF : FK :: FA : ^X; 
 also F^F : FK, :: F^A : AX,, 
 
 .♦. FP + FF,: FK+PA\:: FA + F,A : AX + AX, 
 : XX, :: AA, : XX„ 
 i.e. FF + FF, = AA,. 
 
 To draw the tangent at any point of the curve. 
 If Q, and Q^ (fig. 5o) be any two adjacent points of the curve, 
 and the straight line drawn through them meets a directrix in 
 f draw QJc,, QJc^ perpendicular to the directrix and draw fF^ to 
 the corresponding focus. 
 
 Then F^Q, : F^Q,^ :: QJc, : QJc^ 
 
 :: QJ : QJ, 
 therefore F^f bisects the exterior angle between Q,F, and Q F,. 
 (Euc. VI. Prop. A.) 
 
 Hence, exactly as in the case of the parabola (p. 59), when 
 (?2 moves up to and coincides with Q, so that the line through 
 Q1Q2 becomes the tangent at Q, (Def. p. 30), the line F,f becomes 
 perpendicular to the line joining the focus to the point of contact 
 of the tangent. The tangent at any point Q, of an ellipse may 
 therefore be drawn by drawing a line from Q, to either focus, 
 erecting a perpendicular to this line at the focus meeting tlie 
 directrix, and drawing the tangent through this point and the 
 proposed point of contact. It may also be drawn by using the 
 known property that the normal bisects the angle between the focal 
 distances, which may be proved thus. In fig. 56 Q is any point of 
 the curve, i^ is a focus, and FS is perpendicular to QF meeting the 
 corresponding directrix in S so that QS is the tangent at Q. 
 Draw the normal QG perpendicular to QS meeting the major axis 
 in G, and draw FD perpendicular to the major axis meeting QS in 
 J9, and QK perpendicular to the directrix. Join FK. 
 
 The angle QFG is the complement of QFD 2iwd. is therefore 
 equal to the angle SFD ; the angle FQG is the complement of 
 SQF and is therefore equal to the angle FSD^ and therefore the 
 triangle QFG is similar to the triangle SFD. 
 
THE ELLIPSE. 
 
 103 
 
 Hence FG : FQ :: FD : FS (1). 
 
 But since 8FQ^ SKQ are right angles, a circle can be described 
 round FSKQ, and therefore the angle FSQ = the angle FKQ. 
 
 Also the angle ^T^'^r^the angle FQK since GF is parallel 
 to QK, therefore the angle FQK = SFD, therefore the triangle SFD 
 is similar to the triangle KQF, and 
 
 i< 
 1 
 
 bi 
 
 Hence SQT being the tangent the angle SQF is equal to the 
 angle TQF^ or the tangent is equally inclined to the focal distances 
 of the point of co7itact. It follows that if F^Q be produced to L 
 the tangent bisects the angle FQL. 
 
 Problem 61. To describe approximately by means of circular 
 arcs, an ellipse having given axes. 
 
 First Method (fig. 57). CA, GB, CA^, CB^ are the semi-axes. 
 Draw A^M parallel to GB and BM parallel to CA meeting in M. 
 
 .'. FD : FS :: FQ : QK :: FA : 
 
 AX, 
 
 
 .-. FG : FQ :: FA : AX 
 
 
 from (1); 
 
 similarly F,G : F,Q :: FA : AX, 
 
 
 
 and .-. FG : Ffi :: FQ : F,Q, 
 
 
 
 or the angle FQF^ is bisected by the normal QG. 
 
 
 (Euc. VI. 3.) 
 
 Fig.57. 1^'p^ 
 
 
 ^ 
 
 3 
 
 V'05 
 
 L— -= 
 
 
 M 
 
 //' 
 
 
 \° 
 
 K 
 
 V 
 
 
 c 
 
 
 / 
 
 'O, 
 
 Bisect A^M in D. Join BD and draw MB^ cutting BD in P. P 
 will be a point on the true ellipse with axes AA^ and BB^, Bisect 
 
104 
 
 APPROXIMATE ELLIPSE BY CIRCULAR ARCS. 
 
 PB in E. Draw EO^ perpendicular to PB meeting BB^ in 0^, and 
 with centre 0^ and radius to ^ or P draw the arc PBF meeting in 
 i^a line through 0^ parallel to A Ay Draw FA and produce it to 
 meet the arc in G. Draw GO^ cutting ^^j in 0^, and with centre 
 0^ and radius Ofi draw an arc which will be found to pass 
 through A, since by the similar triangles GO^A, GO^F, 
 GO^ : AO^ :: GO, : FO,, 
 i.e. Ofi=O^A. 
 The two arcs AG and GB form one quadrant of the approximate 
 ellipse and the remainder can of course be put in by symmetry, 
 taking centres 0^ and 0^ in corresponding positions to those 
 already obtained. 
 
 Second Method (fig. 58). Draw AM, ^i/ parallel respectively 
 to BG, AG, meetiug in M. Draw MO, perpendicular to AB, 
 
 Fig.58. 
 
 cutting BB, in 0, and AA^ in 0^. Find a mean proportional 
 {BD) between GA and CB. (This may conveniently be done by 
 making Be on MB produced equal to BG, and describing a semi- 
 circle on Mc cutting BC in D.) Make AE equal to BD. With 
 centres 0,, 0^ and radii 0,D, O^E describe arcs intersecting in 0^. 
 Then 0^, 0^, 0.^ are points which can be used as centres for 
 
THE ELLIPSE. 105 
 
 successive arcs of the required curve. The arc struck from Oi will 
 pass through B and extend of course to F on the line Ofi^, that 
 from 0^ will pass through F and extend to G on 0^0.^, and that 
 from Og will start from G and pass through A. Thus each 
 quadrant will consist of three arcs, and the centres for the 
 other three quadrants can be taken by symmetry. 
 
 The arc struck with centre 0^ and radius Ofi will evidently 
 pass through A, since GO^ = FO^ = BD = AE and 
 
 GO^ = GO^-Ofi^ = GO^-O^E = AE-O^E-^AO^. 
 It will be shewn hereafter that the points 0^ , 0^ are the centres of 
 curvature at B and A respectively ; the circular arcs struck with 
 these centres through B and A coincide therefore more nearly with 
 the true ellipse at those points than any others which can be 
 drawn. 
 
 Definition. Any line drawn through the centre of the 
 ellipse and terminated both ways by the curve is called a dia- 
 meter, and a semi-diameter CD parallel to the tangent at the 
 extremity of a semi-diameter GP is said to be conjugate to CP. 
 Every diameter is evidently bisected by the centre. 
 
 The following important properties of the ellipse should be 
 carefully noticed. 
 
 Prop. 1. Tangents drawn at the extremities of any chord sub- 
 tend equal angles at the focus. 
 
 Let PP^ (fig. 69) be any chord of an ellipse, and let the 
 tangents at P and P^ meet in T. Let F be the focus, and from T 
 draw TM, TM^ perpendicular to FP, FP^, and draw TN perpen- 
 dicular to the directrix XS. Let the tangent at P, meet the 
 directrix in S, then FS is perpendicular to FP^ and therefore 
 parallel to TM^, 
 
 .-. FM^ :FP^ ::ST:SP^ 
 :: TI^ : P^K, 
 where P^K is perpendicular to the directrix ; 
 
 .-. FM^ : TN y.FP^ .PJC 
 :: FA -.AX, 
 
106 
 
 Similarly 
 
 CONGUGATE DIAMETERS. 
 
 FM: TN::FA : AX, 
 
 
 
 Fig.59. 
 
 
 
 
 Q 
 
 q 
 
 
 
 ^■rii^ 
 
 
 
 
 T, 
 
 V-'M, 
 
 
 ^c \ 
 
 
 
 Hence in the right-angled triangles TFM, TFM^, FM = FM^ and 
 TF is common, therefore the triangles are equal in all respects, 
 ie. the angle TFP equals the angle TFP^ and TM= TM^. 
 
 Prop. 2. A diameter bisects all chords parallel to the tangents 
 at its extremities, i.e. all chords parallel to its conjugate. 
 
 Let QQ^ (fig. 59) be any chord of an ellipse meeting the 
 directrix in R and let be the centre point of QQ^ and F^ the 
 focus. Join F^Q, Ffi^ and draw -F^ F perpendicular to QQ^, then 
 
 FQ\' -tq;]' ='QY\' - QJ\" 
 
 = '2QQ^^0Y (1); 
 
 but since Q and Qi are on the ellipse 
 
 F,Q : Ffi, :: QE : Q^E, 
 F,Q'-F^Q^' ^ QR'-Qfi' ^ WE. QQ^ 
 
 ' ' T\Q\' 'QE\' Q^' * 
 
 therefore from (1) and (2), 
 
 (2); 
 
 0E~ Qe\ "yljr 
 where ^jlFis drawn through the vertex parallel to QE, meeting 
 the directrix in W j i.e. OY : OE in a constant ratio. 
 
THE ELLIPSE. 
 
 107 
 
 Take any second chord qq^ parallel to QQ^^ meeting T^^Fin Y^ 
 and the directrix in i?j, let 0^ be its centre point, then since 
 
 OY Y 
 
 — ^ = .^^ , it follows that the line 00 must pass through the 
 OK O^Jti^ 
 
 point T^ in which F^ Y meets the directrix and is therefore fixed 
 
 for all chords parallel to QQ^. This line T^O will pass through 
 
 the centre (i.e. will be a diameter), because the chord through 
 
 the centre parallel to QQ^ is bisected by the centre and also by 
 
 Tfi. Let T^O meet the ellipse in P^ and suppose qq^ to move 
 
 parallel to itself till it approaches and ultimately passes through 
 
 Pg, Since O^q = O^q^ throughout the motion the points g, q^ will 
 
 evidently approach P^ simultaneously, and in the limiting position 
 
 qq^ will be the tangent at P^. It follows that if P^ be the other 
 
 extremity of the diameter through P^^ the tangent at P^ is 
 
 parallel to QQ^^ and therefore to the tangent at P^. 
 
 Corollary. The 'perpendicular on the tangent at any point 
 
 from the focus meets the corresponding diameter in the directrix. 
 
 Prop. 3. If PCP^ he a diameter and Q VQy^ a chord parallel 
 to the tangent at P and meeting PP^ in V, and if the tangent at Q 
 meet PP, produced in T, then GV . CT=UF\^ (fig. 60). 
 
 Let TQ meet the tangents at P and P^ in R and r, and F 
 being a focus draw RN perpendicular to the focal distance FP 
 
 Tn. 
 
 
 Fig.60. 
 
 m 
 
 K 
 
 \ 
 
 
 ^" 
 
 -5^^^ 7^ 
 
 > 
 
 / 
 
 4 
 
 
 
 x^^ 
 
 '^ 
 
 / 
 
 -^ 
 
 meeting FP in JV, rn perperpendicular to FP^ meeting it in 
 and RM, rm perpendicular to the focal distance FQ. 
 
108 CONJUGATE DIAMETERS. 
 
 Let F^ be the other focus and join F^P, F,P^. 
 
 Since CF^CF,, CP^CP,, and the angle i^CP = the angle 
 F^CP^, therefore the triangles FGP, F^CP^ are equal in all respects, 
 and therefore the angle C7Pi^=the angle GP^F^ ; 
 similarly CPF, = the angle CP,F, 
 
 and therefore the whole angle i^Pi^'j^the whole angle F.P^F 
 But the tangents are equally inclined to the focal distances, and 
 therefore also the angle FPP = the angle F^P^r, 
 
 .'. the angle FPP = the angle FP,r, 
 i.e. the right-angled triangles RPN, rP{ti are similar, and there- 
 fore RP : rP, :: RN : rn. 
 But RN= RM and rn = rm (Prop. 1), 
 
 But 
 
 or CT-CP 
 
 .'. CT . CV=--CP\ 
 
 Cor. 1. Since CFand GP are the same for the point (^d the 
 tangent at Q^ passes through T or the tangents at the extremities 
 of any chord intersect on the diameter which bisects that chord. 
 
 Gor. 2. Since TP, : TP :: P,V : VP, it follows that TPVp is 
 harmonically divided (p. 13). 
 
 The above proposition has been proved generally; it therefore 
 holds when the diameter GP coincides with the major axis. Let 
 Pj be any point on an ellipse (fig. 56) and draw the ordinate P^j^ 
 perpendicular to GA, producing it to meet the auxiliary circle in 
 jt), and draw the tangent at P, meeting GA in T, then 
 
 G^.GT^GA' = G2A 
 and .'. (7/»P is a right angle, 
 
 and therefore pT is a tangent at p to the auxiliary circle : hence 
 
 RP : rP, :: 
 
 RM : rm 
 
 
 RQ : rQ. 
 
 TR : Tr :: 
 
 RP : rP,, 
 
 TP : TP, :: 
 
 PY : PJ, 
 
 : GT+GP :: 
 
 GP-GV : GP + PV, 
 
 .-. GT : GP 
 
 :: GP : CV, 
 
THE ELLIPSE. 109 
 
 Cor. 3. The tangents at the extremities of corresponding 
 ordinates of the ellipse and auxiliary circle intersect on the 7najor 
 axis. 
 
 Draw CD (fig. b&) the diameter conjugate to CP^, dDn the 
 corresponding ordinate meeting the auxiliary circle in d, and the 
 tangents at D and d meeting the major axis in t. 
 
 Then F,N : pN :: BC : AC '.: Dn : dn, 
 
 and P,N -.NT.-.Dn: Cn, 
 
 since CD is parallel to P^T, 
 
 .'. pN : NT y. dn : Cn, 
 and therefore Cd is parallel to pT, i.e. pCd is a right angle, or 
 
 Cor. 4. Conjugate diaineters in the ellipse project into dia- 
 meters at right angles to each other in the auxiliary circle. 
 
 If the tangent at d meet the major axis in t, since dt is 
 parallel to Cj), Dt (the tangent at D) will be parallel to CF^, or, 
 
 Cor. 5. If CD he conjugate to CP^ , CP^ is also conjugate to CD. 
 
 Since pCd is a right angle, the angle dCn is the complement 
 of the angle pCN, and therefore equals the angle CpN, therefore 
 the triangles CpN^ dCn are equal in all respects, i.e. Cn-pN 
 and dn = CN, 
 
 CP,^ = P, N' + CN' and CD' = Cif + Dn\ 
 ^ .'. Cl\\' + UD\'^CN\'+^\'^I\N\'+~M\' 
 
 ^CA\' + P;Nf + Dn^ (1). 
 
 But P,N : pN :: Dn : dn :: JBC : AC, 
 
 .'. P,N: Dn r.pN -.dn, 
 . •. P,N' + Dn' : P,N' y. pN' + dn' : pN', 
 and pN' + d7i' = AC'; 
 
 or P^N' + Dn' {Jv\' :: JBC : AC, 
 
 .-. P;N\' + Dn' = JBC\ 
 Therefore, from (1) 
 
 Cor. 6. CPr- + CD? =^~CA\ ' +C1B,\ 
 
110 
 
 AXES FROM CONJUGATE DIAMETERS. 
 
 Prop. 4. If PCP^^ DGD^ be conjugate diameters and QVhe 
 drawn parallel to CD meeting the ellipse in Q and CF in F, then 
 QV : PV. VP, :: CD' : CP\ 
 [^F is called an ordinate of the diameter PCP^.] 
 
 Let the tangent at Q (fig. 60) meet CP, CD in T and t, and 
 draw ^t/" parallel to CT meeting CD in U. 
 
 Then CV. CT= CP' and CU . Ct = CD' (Prop. 3). 
 But CU=QV, 
 
 .-. CD' : CP' :: QV.Ct : CV.CT 
 :: Wr : C'F. Fr. 
 Since C^ : (?F :: Cr : Vl\ 
 
 and CF. Fl^^CF. Cr-CF^ 
 
 = CP^-CF^ = PF. FP„ 
 .-. ^F' : PF. VP, :: CD' : (7P^ 
 
 Problem G2. Given a pair of conjugate diameters to determine 
 the axes (fig. 61). 
 
 F«g.6I, 
 
THE ELLIPSE. Ill 
 
 PCP^, BCD^ are tlie given conjugate diameters. Tlirongh D 
 draw Q^DQ perpendicular to CP. Make DQ and DQ^ each equal 
 to CP and draw the lines C(?, CQ^. Then the major axis ACA^ 
 bisects the angle QCQ^ and the minor axis (BCB^) is of course 
 a line through G perpendicular to ACA^. The axes are therefore 
 determined in direction. To determine them in magnitude : — 
 On Qfi on opposite sides of C make Cq and Cq^ each equal to CQ, 
 then Q^q will be the length of the major axis AA^ and Q^q^ will 
 be the length of the minor BB^. Bisect each of these lines and 
 CA, CB will be given respectively. 
 
 Proof. Since CQ--Cq and BC bisects the angle QGq, therefore a, 
 the point in which Qq cuts BC, bisects Qq and therefore Da is parallel 
 
 to Q^q and = -^ = CA. Similarly h the point in which Qq^ meets 
 
 CA bisects Qq^ and Dh is parallel to Q^q and = — ]y - = CB, and 
 
 D, h, a are in the same straight line. Hence D is a point on 
 the ellipse described with CA, CB as semi-axes. Also DQ is 
 the normal at D, since Q is the instantaneous centre of rotation 
 for the line ab moving along the axes. Therefore the tangent at D 
 will be parallel to CP. Lastly, to shew that P will also be on the 
 curve, 
 
 CQ'-^CQ,' = 2CD' + 2DQ' (Euc. ii. 12 and 13), 
 
 = 2(CD'+CP'). 
 But CQ, = AC + BC, 
 
 and * CQ = AC-BC, 
 
 .-. CQ' + CQ^' = 2{AC' + BC% 
 
 .-. CD'+CP' = AC' + BC% 
 a known property of conjugate diameters. (See Cor. 6, p. 109.) 
 
 Problem 63. 2^o describe an ellipse having given conjugate 
 diameters PCP^ , DCD^ . 
 
 This might of course be done by the last problem : the curve 
 may however be drawn inde{)endently, though none of the follow- 
 ing constructions give any information as to the position of the 
 axes, foci, or directrices. 
 
112 
 
 GIVEN CONJUGATE DIAMETERS. 
 
 First Method. By contimioiis motion (fig. 62). From C draw 
 Ca perpendicular to CD and through P draw Pa parallel tc» 
 
 CD meeting Ca in a 
 
 On CP make Cd=GD and on Ca make Cp = CP. Through 
 a draw ah parallel to pd meeting GP in h. If a triangle equal 
 and similar to the triangle ahC be moved round so that the 
 angle a is always on the diameter DCD^ and the angle h on PCP , 
 the angle C will be on the curve. The most convenient way of 
 proceeding practically is to cut a strip of paper of breadth equal 
 to the perpendicular distance between C and ah. The points 
 a and h can then be marked off on one edge (as at aj)^ and the 
 point C on the other edge (as at CJ. The slip can easily be 
 adjusted in any number of positions and the corresponding 
 positions of C^ marked. Any number of points on the curve may 
 thus be determined"*. 
 
 Second Method (fig. 63). Draw PM, P^M^ parallel to CD and 
 DMM^ parallel to CP meeting PM in AT and P^M^ in M^. 
 Divide MD into any number of equal parts 1, 2, 3... and CD 
 into the same number of equal parts. Then lines drawn from P 
 to any of the points on MD intersect lines drawn from P^ through 
 the corresponding points on CD in points on the curve, and thus 
 any number of points in the quadrant PD can be determined. 
 
 * I am indebted to Prof. Minchin for this construction. 
 
THE ELLIPSE. llo 
 
 Similarly if M^D be divided into any number of equal parts 
 r, 2', 3'... and CD into the same number 1, 2, 3... lines drawn 
 from Pj to the points on M^D intersect lines drawn from P to 
 
 the corresponding points on CD in points on the required curve, 
 and thus the quadrant DP^ can be determined. When half the 
 curve is drawn the remainder can be put in by symmetry, since 
 every diameter is bisected by the centre ; thus if QCQ^ be drawn 
 and CQ^ be made equal to CQ, Q^ will be a point on the curve 
 and similarly for any other points on the semi-ellipse P, D^, P^. 
 
 Third Method (fig. 64). PCP^ , DCD^ are the given conjugate 
 diameters. Draw PM, P^M^ parallel to CD and MDM^ parallel 
 to CP meeting PM in if and P^M^ in M^. 
 
 Draw the line PD and take on it any number of points 1, 2, 3... 
 
 Draw the lines la, 26, 3c... parallel to CD meeting DM in 
 a, 6, c... ; and the lines J/^l, J/j2, M^Z... meeting MP in a\ b', c .... 
 Then the lines aa\ hh\ cc ... will be tangents to the curve, which 
 must be drawn in to touch these lines, so giving the quadrant PD. 
 A similar construction will give a second quadrant DP^^ and 
 the remaining semi-ellipse can of course be put in similarly or by 
 drawing any number of diameters. 
 
 Fourth Method (fig. 64). Draw PM^, P,^,^ parallel to CD^ 
 
 and MJ)^M^ parallel to PP^ meeting PM^ in J/^ and PM^ in M^. 
 
 Make M^E^P^M^ and divide PM^ into any number of equal 
 
 parts as at 1, 2, 3.... Draw E\, El\ E7>... cutting D^M^ irif,g,h 
 
 E. 8 
 
114 
 
 GIVEN AXIS AND POINT. 
 
 respectively. Then tlie lines joining corresponding points on PM_^ 
 and i/gZ^i, as/3, g1 and so on, will be tangents to the curve, which 
 must therefore be drawn touching these lines. 
 
 Similarly for the remaining quadrants, or as before, when half 
 the ellipse is obtained the other can be put in by symmetry. 
 
 To draw a tang&nt at any point of an ellipse having a given 
 pair of conjugate diameters. 
 
 Let Q (fig. 64) be the point, PCP^, DCD^ the given conjugate 
 diameters. Draw ^iV^ parallel to CP meeting CD^ in A", so that 
 
 CN is the abscissa and QN the ordinate of Q referred to the given 
 conjugate diameters as axes. Make Cn on CP^ equal to GN and 
 Cd = CJDi and draw through d a line dT parallel to nDi cutting 
 CD^ in T. The line QT will be the tangent at Q, for by similar 
 
 triangles 
 
 I.e. 
 
 CTiCd ::CD, : Cn, 
 CT '.CD,:: CD, : CN (Prop. 3, p. 107). 
 
 Problem 64. To describe an ellipse, one axis and a point (P) 
 on the curve being given (Fig. 55). 
 
 The axis is of course given in direction and magnitude, and 
 this really involves the centre of the curve and the position of 
 the other axis. 
 
 First, suppose the major axis AA^ given. Bisect it in C and 
 draw BCD, perpendicular to AA^. From P with AC as radius 
 
THE ELLIPSE. 
 
 115 
 
 w 
 
 mark the point a on BB^ and draw Fa cutting AA^ in b. Ph will 
 be the length of the semi-minor axis, which can therefore be 
 marked off from G to B and B^ . 
 
 Second, if the minor axis BB^ is given. Bisect it in C, through 
 C draw ACA^ perpendicular to BB^. From P with radius BO 
 mark off the point b on AA^ and draw Fb, producing it to meet 
 BBj^ in a. Then Fa will be the length of the semi-major axis, 
 which can be set oflf from (Tto A and A ^ . 
 
 The construction is obvious from the original method of draw- 
 ing the curve. 
 
 Problem 65. To describe an ellipse, an axis ACA^ and a 
 tangent Tt being given (Fig. 65). 
 
 T, t are the points in which the given tangent cuts the axes. 
 
 Draw the second axis BCB^. 
 
 Take ON on (7J, a third proportional to CT, CA (i.e. on CB 
 make Ca^CA, draw ^t^ parallel to Ta, cutting CB in w, and 
 
 Fig.GS 
 
 make CN=^ Cn). Then N is the foot of the ordinate of the point 
 of contact of the given tangent (Prop. 3, p. 107), and therefore by 
 drawing NF perpendicular to CA meeting the tangent in P, a 
 point on the curve is determined and the problem reduces to 
 Problem 64. 
 
 8—2 
 
116 GIVEN CENTRE, 2 POINTS & DIRECTIONS OF CONJ. DIAM^. 
 
 Peoblem 66. To describe an ellipse, the directions of a pair 
 oj conjugate diameters CA, CB, a tangent FT and its point of 
 contact P being given (Fig. 65). 
 
 In the figure the given conjugate diameters are the axes, 
 but the construction holds in any case. 
 
 Through P draw PN parallel to GB meeting CA in N. Take 
 OA a mean proportional between CN and CT, which determines 
 the length of the semi-diameter CA. Similarly determine the 
 length CB. 
 
 Problem 67. To describe an ellipse, the centre (C), two points 
 on the curve (P and Q), and the directions of a pair of conjugate 
 diameters {CA, CB) being given. The lengths CA, CB are not 
 given (Fig. 66). 
 
 From P and Q draw PM, QN parallel to BC meeting CA in 
 M and iV. [In order that the problem may be possible, if PM is 
 
 Fig.66. 
 
 less than QN, CM must be greater than CN.^ Produce PM to 
 E and P^, making J/Pj equal to MP; P, will evidently be a 
 point on the curve. Similarly, drawing EQnQ^ parallel to CA 
 meeting PM in E and CB in n, and making nQ^ — nQ, Q^ will be 
 a point on the curve. Through M draw MX parallel to PQ and 
 Jt/'Z parallel to P^Q^ meeting EQ^ in X and Y respectively. Find 
 ND a mean proportional between EX and EY and set it up from 
 X on a perpendicular to CA. [The mean proportional may con- 
 
THE ELLIPSE. 117 
 
 veniently be found by producing XE to y, making Ey = EY, and 
 on Xy describing a semi-circle cutting Ed perpendicular from E 
 to Xy in d. Ed is the required mean proportional.] Then a 
 circle described with centre C and radius CD cutting CA in A 
 and A^ will determine A and A^ the extremities of that diameter, 
 and if (7o?i be made = ND on CA j, and a parallel to d^n be drawn 
 through A^ cutting CB in B, this will determine an extremity of 
 the other. The curve can then be completed by preceding problems. 
 
 Proof. The construction depends on the known proposition 
 that EP. EP, : EQ . EQ, :: CB' -.CA'-, PP, and QQ, being any 
 chords parallel to the conjugate diameters CB, CA and inter- 
 secting in E. Admitting this, then by Prop. 4, p. 110, 
 QN' '.AN.NA, :: EP.EP, : EQ.EQ,. 
 
 By the construction 
 
 EX : EQ :: QX : EP, 
 and EY : EQ, :: QX : EP, , 
 
 .-. EX.EY : EQ.EQ, :: QX^ : EP.EP,; 
 but EX . EY^-XD' = AX. X'A,, 
 
 .'. AX.XA, : QX' :: EQ . EQ, : EP.EP,, 
 which proves that A A, is the diameter parallel to EQQ, . 
 Also by construction 
 
 CB : CA, :: QX : XB; 
 .'. CB' : CA,' :: QX' : AX. X'^A,, 
 or CB is the semi-diameter conjugate to CA,. 
 
 That ^/^ . EP, : ^(^ . EQ, :: Ci?' : C^' may be proved thus :— 
 Through E draw the diameter EBP, and draw the ordinate 
 7? CT parallel to PP, or to CB, then by Prop. 4, p. 110, 
 
 CB'-RU' : 06^^ :: CB' : CA\ 
 and PM' : CA' - CM' :: CB' : CA' ; 
 
 .'. CB'-PM' : CM' :: CB' : CA', 
 so that CB'-RU' : CZ7'' :: CB'-PM' : CM'; 
 
 but PU' : (7^' :: ^i/' : CM'; 
 
 .'. CB' : CU' :: CB'^PM' + EM' : CM', 
 or C^' : CB'-PM' + EM' :: CU' : CM' :: CE' : CE' ; 
 
lis GIVEN CENTRE, DIRECTION OF MAJOR AXIS & 2 TANGENTS. 
 
 .-. CB' : EM'-P3P :: CB' : CE'-CR\ 
 
 or CB' : EP.EP, :: Cli' : ER.EB,. 
 
 Similarly CA' : EQ . EQ, y. CR^ : ER.ER,, 
 
 or EP.EF, : EQ.EQ, :: CB' : C^\ 
 
 Problem 68. To describe an ellipse, the centre (C), direction of 
 major axis CT, and two tangents {PT, Pt) being given (Fig. 67). 
 
 Bisect the angle TPt between the given tangents by PR 
 meeting CT in R^ and draw PU perpendicular to PR meeting C2' 
 in U. 
 
 Fig.67. 
 
 Describe a circle round the triangle RPU and draw a tangent 
 from G to this circle meeting it in K. CK will be the distance 
 of either focus from C, i.e. make CF=GF^ = CK, and F and F, 
 will be the foci of the required ellipse. From F draw FY per- 
 pendicular to Pt meeting it ''in Y, and make YL on FY produced 
 ^ YF. Draw F^L cutting Pt in Q, and Q will be the point of 
 contact of Pt, i.e. Q will be a point on the ellipse, which can 
 therefore be completed by preceding problems. 
 
 Proof. Since CK is a tangent to the circle RPUf 
 CK : CR :: CU : CK; 
 .-. CK+CR : CK^CR :: CU+CK : CU-CK, 
 or i^i? : i?i^i :: i^^:^ : F,U, 
 
 i.e. J^t^ is divided harmonically in R and i^^ or P{FRF^U} is 
 
THE ELLIPSE. 119 
 
 a harmonic pencil. But the angle RPU is a right angle, and 
 therefore PF and PF^ make equal angles with PR (p. 15). 
 Therefore also the angle FPQ = the angle F^PT since PR 
 bisects the angle QPT ; or the tangents from P make equal angles 
 until the focal distances of P: a known property of the ellipse. 
 
 F^L is evidently the length of the major axis, for, by the con- 
 struction QL = FQj and therefore F^L = F^Q + QF, the sum of 
 the focal distances (Prob. 60, p. 101). 
 
 It follows that Y is on the auxiliary circle, for CF = CF^ and 
 FY^YL; therefore CF is parallel to and equal to ^F^L - CA : 
 and similarly if F^ F, , FZ and i^i^i are perpendiculars from tlie 
 foci on the tangents, Y^, Z and Z^ are all on the auxiliary circle. 
 Produce YF to meet the auxiliary circle in F^, then FY ^ is equal 
 to F^ Fj , and therefore 
 
 FY.FJ^^FY .FY^^^AF.FA^. (Euc. iii. 35.) 
 Similarly FZ . F^Z^ =AF.FA, = FY.FJ^, 
 
 i.e. FY : FZ :: F^Z^ : F^Y^ ; 
 
 and since the angle YFZ is equal to the angle Y^FiZ^ , therefore 
 the triangles YFZ, Z^F^ Y^ are similar (Euc. vi. 6), i.e. the angle 
 FZY^FJ^Z^. 
 
 Circles can be described about the figures 
 YFZP and F.Z^PY^, 
 and therefore the angle ^PF=the angle FZY, 
 
 ' „ „ F^PZ^ = „ F^ Y,Z, ; (Euc. iii. 21.) 
 
 therefore the angle i^PF= the angle i^iPiTj , which proves the 
 property above refeiTed to. 
 
 Problem 69. To describe an ellipse^ the centre C, the direc- 
 tions of a pair of conjugate diameters CT, Ct, a tangent Tt, and a 
 point P being given (Fig. Q^). 
 
 [P must lie between the line Tt and a parallel corresponding 
 line on the other side of (7.] 
 
 Draw PCL meeting Tt in Z, and make CP^ = CP. P^ is a point 
 on the curve. 
 
 Take a mean proportional (Zm) between LP and LP^ and 
 
120 GIVEN CENTRE, DIREC. OF CON J. DIAM. TANGENT & POINT. 
 
 make LM on LT equal to Lm. On Tt describe a semicircle 
 Tqq^t) draw Mn perpendicular to LM and make Mn^GP. 
 Draw Ln cutting the semicircle in q and q^. From q draw qQ 
 
 Fig.68. 
 
 ~ — 1P, 
 perpendicular to LT meeting it in ^ ; then Q will be the point of 
 contact of Tt^ and Qq will be the length of CD the semi-diameter 
 conjugate to GQ ; the curve can therefore be completed by 
 Problems 62 or 63. 
 
 Since Ln cuts the semicircle in two points, there are two 
 solutions. 
 
 Proof. The construction depends on the property of the 
 
 ellipse proved in Problem 67, that the rectangles contained by 
 
 the segments of intersecting chords are in the ratio of the squares 
 
 of the parallel diameters ; and on the further property that if the 
 
 tangent at Q meet a pair of conjugate diameters in 7' and t, and 
 
 GI> be conjugate to CQ, 
 
 QT.Qt = GD\ 
 
 If Q be the point of contact of Tt, it follows that 
 LP.LP^ : LQ' :: GP' : GB'; 
 but LP. LP^ = IjM^ by construction, 
 
 .-. LM : LQ :: GP \ GD; 
 but by construction 
 
 LM : LQ '.: Mn : Qq, 
 
 and Mn = GP, so that Qq = GD; also Qq" - QT. Qt, since Tqt is a 
 semicircle, therefore GD^ - QT. Qt in the figure as drawn. 
 
THE ELLIPSE. 
 
 121 
 
 To prove that it does so in the ellipse, draw the ordinates 
 QK, DK, parallel to Ct, and let the tangent at D meet CT in 7t, 
 then by similar triangles, 
 
 QT : QN :: CD : DK, 
 and Qt : GN :: CD : CK; 
 
 .'. QT.Qt : QN.CN :: CD'' : DK.CK. 
 But CN. CT=^ CA'^ CK. CR (Prop. 3, p. 107), 
 
 .-. CN : CK :: CE : CT 
 CD : QT 
 DK : QN; 
 .'. CN.QN=CK.DK, 
 md .-. QT.Qt = CD\ 
 
 Problem 70. To describe an ellijjse, the centre C, two tangeiitit 
 *T, QT, and a point on the curve (R) being given (Fig. 69). 
 
 Fig.69. 
 
 [It is of course possible to draw at once two more tangents by- 
 producing TC to Tj, making CT^ = CT, and drawing through 2\ 
 parallels to TP, TQ. The point R must lie within the quadri- 
 lateral thus formed. Let the parallel {T^t) to TP meet TQ in t.'\ 
 
122 GIVEN CENTRE AND THREE TANGENTS. 
 
 Draw RCR^ and produce it to meet TQ in L; make CR^ = CR. 
 Take a mean proportional [Lm) between LR and LR^ and make 
 LM on TQ = Lm. Draw i/Tj perpendicular to TQ and equal to 
 C^, and join Lr^, cutting the circle described on Tt as diameter 
 in q and q^ ; from 5' or ^^ drop a perpendicular {qQ) on ^if, and Q 
 will be the point of contact of Tt. CD drawn parallel to Tt and 
 equal to Qq will be the semi-diameter conjugate to CQ. 
 Proof. By construction, 
 
 LM' : LQ' :: Mr^ : Qq\ 
 i. e. LR . LR, : LQ' :: CR' : Qq' ; 
 therefore if Q is the point of contact of Tt, Qq must be the length 
 of the semi-diameter parallel to Lt : and since 
 
 QT.Qt=Qq'=CD% 
 Q is such point of contact. (See last problem.) 
 
 Problem 71. To describe an ellipse, tJie centre G and three 
 tanjents {SV, SW, VW) being given (Fig. 70). 
 
 Through G draw TCT^ meeting ^F in ^ and SW in T, so that 
 TT^ is bisected in G (Prob. U, p. 19). GT will be conjugate to GS. 
 Draw T V parallel to CT meeting VW in v, then T'v will be an 
 ordinate of the diameter CF, for if it meets GV in m, Tm — mv, 
 Bince TC=GT,. 
 
THE ELLIPSE. 123 
 
 Similarly, if Tw be drawn parallel to CW meeting VW in w, 
 T^w will be an ordinate of the diameter CW. 
 
 Let Tv, T^io intersect in E. Draw -S'^ cutting YWinP, and 
 P will be the point of contact of VW. Also PQ parallel to Tv 
 meeting ST in Q will be the chord of contact of the pair of tangents 
 VT, VW, i.e. P and Q are points on the curve; and the problem 
 reduces to one of several previously given, or may be completed 
 thus : — Draw Qy i)arallel to CS meeting CT in iV^. QJ^ is an or- 
 dinate of the diameter CT, and therefore CA the length of the 
 semi-diameter is a mean proportional between CiVand C2^ (Prop. 3, 
 p. 107). Similarly if Qn be drawn parallel to CT meeting CS in 
 it, CB must be taken a mean proportional between Cn and CS. 
 
 Proof. The only point in the construction requiring proof is 
 that SE cuts FTT in its point of contact; 
 
 Now the chords of contact PQ, PR, RQ of the given tangents 
 are parallel respectively to TE, ET^, T^T, which is impossible 
 unless EP passes through S the intersection of TQ and TR. 
 
 Problem 72. To describe an ellipse, the centre C, two points 
 (A and B) of the curve and a tangent Tt being given (Fig. 71). 
 
 [A second tangent can at once be drawn parallel to Tt on the 
 opposite side of C, and at the same distance from it; A and B 
 must lie between these lines.] 
 
124 
 
 GIVEN CENTRE AND THREE POINTS. 
 
 Through C draw CT parallel to AB meeting the given tangent 
 in T. Bisect ^i? in iV^ and draw NCt meeting Tt in t. CT, Ct are 
 a pair of conjugate diameters, and the problem reduces to Prob. 69. 
 Draw the diameter ACA^ meeting Tt in L. Take Lm a mean 
 proportional between LA and LA^. Make L^f on Lt equal to Lm, 
 draw Mn perpendicular to Lt and equal to CA and Ln cutting a 
 circle on l''t as diameter in p and /?,. Perpendicultirs from p and 
 /^j on Tt will determine two points, either of which can be taken 
 as the point of contact of Tt, and the length pP will be the corre- 
 sponding conjugate diameter CQ. 
 
 The construction is obvious from preceding problems. 
 
 Problem 73. To describe an ellipse, the centre C and three 
 2)oints P, Q, R being given (Fig. 72). 
 
 [Any one of the three points, as R, must lie between one pair of 
 the parallel lines furnished by the remaining points and their cor- 
 responding points on the other side of the centre, and outside the 
 other pair.] 
 
 Bisect PQ in p, QR in q, and RP in ?•, and draw Cp, Cq and 
 Cr, producing each indefinitely. PR is a double ordinate of the 
 
 Fig.72 
 
 diameter Cr, and therefore the tangents at P and R will intersect 
 on Cr produced ; similarly the tangents at P and Q will intersect 
 
THE ELLIPSE. 125 
 
 on Cj) and those at Q and R on Cq. If therefore a triangle be 
 drawn the sides of which pas:s through P, Q, R and the vertices of 
 which lie on Cp, Cq, and Cr respectively, the sides of this triangle 
 will be the tangents at P, Q and R. This can be done by Prob. 
 15, p. 20 : — Take any point a on Cr, draw Pa, Ra cutting Cp, Cq 
 in h and c respectively ; join ho cutting PR in X, and draw XQ 
 cutting Ch in T and Cc in t : P2\ Tt, and Rt will be the tangents 
 at P, Q, and R respectively, and the problem may be completed 
 by preceding problems, or thus ; through C draw DCD^ parallel 
 to Tt so that CD is conjugate to CQ ; let TP meet CD in 1\^ 
 draw PX parallel to CQ meeting CD in X. Take CD a mean 
 proportional between CN and CT^, and CD will be the extremity 
 of the diameter CD (Prop. 3, p. 107). 
 
 The construction is obvious. 
 
 The given data are evidently equivalent to a diameter and two 
 points of the curve. 
 
 Problem 74. To describe an ellipse, the foci F arid F^ and a 
 point Q on the curve being given (Fig. 56). 
 
 It has been shewn already that the foci lie on the major 
 axis and that FP + PF^ = the major axis (p. 101). 
 
 Bisect P^F^ in C, and through C draw RCR^ perpendicular to 
 
 FF^. On CF, CF, make CA = CA^^^^^~', and make 
 
 F£ = F£, = CA. AA^, BB^ will be the axes of the required 
 ellipse. 
 
 Problem 75. To describe an ellipse, the foci F and F^ and a 
 tangent {PQ) to the curve being given (Fig. 67). 
 
 [PQ must not lie between F and F^.] 
 
 From F draw FY perpendicular to PQ and produce it to L 
 making Yfj = P^Y. Draw F^L cutting PQ in Q, which will be 
 the poiut of contact of PQ and the problem reduces to the pre- 
 ceding. 
 
 The construction is obvious from Prob. 68. 
 
126 
 
 GIVEN FOCUS, TANGENT AND TWO POINTS. 
 
 Problem 76. To describe an ellipse, a focus F, a tangent R2' 
 with its point of contact R, and a second point P on the curve 
 being given (Fig. 73). 
 
 From F draw FY perpendicular to RT meeting it in Y, and 
 produce FY to / making Yf= YF. 
 
 [F and F must lie on the same side of RT and the distance 
 of F from F must be less than its distance from a line drawn 
 through /perpendicular to fR. See Problem 106, Chap, v.] 
 
 '^'2.73. 
 
 DrSiwfR, -which will be a locus of the second focus. On fR 
 towards R make fP^ = FP. Draw PP^ and bisect it in r ; through 
 r draw rF^ perpendicular to PP^ intersecting fR in F^^ which 
 will be the second focus. Hence both foci being known the pro- 
 blem may be completed by Probs. 74 or 75. 
 
 Proof That fR is a locus of the second focus has been 
 shewn in Prob. 68 ; that the second focus lies on rF^ is evident 
 thus : it must be so situated that 
 
 FR + RF^=^FP + PF^^fF^=fP^+P^F^. 
 
 But FP=fP^, therefore PF, must be equal to P^F^, which 
 by construction it is ; therefore i^j is the second focus. 
 
 If fP^ be made = FP on Rf produced (i.e. on the side remote 
 from R)j and a perpendiciilar to PP^ be drawn through the centre 
 point of PP^ meeting Rf in F^, F and F^ will be the foci of an 
 hyperbola fulfilling the given conditions. 
 
THE ELLIPSE. 127 
 
 Problem 77. To describe an ellipse, a focus F, a tangent R2\ 
 and two points P and Q of the curve being given (Fig. 73). 
 
 [F, P, and Q must all lie on the same side of RTJ\ 
 
 Let FQ be greater than FP, and on FQ make Fp = FP. With 
 P as centre and radius =pQ describe a circle DG, then evidently 
 the second focus must be equidistant from this circle and from 
 the point Q, since the sum of the focal distances is constant. 
 From F draw FY perpendicular to the given tangent PT, produce 
 it tof make Yf^ YF, and with/ as centre and radius FQ describe 
 a circle FG : the second focus will evidently be equidistant 
 from this circle and from the point Q, for it has been shewn 
 (Prob. 68) that the distance of/ from the second focus is equal 
 to the major axis, and therefore equal to the sum of the focal 
 distances of any point on the curve. 
 
 The problem therefore is reduced to finding the centre of a 
 circle to touch externally two given circles (BG, FG) and pass 
 through a given point {Q), which is always possible since the 
 circles must cut each other and Q lie outside both, i. e. the 
 problem reduces to Prob. 32. 
 
 [Draw a common tangent EDM to the two circles meeting 
 fP in if. Take MN on J/^ such that 
 
 MN : MD :: ME : MQ, 
 and the second focus F^ will lie on the line perpendicular to NQ 
 and passing through the centre point of NQ."] 
 
 If the centre of the circle touching the above two circles 
 internally be found (as F^, F and F^ will be the foci of an 
 hyperbola which can be drawn through P and Q and touching RT. 
 (See Prob. 107.) 
 
 Problem 78. To describe an ellipse, a focus F, a point P on the 
 curve, and two tangents TQ, TR being given (Fig. 74). 
 
 [The points F and P must not lie on opposite sides of either 
 tangent.] 
 
 From F draw FYf perpendicular to ^:Z^ and FY^f 2)erpen- 
 dicular to RT, meeting them respectively in Y and Y^, Make 
 }/= ri^ and r,/-7iF. 
 
128 
 
 GIVEN FOCUS, POINT AND TWO TANGENTS. 
 
 With centre P and radius PF describe the circle GU. De- 
 termine the centre (i^,) of a circle to touch this circle internally 
 
 Fig.74. 
 
 i"^ 
 
 and to pass through / and /, (Problem 27): F^ will be the second 
 focus, and the axes can at once be determined by preceding 
 problems. 
 
 Proof. It has been shewn (Prob. 68) that if F^ is the second 
 focus,/i^i =/i^^i = tlie major axis =FP + PF^^ which by construction 
 it does. 
 
 Referring to Problem 27 it will be seen that if the line ff^ 
 cuts the circle GU and y* and f^ lie on opposite sides of it a 
 second ellipse can be drawn with foci F and F^. If this second 
 solution is impossible, a circle can generally be drawn passing 
 through f and f^ and touching the circle Gil externally. F and 
 the centre of this circle will be the foci of an hyperbola fulfilling 
 the conditions of the problem. 
 
 Hence either two ellipses or an ellipse and hyperbola can 
 always be drawn to satisfy the given conditions. 
 
THE ELLIPSE. 
 
 129 
 
 Problem 79. To describe an ellipse, a focus F and three 
 tangents TP, TQ and SR being given (Fig. 74). 
 
 [The point F must lie within one of the three angles of the 
 triangles (as PTQ), i. e. it must not lie within either of the angles 
 (as PSY^) where Y^ is on PS produced.] 
 
 From F drop perpendiculars FY/, FY^f , FY^f on the given 
 tangents meeting them respectively in Y, Zj , Y^, and make Yf- YF, 
 Y,f = Y,F, YJ^= Y„F; then/,/,/^ must all be equidistant from 
 the second focus (Prob. 68) and the problem therefore reduces to 
 finding the centre [F^ of a circle which will pass through three 
 given points. (Prob. 20.) To do this it is not really necessary 
 to bisect j/^i and ff^ because it will be found that the perpen- 
 diculars through their points of bisection will pass through the 
 points aS' and K in which the given tangents intersect, so that it is 
 only necessary to draw through S and K perpendiculars to ff and 
 /1/2' '^^hic^ "^^^1 intersect in F^ the second focus. The major axis 
 is of course known since it is equal to F^f. 
 
 Problem 80. To describe an ellipse, a focus F amd three points 
 P, Q, R on the curve being given (Fig. 75). 
 
 [The point F must lie within one of the three angles PQR, 
 QRP, RPQ, and if circles be described with two of the given 
 
 Fig.75. 
 
 E. 
 
130 GIVEN FOCUS AND THKEE POINTS. 
 
 points as centres passing through F and common tangents be 
 drawn, the third point must be nearer to i^than it is to the tangent 
 more remote from F?^ 
 
 First Method. It is a known proposition (Prop. 1, p. 105) that 
 tangents drawn to an ellipse from any point subtend equal angles 
 at the focus. The tangents at P and Q will therefore intersect 
 on Fp the line bisecting the angle PFQ, those at R and Q will 
 intersect on Fr the line bisecting the angle RFQ, and those at 
 P and R will intersect on the line Fs bisecting the angle PFR. 
 If therefore the three concurrent lines Fp, Fr, Fs be drawn and 
 a triangle be constructed with its sides passing through P, Q and 
 R and with its vertices on the corresponding lines respectively 
 (Prob. 15), these sides will be tangents to the curve at those points. 
 
 On Fs take any point s. Draw Rs cutting Fr in r, and Ps 
 cutting Fp in p. Let RP and rp meet in £c, and draw xQ cutting 
 Fp in T and Fr in t. PT, QT, Rt will be the tangents at P, Q, R 
 respectively, and the second focus can then be easily determined 
 and the problem completed by preceding problems. 
 
 Although there are generally six solutions to Prob. 15, one only 
 is available here, since the sides through the points have to terminate 
 on definite pairs of lines. 
 
 Second Method (same fig.). 
 
 Draw FP, FQ and FR and let FP be greater than FQ or FR. 
 
 Draw PQ and produce it to Z so that PZ : QZ :: FP : FQ, 
 i. e. on FP make FP, = PQ and PQ, = FQ. 
 Through P, draw P^Z, parallel to QQ, meeting FQ in Z„ and on 
 PQ produced make QZ—FZ,. Z will be a point on the directrix. 
 
 Similarly on PR produced take a point W such that 
 PW \ RW wFP : FR. 
 W will be a second point on the directrix, which is therefore 
 determined. 
 
 From F draw a perpendicular FX to WZ meeting it in X, and 
 on FX take points AA, such that FA : AX :: FA, :A,X:: FP 
 is to the perpendicular distance of P from XZ. A A, will be the 
 major axis. 
 
THE ELLIPSE. 131 
 
 The second focus and consequently the length of the major 
 axis may perhaps be more easily determined thus. It is a known 
 proposition (p. 102) that the tangent at any point, say Q, 
 meets the directrix in a point K such that KFQ is a right angle. 
 Therefore draw FK perpendicular to FQ meeting the directrix 
 in K, and draw the tangent KQ. From F draw FY/ perpen- 
 dicular to KQ meeting it in Z, make Yf= YF, and draw fQ 
 meeting XF in F^ , the second focus. /F^ is of course the length 
 of the major axis. 
 
 Proof. Since by construction 
 
 PF .PZ :: QF: QZ, 
 therefore evidently PF: dist. of P from WZy.QF: dist. of Q from 
 WZ, and since PF -. PW w RF -. PW, 
 
 .\ PF: dist. of P from WZ :: PF : dist. of P from WZ; 
 therefore the distances of the given points from the focus are in a 
 constant ratio to their distances from WZ^ which is therefore the 
 directrix. 
 
 If the lines PQ, PR are divided internally in the same ratio 
 as above, two points are determined which being joined, either 
 to each other or to the opposite points of the first pair, give three 
 lines, either of which may be taken as the directrix of an 
 hyperbola passing through the three given points and having F 
 as focus. In each case one of the given points will lie on one 
 branch of the curve and two on the other. 
 
 Thus generally four conies can be drawn fulfilling the given 
 conditions, one of which is an ellipse. 
 
 Problem 81. To describe an ellipse, two tangents TQ, TR 
 with their points of contact Q and R, and a point P on the curve 
 being given (Fig. 76). 
 
 [The point P must lie within the parabola which can be 
 described touching TQ, TR at Q and i^.] 
 
 This is of course a simple case of the more general problem 
 to describe an ellipse to touch two given lines and to pass through 
 three given points. 
 
 9—2 
 
132 A POINT AND TWO TANGENTS WITH POINTS OF CONTACT. 
 
 First Solution. Let QP produced meet the tangent TR in S. 
 From S draw a line passing through the intersection of PR and 
 
 Fig.76. ^.^k 
 
 the other tangent QT (Prob. 4) and meeting QR in W ; then W 
 will be a point on the tangent at P, which can therefore be drawn. 
 Let it intersect TR in X. Bisect QR in V and draw TVC, which 
 will evidently be a diameter of the curve, i.e. is a locus of the 
 centre. Bisect PR in Fj and draw XV fi, which will similarly 
 be a locus of the centre. The centre is therefore at C, the inter- 
 section oi TV and XFj, and the centre being known the problem 
 can be completed by Probs. 70, 71, &c. 
 
 Second Solution. Bisect QR in V and through T draw 
 TD^ VD, which will evidently be a diameter of the ellipse, i. e. will 
 pass through the centre. Through P draw LPNL^ parallel to 
 QR, meeting TR in Z, TD in X and TQ in L^. Take Pk a mean 
 proportional between PL and PL^, and from L and Zj towards X 
 make LK=L^K,^Pk', then RK or QK^ will intersect TD in D, 
 
THE ELLIPSE. 133 
 
 the extremity of the diameter. On TD take a point C such that 
 TO :CD :: CD : CV, 
 i.e. TC+CD : TO :: CV+CD : OT, 
 or TD'. TO :: Fi> : CD, 
 G will be the centre of the ellipse. 
 
 [The point C can easily be found by drawing any two parallels 
 through T and D (as Td, Dv), making Td=TD and Dv= VD, 
 and joining dv cutting TD in C] 
 
 The direction of the diameter CB conjugate to CD is known, 
 since it is parallel to QE; its length can easily be determined by 
 taking a mean proportional between Cn and Ct^ where n is the 
 foot of the ordinate from Q on CB and ^ the intersection of CB 
 and of the tangent at Q. 
 
 Proof. Let DM be the tangent at D meeting TE in M, and 
 let LF meet the curve again in p, so that L^P = Lp. 
 
 Then LP . Lp : LR^ is the ratio of the squares of the parallel 
 diameters (p. 117); but MD^ : MR^ is the same ratio, 
 .-. LP . Lp '. LR' :: MD^ : MR^ 
 
 :: LK^ : LR' hy similar triangles, 
 .*. LP . Lp — LK^^ which justifies the construction. 
 
 Problem 82. To describe an ellipse, two tangents TP, TQ and 
 three points A, Bj C on the curve being given (Fig. 77). 
 
 [The points ABC must not lie on opposite sides of either line.] 
 
 Draw the line AB cutting the given tangents in P and Q. 
 Find X the centre, and LJ, LJ^, the foci, of the involution A, B 
 and P, Q (Prob. 13). 
 
 [In the figure, Pb on TP^PB, J^j on a parallel to TP drawn 
 through A is equal to ^^; then qj^ cuts AB in X, the required 
 centre. XE is a mean proportional between XA and XB.^ 
 
 E or E^ will be a point on the chord of contact of the given 
 tangents. 
 
 Similarly draw BC cutting the given tangents in p and q, and 
 find Xj the centre, and F, F^ the foci of the involution B, C and p, q\ 
 then F or F^ will be a second point on the chord of contact of the 
 
184 
 
 GIVEN THREE POINTS AND TWO TANGENTS. 
 
 given tangents, the points of contact of which i?, R^ are therefore 
 determined, and the problem reduces to the preceding"^. Since 
 
 Fig.77. 
 
 E and j&\ can be joined to either F or F^ four chords of contact 
 can in general be drawn, but one at least of the corresponding 
 conies will be an hyperbola. 
 
 For proof that E and F are points on the chord of contact see 
 Prop. 7, p. 143. 
 
 Problem 83. To describe an ellipse, two points A, B on the 
 curve, and three tangents PQ, QR, RP being given (Fig. 78). 
 
 \A and B must not lie on opposite sides of either line.] 
 
 Draw a line through AB cutting the tangents through P in L 
 and 31 and the remaining tangent in JV. 
 
 Find X the centre, and J), D^ the foci of the involution AB and 
 LM (Prob. 13). D or D^ will be a point on the chord of contact 
 of the tangents PQ, PR. 
 * In the figure the point of bisection of RR^ accidentally coincides with F. 
 
THE ELLIPSE. 
 
 135 
 
 [In tlie fig. La on PR — LA^ Bm on a parallel to PR drawn 
 through B = B2I, and ma cuts AB in X, the required centre. XD 
 is a mean proportional between XA and XB.^ 
 
 F'g.78. 
 
 Similarly find X^ the centre, and U, ^, the foci of the involution 
 I, B and M, X (Prob. 13), and £J or ^i will be a point on the 
 bhord of contact of the tangents QP, QR. 
 
 [In the fig. Mb on Qp — MB, ^t^ on a parallel to QP through 
 is equal to AX, and bn cuts AB in X^, the required centre, 
 r,^ is a mean proportional between X^if and X,ir.] 
 
 Find MV, the harmonic mean between MB and J/7), M being 
 
 le point on the given tangents which has appeared in each of the 
 
 ibove involutions (Prob. 11); then ^Fwill cut the opposite tan- 
 
 mt PQ in its point of contact (p) with the curve, and therefore 
 
 )Eq will be the chord of contact of the tangents QP, QR and pDr 
 
 that of PQ, PR. The problem therefore reduces to No. 81. 
 
 The construction depends on the property made use of in the 
 last problem and proved in Prop. 7, p. 143, that the chord of contact 
 of PQ, PR must pass through D or D^, the foci of the involution 
 AB and LM, and similarly that the chord of contact of QP and 
 
136 
 
 GIVEN FIVE TANGENTS. 
 
 QR must pass through E or E^, the foci of the invohition AB 
 and MN. 
 
 Also if rq meets VR in n and the tangent PQ in T, 
 since BY EM is harmonic (by construction) so also is Truq^ and 
 therefore ?tFis the polar of ^and therefore determines p, the point 
 of contact of PF (Prop. 5, p. 141). 
 
 Since either D or B^ may be taken with E or E^ there are in 
 general four solutions. 
 
 Problem 84. To describe an ellipse to touch five given lines 
 AB, BC, CD, DE, EA. 
 
 [The lines must form a pentagon without a re-entering angle 
 and the vertices are supposed to be lettered consecutively.] 
 
 Draw AC and BD intersecting in F. Then EF will intersect 
 BC in P, the point of contact of BC, Similarly ii BD and CE 
 
 intersect in G, AG will intersect CD in Q, the point of contact of 
 CD; and continuing the construction R, T and F, the points of 
 contact of DE, EA and AB may be determined. 
 
 The centre of the curve can easily be found and the curve 
 completed by preceding problems. 
 
 The construction depends on Brianchon's well-known theorem : 
 
THE ELLIPSE. 
 
 187 
 
 " The three opposite diagonals of every hexagon circumscribing a 
 conic intersect in a point." 
 
 For if T be the point of contact of AB the pentagon may be 
 considered as a hexagon AT, TB, BG, CD, DE, EA, and therefore 
 AC, BE and DT must meet in a point L; and conversely if X is 
 the point of intersection of ^C and BE^ DL must pass through T, 
 the point of contact of AB, and similarly for the remaining sides. 
 
 Problem 85. To describe an ellipse, four tangents AB, BC, 
 CD, DA and a point E on the curve being given (Fig. 80). 
 
 [The point E must lie within the quadrilateral A BCD, which 
 must not be a parallelogram.] 
 
 Let BE, CE meet AD in B^diud. C^ respectively. 
 
 Fig.80. / 
 
 Find X the centre, and P and P^ the foci of the involution 
 AC, SiudDB,. Prob. 13. 
 
 Then the tangent at E must pass through P or P, and the 
 problem reduces to the preceding. 
 
 There are two solutions. 
 
 In the figure B^c on B,B = B,C,; Ad on a parallel to B^B is 
 equal to AD and cd intersects AB, in X, the required centre. XP 
 is a mean proportional between XA and XC ^ . 
 
 Also BP and AF intersect in L and CL will pass through T, 
 the point of contact of A P. 
 
138 GIVEN FIVE POINTS. 
 
 Problem 86. To describe an ellipse to pass through Jive given 
 points ABODE (Fig. 81). 
 
 [No point must lie inside the quadrilateral formed by the 
 other four.] 
 
 Let AB, DC meet in F and AG, BE in G. 
 
 Fig.81 
 
 4- 
 
 Draw FG meeting DE in P. P will be a point on the 
 tangent at A. 
 
 Similarly if BG and ED meet in H and AG, BD in K, HK 
 will meet EA in Q, a point on the tangent at B. 
 
 The problem can evidently be completed in various ways by 
 preceding problems. 
 
 The construction depends on Pascal's well-known theorem : 
 " The three intersections of the opposite sides of any hexagon in- 
 scribed in a conic section are in one right line." For the tangent at 
 A may be considered as meeting the curve in two consecutive points 
 A and a, and therefore P, the intersection of Aa and DE, must lie 
 on FG, the straight line through the intersections oi AB and DG 
 and of BE and Ga. 
 
 This line is known as the Pascal line. 
 
 There is only one solution. 
 
THE ELLIPSE. 
 
 139 
 
 Problem 87. To describe an ellipse, four points on the curve 
 A, B, C, B and a tangent ad being given (Fig. 82). 
 
 [All the points must lie on the same side of the tangent.] 
 Draw AB meeting ad in a, BG meeting ad in 6, DO meeting 
 it in c, and ^i) meeting it in d. 
 
 Fig.82 
 
 / ^p 
 
 -AC, 
 
 Find X the centre, and F and P^ the foci of the involution ac 
 and bd. 
 
 P or Pj will be the point of contact of the given tangent and 
 the problem may be completed by several preceding ones. 
 
 In the fig. ab^ on aA = ab ; dc^ on a parallel to aA = dc, and 
 6jCj intersects ad in X, the required centre. XF = Xp, a mean 
 proportional between Xc and Xa. 
 
 If DC, BP meet in F, and BC, PA in G, then FG and DA 
 will intersect in H, a point on the tangent at B. 
 
 There are of course two solutions, as either P or P^ may be 
 taken as the point of contact. 
 
140 
 
 POLE AND POLAR. 
 
 That P, the point of contact, is a focus of the involution is 
 proved in Chapter 8. 
 
 POLE AND POLAR. 
 
 It has been shewn in the case of the circle (Cor. 3, p. 31) 
 that the ]>airs of tangents drawn at the extremities of any chord 
 through a fixed point intersect in a straight line. 
 
 This is also true in the case of any conic section, for let V 
 (fig. 83) be any point in a conic and C the centre, and let CV 
 
 Fig.83. 
 
 meet the curve in P. Take jT in CV produced such that 
 GV :CP ::CP :CT, and through Fdraw the chord QVQ, parallel 
 to the tangent at P. 
 
 QQ^ will be the chord of contact of the pair of tangents drawn 
 from T to the conic, and will be bisected in V. 
 
THE ELLIPSE. 141 
 
 Through V draw any chord AVB and let the tangents at A 
 and B intersect in 2\ . 
 
 Join CT^^ and draw PN parallel to AB, meeting CT^ in N. 
 Then if GT^ meet AB in K and the tangent at P in L, 
 
 CK . CT^ = CX . CL. (Prop. 3, p. 107.) 
 .-. CT^ :CL y.CN: CK 
 CP'.CV 
 CT: CP; 
 
 hence TT^ is parallel to PL, and therefore T^ , the intersection of 
 the tangents at the extremities of ani/ chord through V, lies on a 
 fixed line. 
 
 Def. As in the circle, the line TT^ is called the polar of the 
 point V with respect to the conic and the point V is called the 
 pole of TT^ with respect to the conic. 
 
 If the pole lies without the conic (as T), its polar is the line 
 QQ^ parallel to the tangent at the point (P) where CT meets the 
 conic, and meeting C^' in a point V such that 
 
 CV:CP:: CP : CT, 
 
 i. e. is the chord of contact of tangents from the pole. 
 
 If the conic be a parabola, since the centre may be considered 
 as at an infinite distance, the line VT must be drawn parallel to 
 the axis meeting the curve in P and PT be made equal to PV, 
 the polar of V will then be parallel to the tangent at P and will 
 pass through T. 
 
 If the pole be on the curve, the polar is the tangent at the 
 point. 
 
 The directrix is the polar of the corresponding focus. 
 
 If a point (as T^) lies on the polar of V, the polar of T^ passes 
 through V. 
 
 The following important harmonic properties should be noticed. 
 
 Prop. 5. A straight line drawn through any point is divided 
 harmonically by the point, the curve, and the polar of the point. 
 
142 
 
 HARMONIC PROPERTIES. 
 
 a. Let the point be without the curve, as T (fig. 84), and let 
 the line meet the curve in ^^ and the polar ol T in C. Draw 
 
 Fig;.84 
 
 the tangents TP^ TQ meeting the curve in P, Q. C of coiu-se 
 lies on the line PQ. Through A and B draw DAEF, GBHK 
 parallel to PQ meeting the tangents respectively in D, F and (?, K 
 and the curve in F and H. 
 
 Then the diameter through T bisects AE and PQ, and there- 
 fore also bisects BF ; 
 
 hence DA = EF and similarly GB - KH, 
 Also GB'.BK::DA :AF; 
 
 .'. GB . BK : GB\' :: DA . AF : Da\% 
 or GB .GH'.DA. DE •.'~GB\' : DA' 
 :: GT' .DT"; 
 
 but GB . GH : DA . DE :: GP" : DP' (p. 117); 
 
 .-. GP :PD :: GT : DT, 
 
 and .-. TA : TB w AC : CB, 
 
 i. e. TAGB is divided harmonically. 
 
 p. Let the point be within the curve, as V (fig. 83), then 
 drawing any chord A VBG meeting in G the polar of F, the polar 
 of (r passes through V and therefore A VBG is harmonically 
 divided. 
 
THE ELLIPSE, 143 
 
 Prop. 6. If two tangents be drawn to a conic, any third 
 tangent is harmonically divided by the two tangents, their chord 
 of contact, and the point in which it touches the curve. 
 
 Let LMAN (fig. 84) be the third tangent meeting FQ in Z, 
 and TP, TQ in M and N. Through N draw Nach parallel to TP 
 meeting the curve in a, h and PQ in c. . 
 
 Then Na . M : NQ^ :: tF\' : TQ^ (P- H^) 
 
 but LN' : LM\' 
 
 iVcj* : Pif|* by similar triangles, 
 Na.M:PM\' 
 IiA\' lAM'i' (^. 117), 
 i.e. LMAN is divided harmonically. 
 
 Prop. 7. If a straight line meet two tangents to a conic in 
 PQ and the curve in AB, the chord of contact of the tangents 
 will pass through one of the foci of the involution P, Q and A, B 
 (fig. 77). 
 
 Since X is the centre and £J, E^ the foci of the involution P, Q 
 and A^ B, 
 
 XP :XA::XB :XQ; 
 .'. XA-XP :XA ::XQ-XB : XQ, 
 or PA :XA :: BQ : XQ. 
 Similarly PB : XB :: AQ : XQ, 
 
 .-. PA . PB :AQ . BQ :: XA . XB : XQ\' :: XB\' : Z^|^; 
 but (p. 18) BP : BQ :: PB^ : B^Q, since ^P^,(? is harmonic; 
 .-. BP : BP + PB^ :: j^^ : BQ + QB^, 
 or BP :BQ :: 2X^ : 2X^; 
 .-. PA. PB :AQ . BQ :: ZiPp : ^p ,. (1). 
 
 Draw the tangent ghkl parallel to PQ meeting TP^ TQ in g 
 and k, the chord of contact in I, and touching the curve in h; and 
 
144 HARMONIC PROPERTIES. 
 
 PB . PA '. ghY 
 and QA . QB -JchV 
 
 if the cliord of contact does not pass through E let it meet PQ in G. 
 TTiY : %1^ (p. 117) 
 PGy : IgY by similar triangles, 
 
 but Ikhg is harmonic (Prop. 6), 
 
 and .-. Uc^" :'lg\' -.-.khY '.~^i\\ 
 
 .-. PB.PAiQA. QB :: GP^' : ^j^ (2); 
 
 but (1) and (2) cannot be simultaneously true unless the points U 
 and G coincide. 
 
 Prop. 8. If a quadrilateral be inscribed in a conic, its opposite 
 sides and diagonals will intersect in three points such that each is 
 the pole of the line joining the other two. 
 
 This follows at once from the harmonic properties of a complete 
 quadrilateral, p. 16, combined with Prop. 5, p. 141. For since EGfA 
 (fig. 11) is harmonic it follows that /is a point on the polar of E 
 with respect to any conic passing through A and C, and since 
 EDf^B is harmonic/ is a point on the polar of E with respect to 
 any conic passing through BD. Therefore j/,, i.e. OF, is the polar 
 of E with respect to a conic passing through ABCD. Similarly 
 OE is the polar of F. Also since is on the polar of E the polar 
 of must pass through E, and since it is also on the polar of F 
 the polar of must pass through F, i. e. EF is the polar of 0. 
 
 The triangle EFG is of course self -conjugate with respect to 
 any conic circumscribing the quadrilateral, Def. p. 32. 
 
 Prop. 9. If a quadrilateral circumscribe a conic its three 
 diagonals form a self-conjugate triangle (fig. 85(x). 
 
 Let ABCD be the quadrilateral and let AB and CD intersect 
 mG,AC and BD in E, and AD and BG in F. Let BD and ^C 
 meet EG in K and L respectively. The triangle EKL is self- 
 conjugate with respect to any conic inscribed in the quadrilateral ', 
 ABCD. 
 
THE ELLIPSE. 
 
 145 
 
 Let the polar of F (i. e. the chord of contact PP^) meet FG in 
 R] then, since R is on the polar of F, it follows that F is on the 
 polar of R. 
 
 Fig.ssa 
 
 H Now F {AEBG) is a harmonic pencil (p. 16), and if PP^ does 
 
 . not pass through E let FE meet PP^ in T ; then PTP^R is a 
 
 harmonic range; hence by (Prop. 5, p. 141) FE is the polar of R. 
 
 Similarly, if the other chord of contact QQ^ meet FG in i?j, GE 
 
 is the polar of R^ . 
 
 .-. ^ is the pole of RR^, i.e. of LK. 
 Again, DEBK is a harmonic range, and if QP meet AC in S 
 and CK in F, QSPY i^ harmonic, and therefore S is on the polar 
 of F; but S is also on the polar of C, therefore CV or C^ is the 
 polar of aS'. Similarly, if P^ Q^ meet ^C in S^ , A K is the polar of S^ . 
 .*. JT is the pole of LS^, i.e. of ^Z; 
 .'. ELK is a self-conjugate triangle. 
 
 Problem 88. To determine the centre of curvature at any 
 point P of a given ellipse (see page 89), fig. 86. 
 
 CA, CB are the semi-axes, and F, F^ the foci Draw PG the 
 normal at P meeting the major axis in G, and draw GK perpen- 
 E. 10 
 
146 
 
 CIRCLE OF CURVATURE. 
 
 dicular to PG meeting PF or PF^ the focal radii through P in K. 
 KO perpendicular to PK will intersect PG in 0, the required 
 centre of curvature. 
 
 Fig.86. 
 
 If AB^ BD be drawn parallel to the axes and DN be drawn 
 perpendicular to AB meeting the major axis in M and the minor 
 in N, M and N will be the centres of curvature at A and B 
 respectively. The evolute of the quadrant AB will therefore 
 touch the axes at these points, and the evolute of the entire ellipse 
 is made up of four curves similar to the chain dotted curve shewn 
 in the figure between M and N. 
 
 As in the parabola, if the circle of curvature at P cuts the 
 curve again in Q, PQ is inclined to the axes at the same angles as 
 is the tangent at P. 
 
 The construction depends on the known value of the radius 
 
 PC 
 
 of curvature, ^:zZi~TP'l^ (Salmon's Conic Sections, Chap, xiii.), 
 
 cos' FPG 
 
 for 
 
 ■pfr 
 
 PG_ 
 PK' 
 
 and therefore rad. of curvature = PO. 
 
ft 
 
 EXAMPLES. 147 
 
 Examples on Chapter TV. 
 
 1. Describe an ellipse to touch a given straight line (QY) 
 and pass through a given point (P) ; a focus F and the length 
 (2a) of the major axis being given. 
 
 [From F draw FY perpendicular to QY and produce it to T, 
 making YT= YF. With T as centre and 2a as radius describe an 
 arc, and with P as centre and (2a — FF) as radius describe a second 
 arc intersecting the former in F^ , which will be the second focus. 
 There are two solutions.] 
 
 2. Describe an ellipse to touch two given straight lines; a 
 focus and the length of the major axis being given (last question). 
 
 3. Describe an ellipse to touch two given lines OP, OQ at the 
 points P and Q ; one focus (F) beiug on the line PQ and the 
 angle I^OQ less than a right angle. 
 
 [The second focus F^ is the point of intersection of lines mak- 
 ing with the given tangents angles equal to OPQ, OQP respec- 
 tively, i. e. 0PF,=7r-0PQ and 0QF,=7r-0QP. Bisect PQ 
 in V; centre lies on OV. Draw i^/iT parallel to PQ meeting OV 
 in IC Bisect VK in C, which will be the centre of required 
 elUipse.] 
 
 4. Given one focus F of an ellipse, the length 26 of the 
 minor axis, and a point P on the curve ; draw the locus of the 
 centre. 
 
 [A parabola with the centre point of FP as focus, FP as axis 
 
 and latus rectum = 2 ,=r^ .1 
 FP -■ 
 
 5. Given one focus F of an ellipse, the length 26 of the minor 
 axis and a tangent to the curve ; shew that the locus of the second 
 focus is a straight line parallel to the given tangent and at a 
 
 distance from it = — , where p is the perpendicular from F on the 
 
 given tangent. 
 
 10—2 
 
148 EXAMPLES. 
 
 6. Any focal radius FP is drawn in an ellipse, and the point 
 Q on the auxiliary circle corresponding to F is joined to the centre 
 G. Shew that the locus of the intersection of FF and CQ, is an 
 ellipse having F and G for foci. 
 
 7. Given the base, and sum of sides of a triangle ; shew that 
 the locus of centre of inscribed circle is an ellipse, having given 
 base as major axis. 
 
 8. AB and BG are two equal rulers of length a, jointed at B, 
 P is a point on BG distant h from B. The end A of one ruler is 
 fixed and the end G of the other moves along a right line AG 
 through A. Shew that the locus of F is an ellipse with semi-axes 
 a + 6 and a — h. 
 
 9. An ellipse slides between two lines at right angles to each 
 other : shew that the locus of its centre is a circle of radius 
 
 ^o? + 6^, where a and h are the semi-axes of the ellipse. 
 
 10. Draw an ellipse, and from any point F on it draw lines 
 P2>, FE equally inclined to the major axis and meeting the curve 
 again in D and E \ draw FG perpendicular to the major axis 
 meeting BE in G. If the tangent at F meet BE in 0, shew that 
 the triangle FOG is isosceles. 
 
 11. Shew by construction that the normal FG at any point of 
 an ellipse is an harmonic mean between the focal perpendiculars 
 on the tangent at F. 
 
 12. Given one focus F of an ellipse, the length 26 of the 
 minor axis, and a point F on the curve ; draw the locus of the 
 other focus. 
 
 [A parabola with focus /*, axis FF and latus rectum =: 4 -^Tp .] 
 
 13. Shew that the locus of intersection of tangents at the 
 ends of conjugate diameters of a given ellipse (semi-axes a and h) is 
 an ellipse, the axes of which coincide in direction with the given 
 ellipse, and the semi-lengths of which are ^2a and J^h. 
 
EXAMPLES. 149 
 
 14. AB is a line cutting in A and B a circle, centre C ; Q in 
 a point on the perpendicular from C on AB on the same side of 
 AB as C and outside the circle. Shew that the locus of the point 
 P moving so that the tangent from F to the circle is in a constant 
 ratio to the distance of F from AB {Q being a point on the locus) 
 is an ellipse touching the circle at A and B. 
 
 15. Given any point P on an ellipse, inscribe in the ellipse a 
 triangle PQB, tlie bisectors of the sides of which shall pass through 
 the centre. 
 
 [Take p the point on the auxiliary circle corresponding to P. 
 In the circle inscribe the equilateral triangle pqr ; the points 
 corresponding to q and r will be the vertices of the required 
 triangle.] 
 
 16. Given two tangents TP, TQ \ their points of contact 
 Pand Q and the radius of curvature (p) at one of them (P suppose) 
 describe the ellipse. 
 
 \TC bisecting PQ is a locus of the centre. Draw the circle 
 circumscribing the triangle TPQ and let d be its diameter. Draw 
 a straight line through P such that p (the perpendicular distance 
 of any point on it from PT) : q (the perpendicular distance of the 
 same point from QP) :: PT.d : QT . p, i.e. determine the ratio 
 
 p PT.d 
 
 - == -j—r. — (p. 10). This line is a second locus of the centre, 
 q Ql\. p ^^ ^ 
 
 which is therefore known.] 
 
 17. Draw an ellipse, a focus P, a tangent PT, its point of 
 contact P and the radius of curvature (p) at P, being given. 
 
 [Reverse the construction of Prob. 88 to determine (?, the 
 foot of the normal at P, and consequently the direction of the 
 major axis.] 
 
 18. If P is any point on an ellipse and the ordinate Pp per- 
 pendicular to the major axis meets the auxiliary circle in p, the 
 angle between the major axis and the radius of the circle through 
 p is called the eccentric angle of P. 
 
150 
 
 EXAMPLES. 
 
 Shew that 
 trie angle of 
 
 if P be 
 which is 
 
 any point 
 a, three 
 
 on an ellipse, the 
 
 points A, i), and C on 
 
 eccon- 
 the 
 
 curve, the eccentric 
 
 angles 
 
 of which 
 
 a 
 are -3 
 
 f 120'^ and 
 
 + 240", are such that the circle of curvature at each passes 
 
 3 
 
 through P\ and verify that a circle can be described through 
 A^ B, 0, and P, and that the bisectors of the sides of the triangle 
 ABC pass through the centre of the ellipse. 
 
 19. Draw in a given ellipse a pair of conjugate diameters; 
 making a given angle with each other, 
 
 [On any diameter of the ellipse describe a segment of a circle 
 containing the given angle (Prob. 3U). If the points where the 
 circle meets the ellipse be joined to the ends of the chosen diame- 
 ters, the required conjugate diameters will be parallel to these 
 chords. The least possible angle between conjugate diameters of' 
 a given ellipse is the angle between the diagonals of the rectangle 
 formed by the axes.] 
 
 I 
 
CHAPTER V. 
 
 THE HYPERBOLA. 
 
 As in the case of the ellipse, the definition of the curve given 
 on page 56 does not immediately exhibit the property of the 
 curve which furnishes the most convenient method of constructing 
 it. It may also be defined as the locus of a point which moves 
 in a plane, so that the difference of its distances from two fixed 
 points in the plane is constant, and that the two definitions are 
 really identical may be shewn thus : — 
 
 In fig. 87 let i^ be the focus and MX the directrix (Definitions, 
 page 56). 
 
 F«g.87. 
 
 h.'-----'-- 
 
 From F draw FXF^ perpendicular to MX meeting it in X, 
 
 FA FA 
 jmd let yi, ^Ij be points on FX such that .-y-^ ~i~v= *^® given 
 
 A.JL A.,2l 
 
152 GENERAL PROPERTIES. 
 
 constant ratio (greater than unity) for all points on the curve : 
 the points A and A^ are called the vertices of the curve, and AA^ 
 the transverse axis, and if ^^j be bisected in C, C is the centre of 
 the hyperbola. 
 
 To shew that the curve can be constructed from a second 
 focus and directrix corresponding to the vertex A^ 
 
 Let P be a point on the curve, i.e. let 
 
 FF : FM :: FA : AX, 
 where FM is the perpendicular from F on the directrix. 
 
 Draw AF, A^P meeting the directrix in G and //, and let 
 FR meet FM in K, 
 
 Then FK : FA, :: FII : AJI 
 
 :: FM : A,X, 
 or FK : FM :: FA^ : A^X :: FA : AX; 
 
 .-. FK= FF and the angle FI{F= the angle FFK 
 
 = the angle KFA^. 
 Similarly FG bisects the angle between FA^ and FF produced, 
 therefore the angle HFG is a right angle. 
 
 In AA^ take a point X^ such that A^X^=AX, and through 
 X, draw a straight line perpendicular to AA,, and in FA^ pro- 
 duced take a point F, such that A,F, = AF. 
 
 Let FA, and FA produced meet the perpendicular through 
 X, in h and g and join F,g, FJi, 
 then (jX, : GX :: AX, : AX 
 
 :: A,X : A,X, 
 :: IIX : hX„ 
 .-. gX, . hX, = GX . X^ = FX' - i^.A",^; 
 .-. ^i^j/i is a right angle. 
 Let FK (parallel to axis) meet gX, in 3f,, gF, in m, and FJi 
 produced in k, 
 
 Fm : FM, :: F,A : AX,, 
 and P^ : FM, :: 7^,^^ : A,X,, 
 
 .'. Fm = Fk; 
 
THE HYPERBOLA. 158 
 
 and mF^ being a right angle, 
 
 F,P = Fin^Pk, 
 .'. F,P : FM, :: F,A, : A,l\, 
 and the curve can therefore be described by means of the focus 
 F^ and the directrix X^^fy. 
 
 It follows that the curve is symmetrical with regard to the 
 centre C, and that it lies wholly without the tangents at the 
 vertices A and A^, which are perpendicular to CA. 
 We have at any point F of the hyperbola, 
 FF : FAf :: FA : AX, 
 F,F : FM, :: F,A, : A,X, 
 :: F,A : AX,; 
 .-. F,F-FF : FM,-FM :: F,A-FA : AX\-AX; 
 but F3I, - FM^ MM, = XX, = AX, - AX, 
 
 .', F,F-FF = F,A-FA = AA,, 
 i.e. the difference of the focal distances is constant and equal to 
 the transverse axis. 
 
 Problem 89. To describe an hyperbola, the foci and a vertex, 
 or the vertices and a focus, or the transverse and conjugate axes 
 being given (Fig. 88). 
 
 Bisect the distance between the given foci F, F, or the given 
 vertices A, A,u).C. 
 
 With centre F and any radius greater than FA describe arcs 
 as at Q and q, and with centre F, and the same radius describe 
 arcs as at Q, and q,. On any convenient line on the paper mark 
 off a length aa, =^AA,, and with centre a and radius FQ mark off 
 a point on this line on the opposite side from a, as at Q'. Take 
 off the distance Q'a, from this line with a pair of dividers or 
 compasses, and with centres F, and F mark off points on the arcs 
 already described about the opposite foci as centres. These points 
 will of course be on the curve, since the difference of the focal 
 distances of each is equal to A A,, and the process may be repeated 
 and as many points obtained as is necessary to define the curve 
 and allow it to be sketched through the points with accuracy. 
 
154 
 
 GIVEN FOCI, AXES, &C. 
 
 Though somewhat tedious, it is the only method for construct- 
 ing the hyperbola which can be recommended. 
 
 Fig.88. 
 
 Since the radii of the intersecting arcs may increase indefinitely, 
 the curve evidently tends to infinity in both directions from C. 
 
 Through C draw BCB^ perpendicular to AA^, and let the 
 circle described on FF^ as diameter intersect the tangent at the 
 vertex in L. Make CB -GB^- AL, then BB^ is called the con- 
 jugate axis; and if a second hyperbola be described with vertices 
 at B and B^ an<l with foci on BB^ at F\ F^' the same distance 
 from C as those of the original hyperbola, each curve is said to 
 be conjugate to the other. 
 
 The eccentricity of the hyperbola (p. 57) is the numerical 
 
 FA 
 value of the ratio -.-^ . It is usually denoted by e, and its 
 
 value in terms of the axes is. = ^?±1\ 
 
 ^here 
 
 CA=a and CB = h. 
 
I 
 
 THE HYPERBOLA. 155 
 
 FA _ FA, __ CF _CL JaF^b ' 
 ^^ AX " A,X ~'GA~ 0A~ a 
 
 The diagonals of the rectangle formed by the tangents to the 
 hyperbola and its conjugate at their vertices are called asymptotes. 
 The axes therefore bisect the angles between the asymptotes. 
 
 In fig. 87 let PX drawn from any point P of the curve per- 
 pendicular to the transverse axis meet it in X, and, as before, let 
 PA and PA, meet the directrix in G and //, 
 
 then PN : AN :: GX : AX, 
 
 and PN : A,N :: IIX : A.X; 
 
 .'. PX' : AX.A.X :: GX.HX : AX . A,X 
 :: FX' : AX . A,X, 
 since GFII is a right angle, 
 
 !• e. Tiir — T~^ IS ^ constant ratio. 
 AN. A,N 
 
 Since FA : ^Z :: T^J^ : A,X, 
 
 .-. i^^l + i^^, : i^^^ :: AX + A,X : ^.Y, 
 
 or Ci^ : CA :: FA : AX (1), 
 
 and FA,- FA : FA :: A,X - AX : AX, 
 
 or GA : GX :: i^.l : AX (2), 
 
 .-. GF : GA :: C^ : GX :: i^.4 : AX (3). 
 
 Also GF '. GX '.: GF' : GF. GX 
 
 :: GF' : GA' (4). 
 
 Let the directrix meet the asymptotes in D (fig. 88) : then by 
 the similar triangles GDX, GLA, 
 
 CL : GA :: GD : GX- 
 but GL = GF, therefore from (3) GD = CA, or the circle on A A, 
 as diameter will cut the asymptote in a point on the directrix. 
 
 Def. The circle on A A, as diameter is called the auxiliary 
 circle. 
 
156 AUXILIARY CIRCLE, &C. 
 
 Since CD=CA, CF= CL, and the angle DCF is common to 
 the two triangles DCF and A CL, 
 
 .'. the angle CDF^ the angle CAL ^ a right angle, 
 
 or the perpendicular from the focus on the asymptote is a tangent 
 to the auxiliary circle at the point of intersection. 
 
 Corollary. DX' = CX . FX. 
 Again, from (4), 
 
 CF'-CA' : CA' :: CF-CX : CX 
 
 :: FX"" : CX {CF-CX); 
 but CF.CX = CA' from (3), and CA'- CX' = AX . ^.A: 
 Also CF'-'CA'^CB', 
 
 .'. GB' : CA' :: FX' : AX.A.X; 
 and comparing this with the constant ratio above given for 
 
 .S^Ift-, we have FX' : AX.XA, :: BC : AC, which may 
 
 also be written 
 
 FX' : CX'-AC :: BC : AC (5). 
 
 Let FX (fig. 88), where F is any ^^oint on the curve and FX 
 the ordinate, meet the asymptotes in F, then 
 
 FX' : CX' :: BC : J C*, by similar triangles (6), 
 
 ... EX'-FX' : AC :: BC : AC, 
 or Fp.FF=^BC^FF.Fe, 
 
 where p is the point in which FX meets the curve again, and e 
 is the point in which it meets the other asymptote. 
 
 Let the ordinate through F meet the conjugate hyperbola in 
 F (same fig.), and let BM be the ordinate of li perpendicular to 
 BC, then of course 
 
 BM' : CM'-BC :: AC : BC, 
 or CM' : BC :: BM' + AC : AC; 
 
 but CM = MX and BM = CX, 
 
 .-. MX' : BC :: CX' + AC : AC (7); 
 
 I 
 
I 
 
 THE HYPERBOLA. 157 
 
 and combining (6) and (7), we get 
 
 RN' - EN^ = JBC = RE. Er = RE . Re, 
 where r is the point on the other branch of the conjugate hyper- 
 bola corresponding to R, and e is the point in which Rr meets 
 the other asymptote. 
 
 To draw a tangent and normal at any point of the curve 
 (Fig. 88). 
 
 Let P, be a point on the curve adjacent to any point P, and 
 let the chord PP^ meet the directrices in K and K^ . Draw KF 
 to the corresponding focus : then FP : FP^ : : PK : Pj K, or PK 
 bisects the exterior angle between PF and P^F produced (Euc. 
 VI. prop. A). Hence, exactly as in the case of the ellipse (p. 102), 
 when Pj moves up to and coincides with P, so that the chord 
 PP^ becomes the tangent at P, the line FK becomes perpendicular 
 to the line FP drawn from the focus to the point of contact of 
 the tangent. The tangent at any point P of an hyperbola may 
 therefore he drawn by drawing a line from P to either focus, 
 erecting a perpendicular to this line at the focus meeting the 
 directrix, and drawing the tangent through this point and the 
 proposed point of contact. It may also be drawn by making use 
 of the known property that it bisects the angle between the focal 
 distances. For in the two triangles PFK, PF^K^ 
 
 FP : PK :: F,P : PK„ 
 and the angle PFK— the angle PFJ{^, each being a right angle, 
 .-. the angle FPK=F^PK^. (Euc. vi. 7.) 
 
 Hence the normal bisects the exterior angle between the focal 
 distances. 
 
 Problem 90. To describe an hyperbola^ an asymptote CD, 
 a focus F, and a 2)oint P being given (Fig. 88). 
 
 From F draw FB perpendicular to CD, then D will be a point 
 on the directrix as has been previously proved. Through P draw 
 iy parallel to CD and make Pf=PF, then/ will be a second 
 point in the directrix, which is therefore determined. Draw CF 
 
158 GIVEN ASYMPTOTE, FOCUS AND POINT. 
 
 perpendicular to Df meeting the given asymptote in C, which will 
 evidently be the centre of the curve. Make CA, CA^ on CF each 
 equal CD and AA^ will be the transverse axis, and the curve is 
 completely determined. 
 
 Since Pf can be measured on either side of P there are gene- 
 rally two solutions. 
 
 Proof. The only step in the construction which is not obvious 
 is taking/ as a point on the directrix. It can easily be shewn to 
 hold in the hyperbola, for draw Ain parallel to the asymptote 
 meeting the directrix in wi, then in the hyperbola 
 FP : FA :: PM : AX 
 :: Pf :Ani, 
 where PM is a perpendicular on the directrix. 
 
 But Am = DL = AF, .'. FP=^Pf: 
 
 and conversely, if Pf be made ^ PF, f will be a point on the 
 directrix. 
 
 Problem 91. To describe an hyperbola, an asymptote CT, 
 a tangent Tt, and a focus F being given (Fig. 89). 
 
 From F draw FD perpendicular to CT meeting it in B, and 
 FY perpendicular to Tt meeting in Y. Then D and Y are points 
 on the auxiliary circle. Bisect BY in K and draw KG perpen- 
 dicular to DY meeting CD in C. C will be the centre of the 
 curve, CF the direction of the transverse axis, and CD or CY its 
 semi-length. 
 
 Problem 92. To describe an hyperbola, an asymptote CD, 
 a directrix DD^ and a point P being given (Fig. 89). 
 
 Prom D draw DF perpendicular to CD. DF will be a locus of 
 the focus. Through P draw P/* parallel to CD meeting DD^ in/, 
 and with centre P and radius Pf describe an arc cutting DF in F. 
 F will be a focus, and FC drawn perpendicular to DD^ will inter- 
 sect CD in C, the centre of the curve; which is therefore com- 
 pletely determined. 
 
 [The problem is exactly the converse of Prob. 90.] 
 
THE HYPERBOLA. 
 
 159 
 
 Problem 93. To describe an hyperbola^ the asymj^totes CD, CD^ 
 and a point F on the curve being given (Fig. 89). 
 
 Bisect the angles between the asymptotes by the lines ACA^, 
 BCB^; then ACA^ in the angle in which F lies is the position of 
 
 Fig.89. 
 
 the transverse axis. Through F draw QFq perpendicular to AA , 
 and meeting the asymptotes in Q and q. Take a mean propor- 
 tional, as Fb, between FQ and Fq. Fb will be the length C£ of 
 the conjugate semi-axis. 
 
 Draw ^^ parallel to ^^4^ meeting the asymptote in U ; then 
 BE is the length CA of the transverse semi-axis and CU = CF, the 
 distance of either focus from C. 
 
 Froof. The only step in the construction requiring demon- 
 stration is that in the hyperbola BC^ = FQ . Fq. 
 
 Let iV^ be the foot of the double ordinate Fp ; by similar 
 triangles CAE, CNQ. 
 
 QN' : AB' :: CJV' : AC and AB = BG, 
 
 .-. QN'-BC : BC :: CN^-AC^ : AC ; 
 
 but CjV'-AC'=AN'.NA^ and (p. 156) FN' : AJV.IfA^ :: BC : AC% 
 
 .'. QJV'-BC : BC :: FJH' : BC : 
 
160 GIVEN ASYMPTOTES AND TANGENT. 
 
 .-. QN'-BC' = Piy' or QN'-PN' = BC\ 
 
 Le. {QN-\-PN){QN'-PN) = BC' = PQ.Pq, 
 
 for by the symmetry of the curve Nq = NQ. 
 
 If qp be made equal to QP, p will evidently be a point on the 
 curve. 
 
 Problem 94. To describe an hyperbola, the asymptotes CT, Ct, 
 aiid a tangent Tt to the curve being given (Fig. 89). 
 
 Bisect Tt in P. P will be the point of contact of Tt, i. e. wdll 
 be a point on the curve, and the problem therefore reduces to 
 Problem 93. 
 
 The proof will be found on p. 163. 
 
 Definition. Any straight line drawn through the centre and 
 terminated both ways either by the original curve or by the con- 
 jugate hyperbola is called a diameter, and by the symmetry of the 
 curve every diameter is bisected by the centre. A diameter CD 
 parallel to the tangent at the extremity of a diameter CP is said 
 to be conjugate to CP. 
 
 The following important properties of the hyperbola should be 
 carefully noticed. 
 
 Prop. 1. If from any point Q in an asymptote QPpq be 
 drawn meeting the curve in P, p and the other asymptote in q, 
 and if CD be the semi-diameter parallel to Qq, 
 
 QP,Pq= CD' and QP^pq (Fig. 90). 
 
 Through P and D draw RPr, DTt ^perpendicular to the trans- 
 verse axis, and meeting the asymptotes in E, r and T, t ; let Rr 
 meet the axis in N, 
 
 RP :: CD : DT\ , • i , • i 
 Pr :: CD : Dt ] ^7 ^""^1^^ ^^^^^S^^^' 
 
 RP.Pr :: CD'' : DT.Dt. 
 = £C' = DT.Dt{i>. 159), 
 . QP.Pq = CD\ 
 
 Then 
 
 QP'. 
 
 and 
 
 Pq: 
 
 
 .♦. QP.Pq 
 
 But 
 
 RP.Pr 
 
THE HYPERBOLA. 
 
 Similarly qp.pQ= CD'' = QP . Pq ; 
 
 or, if Vhe the middle point of Qq, 
 
 QV'-PV'=QV'-pV\ 
 Hence PV^pV, and therefore PQ = pq. 
 
 161 
 
 Fig.QO. 
 
 Cor. If a straight line PP^ p^p meet the hyperhola in P, p, 
 and the conjugate hyperbola in P^, p^, PP^ =PPv 
 For if the line meet the asymptote in Q, q, 
 P^ QP^^Piq and PQ-^qp, .'. PPi=pp^. 
 
 Prop. 2. A diameter bisects all chords parallel to the tangents 
 nt its extremities ^ i. e. all chords parallel to its conjugate. 
 
 This can be proved exactly as in the analogous proposition for 
 the ellipse. 
 
 Let QQi {G^g. 91) be any chord of an hyperbola meeting the 
 directrix in E, and let be the centre point of QQ^ and F the 
 focus. 
 
 Join FQ, FQ^y and draw i^F perpendicular to QQ^. 
 Then FQ' - FQ,' = QY'- Q, Y' 
 
 = {QY+Q,Y){QY-QJ) 
 
 = 2.QQ^.0Y (1); 
 
 but since Q and Q^ are on the hyperbola, 
 
 FQ : FQ^:: QR : Q^E; 
 E. 11 
 
162 
 
 CONJCGATE DIAMETERS. 
 
 therefore F^-FQl _QE' -Q,R^ 2QQ,.0S 
 
 ttieietore ~-^^, '"W ^ ' ^^W^ " 
 
 therefore, from (1) and (2), 
 
 OY FQ' F^ 
 OR " QR' '~ UP ' 
 
 Fig.91. 
 
 (2): 
 
 where AW is drawn through the vertex parallel to QR meeting 
 the directrix in W. 
 
 I.e. OY : OR in a constant ratio. 
 
 Take any second chord qq^ parallel to QQ^ meeting FY in Y^ 
 
 and the directrix in ^j. Let 0^ be its centre point ; then, since 
 
 OY Y 
 
 (YR ^ 7)1^ ' ^* follows that the line 00^ must pass through the 
 
 point T in which FY meets the directrix, and is therefore 
 fixed for all chords parallel to QQ^. This line will evidently pass 
 through the centre (i. e. will be a diameter), for by the last pro- 
 position it bisects all chords of the conjugate hyperbola parallel 
 to QQ^ , i. e. it bisects the diameter Dd^ which is also bisected by C. 
 Let TO meet the hyperbola in P and suppose qq^ to move 
 parallel to itself till it approaches and ultimately coincides with P. 
 Since 0^q=0^q^ throughout the motion, the points q^ q^ will evidently 
 
THE HYPERBOLA. 163 
 
 approach P simultaneously, and in the limiting position qq^ will 
 be the tangent at P. It follows that if P^ be the other extremity 
 of the diameter through P, the tangent at P^ is parallel to QQ^, 
 and therefore to the tangent at P. 
 
 Corollary 1. The perpendicular on the tangent at any point 
 from the focus meets the corresponding diameter in the directrix. 
 
 \ Cor. 2. If the tangent at P meet the asymptotes in E and e, 
 PE = Pe, for by the last proposition the intercept between q and 
 one asymptote is always equal to the intercept between q^ and the 
 other asymptote, and when q and q^ ultimately coincide with P 
 these intercepts become PE and Pe respectively, i.e. the portion 
 of any tangent between the asymptotes is bisected at the point of 
 contact. 
 
 Cor. 3. If PE be the tangent at P meeting the asymptote 
 in Ey PE^ = CD^, where CD is the semi-diameter conjugate to GP. 
 For taking a parallel chord very near the tangent meeting the 
 curve in p, p^ and the asymptote in e, we have, by Prop. 1, 
 
 ep . ep^ = CD\ 
 and therefore when p and p^ coincide in P, 
 
 EP' = CI)\ 
 
 Cor. 4. The asymptotes are the diagonals of the parallelogram 
 formed by the tangents at the extremities of a pair of conjugate 
 diameters. For E and e, which are on the asymptotes, are also 
 angular points of such a parallelogram. 
 
 Prop. 3. Tangents drawn at the extremities of any chord sub- 
 tend equal angles at the focus. 
 
 Let PQ (fig. 92) be any chord of an hyperbola and let the 
 tangents at P and Q meet in R. Let F be the focus, and from R 
 draw RN, RM perpendicular respectively to FP, FQ ; draw RW 
 perpendicular to the directrix and let the tangent at P meet the 
 directrix in E. 
 
 Then EF is perpendicular to FP (p. 157), and therefore 
 parallel to RK 
 
 11—2 
 
164 CONJUGATE DIAMETERS. 
 
 Therefore FN i FP :: ER : EP 
 
 :: RW : PK, 
 where PK is the perpendicular from P on the directrix. 
 
 Fig.92. 
 
 Therefore FN : RW :: FP : PK 
 
 :: FA : AX. 
 
 Similarly FM : RW :: FA : AX; 
 
 therefore FiV^FM. 
 
 Hence in the right-angled triangles RFX, RF3f, FX=FM, 
 and FR is common. 
 
 Therefore the two triangles are equal in all respects, i.e. tlie 
 angle RFP=^ihe angle RFQ, and RN=RM. 
 
 Prop. 4. If PCP^ he a diameter and QVQ^ a chord 2^cirallel 
 to the ta7igent at P and meeting PP^ produced in V, and if the 
 tangent at Q meet PP^ in T, then CV.CT= CP' (Fig. 92). 
 
 Let 2'Q meet the tangents at P and P^ in R and r, and F being 
 a focus draw RN perpendicular to the focal distance FP meeting 
 
THE HYPERBOLA. 165 
 
 it in i\^, rn perpendicular to FP^ meeting it in n, and RM, rm 
 perpendicular to the focal distance FQ. Let F^ be the other 
 focus, and join F^P, F^P^. 
 
 Since CF=CF,, CP^CP,, and the angle i^CP = the angle 
 F[CP^, therefore the triangles FGP, FfiP^ are equal in all re- 
 spects; and therefore the angle CPF — the angle CP^F^. 
 
 Similarly the angle CPF, = the angle CP,F. 
 
 Therefore the whole angle FPF^ = the whole angle F^P^F; but 
 the tangents bisect the angles between the focal distances, there- 
 fore the angle FPP = the angle FP^r ; i.e. the right-angled tri- 
 angles RPN, rP^n are similar, and therefore 
 RP : rP^ :: RN : rn; 
 but RN'=RM SiTid roi = rm (Prop. 3), therefore 
 RP : rP, :: RM : rm 
 :: RQ : rQ. 
 
 But TR : Tr :: RP : rP^ 
 
 :: RQ : rQ; 
 therefore TP : TP, :: PV : PJ 
 
 by similar triangles, 
 
 or CP-CT : CT+GP :: CV-CP : CV+CP, 
 
 i.e. CT : CP :: CP : CV; 
 
 therefore CT.CV=CP\ 
 
 Cor. 1. Since CFand CP are the same for the point Q^, the 
 tangent at Q, passes through 2\ or the tangents at the extremities 
 of any chord intersect on the diameter which bisects that chord. 
 
 Prop. 5. JfPCP,, BCD, he conjugate diameters, and QV he 
 drawn j^drallel to CD meeting the hyperhola in Q and CP in V, then 
 QV : PV.PJ :: CD^ : GP\ 
 
 Let the tangent at Q (fig. 92) meet CP and CD in T and t 
 respectively, and draw QU parallel to CP meeting CD in U. 
 
 Then CV. GT= CP' and CU . Ct = CD' (Prop. 4) ; 
 but CU=QV, 
 
 therefore CD' : CP' :: QV.Ct : CV.CT; 
 
 but Ct : QV :: CT : VT, 
 
 .'. CD' : CP' :: QV : CV. VT, 
 
166 GIVEN AXIS AND POINT. 
 
 and CV. VT =CV {CV - CT) = CV - CP' ^ PV . PJ; 
 therefore QV : PV.PJ '.-. CD' : CP\ 
 
 Peoblem 95. To describe an hyperbola, the transverse axis 
 AA^ and a point P on the curve being given (Fig. 88). 
 
 Bisect AA^ in (7, which will of course be the centre of the 
 curve. Draw the conjugate axis BCB^. Let PJ,, PA cut BB^ 
 in &j and h respectively. Take a mean proportional CH between 
 Cb and Cb^, which will be the length {CB or CB^ of the semi- 
 conjugate axis. The foci can then be determined, since CF=AB. 
 
 Proof. Let PJSf be the ordinate at P. 
 
 Then PN : bfi :: NA^ : CA^, 
 
 and PN : bC :: NA : CA, 
 
 or PJST' : iVM.iT^, :: bC.bfi : CA'; 
 
 therefore bC . bfi = BC (p. 156). 
 
 Problem 96. To describe an hyperbola, the transverse axis 
 ATA^ and a tangent PT being given (Fig. 88). 
 
 Bisect AA^ in C, the centre of the curve. On CA towards 
 CT take CN a third proportional to CT and CA. N will be the. 
 foot of the ordinate of the point of contact of the given tangent ; 
 i.e. if IsP be drawn perpendicular to AA^ meeting TP in P, P 
 will be a point on the curve, and the problem therefore reducea 
 to the preceding. 
 
 It may also be completed by Prob. 19, p. 23, determining 
 two lines PF, PF^ making equal angles with PT and meeting 
 AA^ in points equidistant from C\ since it has been already 
 shewn (p. 157) that the tangent bisects the angle between the 
 focal distances. 
 
 The proof follows from Prop. 4, p. 164, which of course applies 
 to the principal axes. 
 
 Problem 97. To describe an hyperbola, a pair of cov jugate 
 diameters being given (Fig. 93). 
 
 PCP^, DCD^ are the given conjugate diameters. 
 
THE HYPERBOLA. 
 
 167 
 
 First Method. Complete the parallelogram QtqT formed by the 
 tangents at their extremities; then the diagonals of this parallelo- 
 
 gram are the asymptotes (p. 1G3), and the axes therefore bisect 
 the angles between them. Thus CA and CB are determined in 
 <lirection. 
 
 From P draw two lines PF, PF^ making equal angles with 
 PT, the tangent at P, and meeting ^^, in points i^ and F^ equi- 
 distant from C (Prob. 19, p. 23). Then F and F^ are the foci, 
 and the vertices can be determined by dropping perpendiculars 
 on AA^ from the points in which the circle on FF^ as diameter 
 intersects the asymptotes. 
 
 The curve can therefore be drawn by the general method. 
 
 Second Method. Points on the curve can also be determined 
 without finding the foci thus : 
 
 Complete the parallelogram QtqT as before. 
 
 Divide QD into any number of equal parts as at 1, 2, 3. 
 Divide CD^ into the same number of equal parts as at Ij, 2^, 3j; 
 then PI andPjlj will, when produced, intersect in a point on 
 the curve, and similarly with the other corresponding points. 
 
168 
 
 GIVEN CONJUGATE DIAMETERS. 
 
 This method can also of course be applied to the principal 
 axes; it cannot however be recommended, because a slight in- 
 accuracy in the position of either line makes a considerable 
 alteration in the position of the point on the curve, since in 
 producing the lines the error is magnified, and the lines must 
 often be produced to a considerable distance. 
 
 Problem 98. To describe an hyperbola, the centre C, the direc- 
 tions of a pair of conjugate diameters CA, CB^ and two j^oints on 
 tfie curve P and Q being given (Fig. 94). 
 
 Draw PKj Qn parallel to CB meeting CA in N and n. (Let 
 PN be less than Qn ; then CiY must be less than Cn.) Produce 
 
 Qn to q and draw PMP^ parallel to CA, meeting CB in M. 
 Make MP^ = MP and nq = nQ. Then P^ and q are points on 
 the curve. Let PP^ meet Qq in JE. Through n draw nx parallel 
 to QP^ meeting PP^ in x, and through n draw ny parallel to Pq 
 meeting PP^ in ?/. Take Ed a mean proportional between Ex 
 and Ey. On CN describe a semi-circle CDN and make ND — Ed. 
 CD will be the length CA of the diameter parallel to PP, . On 
 CA make CG = Ed, and through A draw BA parallel to MG. CB 
 will be the diameter conjugate to CA, and the problem reduces 
 to the preceding. 
 
i 
 
 THE HYPERBOLA. 1C9 
 
 Proof. The construction, as in the similar problem for the 
 ellipse, depends on the property of the curve, that "the rectangles 
 contained by the segments of any two chords which intersect each 
 other are in the ratio of the squares on the parallel diameters," 
 
 i.e. EP.EP^ \ EQ.Eq :: GA^ : CB\ 
 which may be thus proved : 
 
 Through E draw the diameter EPE^, and draw the ordinate 
 EU parallel to Qq or to CB ; then, by Prop. 5, p. 165, 
 EU' : CU'-CA' :: CB' : CA', 
 .'. CB'^EU' : CB' :: Ca' : GA\ 
 and qn' : Cn'-CA' :: CB' : CA'; 
 
 .'. CB'+qn' : CB' :: Cn' : CA', 
 so that CB' + EU' : CU' :: CB'+qn' : Cn'. 
 
 But EU' : CU' :: En' : Cn', 
 
 .'. CB' : CU' :: CB'+qn'-En' : Cn\ 
 
 or CB' : CB'+qn'-En' :: CU' : Cn' 
 
 :: CR' : CE'; 
 .'. CB' : qn'-En' :: CR' : CE'-CR', 
 or CB' : EQ.Eq :: C7e' : ER . ER^. 
 
 Similarly CA' : EP . EP^ :: CR' : ER.ER^, 
 
 .-. EP.EP^ : EQ.Eq :: CA' : Ci?'. 
 Bat by construction EQ : ^P^ :: En : Ex, 
 and j5'^ : EP :: ^7^ : %, 
 
 EQ.Eq : EP.EP^ :: J5'/i' : Ex .Ey; 
 but ^o; . Ey = Ed' = Niy =CN' -CA' == AN . NA^, 
 
 .-. En' or PxV'' : JiV.iYJ, :: EQ.Eq : EP.EP^, 
 which proves that AA^ is the diameter conjugate to PJS\ 
 Also by construction 
 
 CB : CA :: CM : CG :: PN : ND, 
 .'. CB' : CA' :: PN' : ^iV^.iYi,, 
 or CB is the semi-diameter conjugate to CA. 
 
170 GIVEN DIEECTIOXS OF CONJUGATE DIAMETERS, &C. 
 
 Problem 99. To describe an hyperbola, the centre C, the 
 directions of a pair of conjugate diameters CT, Ct^ a tangent Tt, 
 and a point P on the curve being given (Fig. 95). 
 
 [If a line be drawn parallel to Tt and at an equal distance 
 from (7, it will of course be a second tangent, and P must not 
 
 lie between these lines.] Draw PVp parallel to Ct cutting CT 
 in F, and make Vp = VP. p) will be a point of the curve. One 
 of the two diameters CP or Cp (in the figure Cp) will always 
 intersect Tt in a point (L) outside Tt\ draw such diameter and 
 on it make CP^ = Cp. P^ will be a point on the curve. 
 
 Take Lm a mean proportional between Lp and LP^. On Tf 
 as diameter describe a circle; through L draw LK perpendicular 
 to Lt and on it take a point K such that Cp : Lm :: ^Tt : ZA', 
 i.e. on LK make Lm^—Lm, and on Lt make Lo = \Tt and Lp^ = (7/a 
 Through o draw oK parallel to ni^p^, K will be the point required ; 
 tlien tangents KQM, Q^KM^ from K to the circle on 2't will intei- 
 sect Tt in points (^Q, and Q^) either of which may be taken for its 
 2)oint of contact with the curve. There are therefore two solutions. 
 
 Through G draw DCd parallel to Tt, make CD'^Cd=QM, 
 the tangent from Q to the circle. CQ and CD will be conjugate 
 semi-diameters, and the problem reduces to Problem 97. 
 
THE HYPERBOLA. 
 
 171 
 
 Problem 100. To describe an hyperbola, the centre C, two 
 tangents PT, QT and a point on the curve (R) being given 
 (Fig. 96). 
 
 Through G draw TCT^ and make CT^ = GT. Draw T^t parallel 
 to PT meeting QT in t, and draw T^t^ parallel to QT meeting PT 
 
 in t^. tT^ and t^T^ will be tangents to the curve and TCT^, tCt^ 
 will be the directions of a pair oi conjugate diameters; and the 
 problem therefore reduces to the preceding. 
 
 [The point R must lie outside the parallelogram TtT^t^ and 
 within one of the exterior angles, such as PTQ.^ 
 
 In the figure RC = GR^ and RG meets Tt in L ; Lm is a mean 
 ])roportional between LR and LR^) and LK : Lm :: GT : GR, 
 where 02'= -|7'^, and ZK is perpendicular to Tt. Then a tangent 
 /CM from K to the circle on Tt as diameter cuts Tt in its point 
 of contact (Q) with the curve, and GD drawn through G parallel 
 to QT and equal to ^ J/ will be the semi-diameter conjugate to GQ. 
 
 There are two solutions, as two tangents can be drawn from 
 K to the circle on Tt. 
 
172 
 
 GIVEN CENTRE, TWO POINTS AND TANGENT. 
 
 Problem 101. To describe an hyperbola, the centre C, two 
 points {A and B) of the curve and a tangent Tt being given 
 (Fig. 97). 
 
 [A second tangent can at once be drawn parallel to Tt on the 
 other side of G and at the same distance from it; the points A 
 and B must not lie between these lines.] If the given points lie 
 on opposite branches of the curve, as e.g. A and B^, i.e. if they 
 are on opposite sides of Tt, draw BfiB and make CB = GB^, then 
 B will be on the same branch as A. 
 
 \"^. 
 
 ^ra, 
 
 Fig.97. 
 
 Draw AB and bisect it in V. Draw GV meeting the given 
 tangent in T^ and Gt parallel to AB meeting it in t. Then GT, 
 Gt are the directions of a pair of conjugate diameters, and the 
 problem reduces to Prob. 99. 
 
 In the figure GA^ = GA and AGA^ meets Tt in Z; Lin is a 
 mean proportional between LA and LA^; LK : Lm :: OT : GA^ 
 where LK i^ perpendicular to Tt and OT — ^Tt. 
 
 Then a tangent (Kq) from K to the circle on Tt as diameter 
 cuts Tt in Qf its point of contact with the curve. GJJ parallel 
 to Tt and equal to Qq is the semi-diameter conjugate to GQ. 
 
THE HYPERBOLA. 
 
 173 
 
 Problem 102. To describe an hyyerhola^ the centre C, and 
 three tangents {SV, VW, WS) being given (Fig. 98). 
 
 Through G draw TCT^ meeting SV in T and SW in T,, so that 
 TC=GT^ (Prob. 14, p. 19). CT will be conjugate to CS. Draw 
 
 Fig.98. 
 
 \T V parallel to OF meeting VW i\\ v, Tw parallel to CW meeting 
 VW in ly, and draw vT, loT^ meeting in E. Then ES will cut 
 
 \VW in P, its point of contact with the curve. Also PQ parallel to 
 iT will cut VS in Q, its point of contact, and QR parallel to CI' or 
 
 \PR parallel to T^w will cut WS in R, its point of contact. The prob- 
 lem can be completed by several of those previously given or thus; 
 
 Draw QN parallel to CT meeting CS in ^Y. QN is an ordinate 
 
 fof the diameter CS, and therefore CA, the length of the semi- 
 
 Fdiameter, is a mean proportional between CS and CN (Prop. 4, 
 
 p. 164). Similarly, if Qn be drawn parallel to C'aS' meeting CT in 
 
 n, CB must be taken as a mean proportional between' (7^ and CT. 
 
 Problem 103. To describe an hyperbola, the centre G and 
 three points P, Q, R being given (Fig. 99). 
 
 [Each of the points must lie either between both pairs of lines 
 furnished by the remaining points and their corresponding points, 
 or outside both these pairs of lines.] 
 
174 GIVEN CENTRE AND THREE POINTS. 
 
 Bisect PQ in ^, QR in q, and RP in r, and draw Cp, Cq and 
 Cr, producing each indefinitely. PQ is a double ordinate of the 
 
 Fig.99 
 
 diameter Gp, and therefore the tangents at P and Q will intersect 
 on Cp; similarly the tangents at Q and R will intersect on Cq, and 
 at R and P on Cr. If therefore a triangle be drawn (Prob. 15, 
 p. 20), the sides of which pass through P, Q and R, and the 
 vertices of which lie on Cp, Cq and Cr respectively, the sides of 
 this triangle will be the tangents at P, Q and R. Take any 
 point a on Cr; draw Pa, Ra cutting Cp, Cq in b and c respectively ; 
 join be cutting PR in x, and draw xQ cutting Cb in T. QT, PT 
 will be the tangents at Q and P respectively; and if PT meet Cr 
 in t, Rt will be the tangent at R. The problem may be completed 
 by preceding problems. 
 
 Problem 104. To describe an hyperbola, the foci F, I\ and a 
 point P on the curve being given (Fig. 100). 
 
 It has been shewn already that the difference between the 
 focal distances of any point on the curve is equal to the transverse 
 axis (p. 153). 
 
 Let F^P be greater than FP. On PF^ make Pf= PF. Draw 
 FF^, bisect it in G and make CA = CA^ = ^FJ. AA^ will there- 
 
THE HYPERBOLA. 
 
 175 
 
 fore be the vertices of the curve, and the problem reduces to 
 Prob. 89. 
 
 Fig.lOO. 
 
 Problem 105. To describe an hyperbola^ the foci F, F^ and a 
 ingent Tt being given (Fig. 1 00). 
 
 [The tangent must lie between F and F^.'] 
 
 Bisect FF^ in (7, the centre of the curve. From F or F^ drop 
 a perpendicular (as FY) on the tangent. On CF, GF^ make 
 CA = GA^ = GY. A and A^ will be the vertices of the curve, and 
 the problem reduces to Prob. 89. 
 
 Problem 106. To describe an hyperbola, a focus F, a tangent 
 FT with its point of contact P, and a second point Q on the curve, 
 being given (Fig. 100). 
 
 If F and Q are on the same side of FT, the solution has already 
 been given in the corresponding problem for the ellipse (Prob. 76). 
 Hence the case of F and Q lying on opposite sides of FT need 
 alone be considered here. 
 
 From F draw FY perpendicular to FT meeting it in Y, on 
 FY produced make Yf=^ YF; then Pf will be a locus of the 
 second focus. From / on fF, on either side of f make fq = FQ. 
 Bisect Qq in r, and draw rF^ perpendicular to Qq meeting Ff in 
 F^ , which will be the second focus. Hence, both foci being known, 
 the problem may be completed by Probs. 104 or 105. 
 
 Since q may be taken on either side of f there are in general 
 two solutions. 
 
176 GIVEN FOCUS, TANGENT AND TWO POINTS. 
 
 Proof. That/P is a locus of the second focus has been shewn 
 in p. 157 ; that the second focus is at the intersection of fF 
 and tF^ is evident thus : — it must he so situated that 
 
 F,P~FF = FQ~Ffi; 
 but /Q=FQ and F,q = F^Q, 
 
 and FP=fP, .'. F^P^FP = F^P^JP=fF^, 
 
 i. e. F^ is the second focus. 
 
 If F and Q are on opposite sides of FT, two hyperbolas can in 
 general be drawn. 
 
 If F and Q are on the same side of FT, and the distance of Q 
 from F is greater than its distance from the line drawn through y* 
 perpendicular to Pf two hyperbolas can in general be drawn. 
 
 If F and Q are on the same side of FT, but the distance of 
 Q from F is less than its distance from the above perpendicular, 
 one hyperbola only can in general be drawn, but an ellipse can 
 also be drawn. 
 
 If F and Q are on the same side of Pl\ and the distance of Q 
 from F is equal to its distance from the above perpendicular, a 
 parabola can be drawn fulfilling the required conditions, but no 
 hyperbola or ellipse, since the second focus removes to an infinite 
 distance. 
 
 Problem 107. To describe an hyperbola, a focus F, a tangent 
 RT and two points P and Q of the curve being given (Fig. 101). 
 
 If F, P and Q are all on the same side of RT, the solution 
 has already been given in Prob. 77, the corresponding jiroblem 
 for the ellipse. Hence the cases of one or both of the points P, Q 
 lying on the opposite side of RT to F need be considered. 
 
 Case 1. Let F and P be on the same side of RT, and Q on 
 the opposite side. Produce PF to q, make Fq = FQ, and with 
 centre P and radius Pq describe a circle qG. From F draw FY 
 perpendicular to RT, produce it tof and make Yf= YF. With 
 centre f and radius FQ describe a circle GIf, and find the centre 
 (F^) of a circle touching the circles qG and Gil internally and 
 
THE HYPERBOLA. 
 
 177 
 
 passing through Q (Prob. 32). F^ will be the second focus. The 
 problem is always possible, since the circles must necessarily cut 
 each other and the point Q be inside both. 
 
 FIg.lOI. 
 
 fq 
 
 I 
 
 Proof. The second focus F^ must be equidistant from Q and 
 from the circle qG, since F^P — FP must be equal to FQ — F^Q. 
 But F,P by construction^ P^-i^\^ and Pq = FP + Fq = FP+FQ. 
 
 Also fF^ must be equal to the transverse axis, p. 153, 
 i.e. to FQ—F^Q or to /G — F^Q, i.e. the second focus must be 
 equidistant from the point Q and from the circle GIL 
 
 Case 2. Let P and ^ be on the opposite side of ^7^ to i?^, as 
 P, and Q. Let FP^ be greater than FQ. On FP^ make Fq^ = FQ, 
 and with centre P, and radius P^q^ describe the circle q^ff. Deter- 
 mine the point / as in Case 1, and with centre / and radius FQ 
 E. 12 
 
178 
 
 GIVEN FOCUS, POINT AND TWO TANGENTS. 
 
 draw the circle Gil. Determine F^ , the centre of a circle touching 
 qjl externally and GH internally. F^ will be the second focus, 
 Froof. FF^ - F^P^ = Fq^ + q^P^ - F^P, = Fq^ - (F^P^ - q^PJ 
 
 = FQ-F^Q 
 =fF^, as in Case 1. 
 
 Problem 108. To describe an hyperbola, a focus F, a poijit 1 
 on the curve, and two tangents TQ, TE being given (Fig. 102). 
 
 Fig.l02. 
 
 [F and P must be either both on the same side or both or 
 opposite sides of each tangent.] 
 
 If F and P are on the same side of each tangent, the necessary 
 condition for a possible solution has been explained in the corre 
 spending problem for the ellipse, Prob. 78, p. 127, and the sohi 
 tion given. If they are on opposite sides, as in fig., draw FYj 
 perpendicular to QT meeting it in F, and FY^f perpendiculai 
 to RT meeting it in T,, and make Yf= YF and YJ, = Y^F. 
 
 With centre P and radius PF describe a circle FG, and fine 
 F^ the centre of a circle to touch FG and to pass through y and y*, 
 Prob. 27. Since y and y, will necessarily lie within the circle FG 
 two solutions can generally be obtained. 
 
 Proof. If F^ is the second focus, fF^ "^fi^x ^^^ transvers( 
 axis = FP — F^P, which by construction it does. 
 
THE HYPERBOLA. 
 
 179 
 
 Problem 109. To describe an hyperbola, a focus F and three 
 tangents PT, QT and RS being given (Fig. 103). 
 
 ^- 
 
 Fig.103. 
 
 \^F must not lie within the triangle formed by the tangents.] 
 From F drop perpendiculars FYf, FY^f , FY^f„ on the given 
 tangents, meeting them respectively in F, Y^ and Y^, and make 
 Yf^ YF, YJ^^Y^F, and YJ^ = Y^F; then, as/,/, and/ must all 
 be equidistant from the second focus (p. 153), and the problem 
 therefore reduces to finding the centre [F^ of a circle passing 
 through three given points (Prob. 20), F^ will be the second 
 focus, and the transverse axis is of course known, since it is equal 
 to FJ. 
 
 Problem 110. To describe an hyperbola, a focus F and three 
 points P, Q, R on the curve being given (Fig. 104). 
 
 [With two of the points as centres describe circles passing 
 through F. The three given points cannot lie on the same branch 
 of an hyperbola, unless 
 
 (1) F lies in one of the three angles PQR, QRP, and RPQ; 
 and (2) the third point is more distant from F than it is from 
 the common tangent to the above circles remote from F. 
 
 12—2 
 
180 
 
 GIVEN FOCUS AND THREE POINTS. 
 
 Whatever the relative position of the points and focus three 
 hyperbolas can always be drawn.] 
 
 Flg.l04. 
 
 
 The points being as in the figure, the above conditions for the 
 points lying on the same branch are not complied with; an ellipse 
 and three hyperbolas can be drawn by the first solution of the 
 corresponding problem in the preceding Chapter, Problem 80. 
 
 The second solution there given can be adapted to the present 
 case thus: Let F and Q lie on one branch of the required hyper- 
 bola and E on the other. 
 
 Bisect the angle FFQ by the line FC, bisect the angle between 
 FF and FF produced by FF, and the angle between QF and FF 
 2jroduced by FF. 
 
 Determine the triangle whose sides pass through F termi- 
 nating on FC and FF, through Q terminating on FF, FC, and 
 through F terminating on FF and FF (Problem 15). The sides 
 of this triangle will be tangents to the required curve at F, Q 
 and R respectively. 
 
THE HYPERBOLA. 181 
 
 To determine the triangle. Take any point G on FC, draw 
 CP, CQ cutting CD, CE in D and E, and draw ED, QP intersect- 
 ing in X. RX will be the tangent at R, and if it meets FE in T^ 
 QT will be the tangent at Q ; similarly, if QT meets FG in T^ , 
 PT^ will be the tangent at P, passing also through the intersection 
 of RT and DF. 
 
 The construction depends on the well-known property of the 
 hyperbola, that the angles subtended at the focus by a pair of 
 tangents are equal or supplementary according as the tangents 
 touch the same or opposite branches of the curve. 
 
 p Problem 111. To describe an hyperbola, two tangents TQ, TR, 
 with their points of contact Q and R, and a point P on the curve 
 being given (Fig. 105). 
 
 [The point P must lie outside the parabola which can be de- 
 scribed touching TQ, TR at Q and R.] 
 
 The construction is exactly similar to the corresponding 
 problem for the ellipse. Prob. 81. 
 
 Eisect QR in V and through T draw TBV. Through P draw 
 PLL^ parallel to QR meeting QT in L and RT in L^ . 
 
182 
 
 GIVEN TWO TANGENTS AND THREE POINTS. 
 
 Find a mean proportional {Lk) between PL and PL^. (If 
 PjZj be made equal to PL, P^ will be a point on the curve.) On 
 PPj make PK= Lk, then QK will intersect TV inD, the extremity 
 of the diameter TV. 
 
 On TV take a point G such that TC : CD -.: CD : CV ', and if, 
 as in the figure, Q and R are on opposite branches of the hyper- 
 bola, C must be taken between T and V ; i.e. on TV as diameter 
 describe a semi-circle; draw DM making an angle of 45" witli 
 D V and meeting the semi-circle in M, and from M draw MC per- 
 pendicular to TV. Evidently MC is a mean proportional between 
 GT and CV, and is equal to CD. C will be the centre of the 
 hyperbola, and the asymptotes can easily be determined and the 
 curve completed by preceding problems. 
 
 The proof is identical with that for the ellipse. 
 
 Problem 112. To describe an hyperbola, two jjoinfs A and B 
 on the curve and three tangents PQ, QE, RP being given (Fig. 106). 
 
 [Either no one of the three tangents must pass between the 
 
 xi5, 
 
 FIg.106. 
 
 Yd 
 
 \ 
 
 points or all three must do so, and the points must not lie withii 
 the triangle formed by the tangents.] 
 
 Draw a line through AB cutting the tangents through P i^ 
 L and M and the remaining tangent in iV^. 
 
THE HYPERBOLA. 188 
 
 Find X the centre and i>, i>, the foci of the involution A^ B 
 and L, M (Prob. 13). D or D^ will be a point on the chord of 
 contact of the tangents FL, PM, 
 
 [In the figure La on PQ = LA , and Bm on a parallel to PQ=BM; 
 am cuts AB in X, the required centre, and XD = XD^ = Si mean 
 jjroportional between XAI and XL.] 
 
 Similarly, find X^ the centre and B, E^ the foci of the in- 
 volution A, B and M, N (Prob. 13), and E or E^ will be a point 
 on the chord of contact of the tangents RM and ^iV^. 
 
 Find MV the harmonic mean between ME and MD, M being 
 the point of intersection of AB with the given tangents which has 
 appeared in each of the above involutions, then QV {Q being the 
 intersection of tangents through N and L) will meet the tangent 
 through M in its point of contact (q) with the curve. 
 
 Therefore qDr will be the chord of contact of the tangents 
 PQ, PR, and Eqp the chord of contact of the tangents RP^ RQ. 
 
 The proof is identical with that for the ellipse, p. 135. 
 
 Problem 113. To describe an hyperbola, two tangents TP, TQ 
 and three points A, B, G on the curve being given (Fig. 107). 
 
 [The points A, B, C being taken together in pairs, each pair 
 
 Fig.107. /f, 
 
184 GIVEN TWO POINTS AND THREE TANGENTS. 
 
 of points must be either both on the same side or both on opposite 
 sides of both tangents. In the figure A and B are both on the 
 same side, and B and G on opposite sides of both TF and TQ, 
 as also C and A.'] 
 
 Draw the line AB cutting the given tangents in F and Q. 
 Find X the centre and U, E^ the foci of the involution A, B and 
 P, Q (Problem 13). 
 
 [In the figure Fa ~- FA and BQ^ parallel to Fa = BQ. Q^a cuts 
 AB in X, the required centre. XE — XE^ = a mean proportional 
 between XA and XB.] 
 
 E or E^ will be a point on the chord of contact of the given 
 tangents. 
 
 Again, draw BC cutting the given tangents in p and q, and 
 find Xj the centre and G, G^ the foci of the involution B, G and 
 p, q. G or G^ will be a second point on the chord of contact of the 
 given tangents, the points of contact of which R, R^ are therefore 
 determined, and the problem reduces to several preceding. 
 
 Since E and E^ can be joined to either G or G^ four chords 
 of contact can in general be drawn, so that there are four solutions. 
 The construction depends on Prop. 7, p. 143. 
 
 Problem 114. To describe an hyperbola, five tangents AB, BC, 
 CD, BE, EA being given (Fig. 108). 
 
 [The pentagon formed by the given tangents must contain 
 a re-entering angle.] 
 
 Draw AC, BD intersecting in F ; and through the remaining 
 angular point E of the pentagon draw EF meeting BC in F. 
 F will be the point of contact of the given tangent BC. Similarly, 
 if BD and CE intersect in G, AG will intersect DC in Q, the 
 point of contact of the given tangent CD ; and if CE and DA 
 intersect in U, BH will intersect ED in R, its point of contact. 
 
 The problem therefore reduces to Problem 111, or the poini 
 of contact S and T of the remaining tangents can easily be deter^ 
 mined. 
 
THE HYPERBOLA. 
 
 185 
 
 The construction depends (as in the corresponding problem 
 for the ellipse, p. 136) on Brianchon s theorem. 
 
 Fig.108. 
 
 Problem 115. To describe an hyperbola, Jive j^oints ABODE 
 Ing given (Fig. 109). 
 
 Draw AB, BE intersecting in F, and BC, EA intersecting in G\ 
 then, if FG meet CD in H, H will be a point on the tangent at J, 
 which can therefore be drawn. 
 
186 
 
 GIVEN FIVE POINTS. 
 
 If a line be drawn through G and the intersection of AB and 
 DO, meeting ED in K, K will be a point on the tangent at B. 
 Hence two tangents with their points of contact being known 
 and also (at least) one other point on the curve, the problem may 
 be completed by Prob. Ill, or the tangents at (7, D and E may 
 also be found by a similar construction to the above. 
 
 [If CD and EA intersect in L, and through L a line LM 
 be drawn passing also through the intersection of BG and DE 
 and meeting BD in M, M will be a point on the tangent at G ; 
 and if LM meet AB in iV, N will be a point on the tangent at D.'\ 
 
 The construction of the tangent at E is left as an exercise 
 for the student. 
 
 The construction (as in the corresponding problem for the 
 ellipse, p. 138) is an adaptation of Pascal's theorem. 
 
 Problem 116. To describe an hypei-hola, four tangents AB, 
 BG, GD, DA and a point E on the curve being given (Fig. 110). 
 
 Join EG and ED, cutting ^^ in c and D^ respectively. Find 
 X the centre and F and F^ the foci of the involution A, c and B, D^ 
 
 Fig.llO. 
 
 (Prob. 13); F or F^ will be a point on the tangent at E, which 
 can therefore be drawn and the problem completed by Prob. 114. 
 
THE HYPERBOLA. 187 
 
 [In the figure Ba on BC = BA, and cd on a parallel to BG 
 = cD^; then ad meets AB in X, the required centre of the in- 
 volution, and XF=XF^ = B. mean proportional between XD^ and 
 XB. 
 
 If FE meets DA in G, the points of contact of the given tan- 
 gents may be determined by drawing GG, DB intersecting in //, 
 when FII will meet GD in P, the point of contact of GB ; 
 GG, DF intersecting in K^ when BK will meet GD in Q^ its point 
 of contact; GF and GB intersecting in X, when DL will meet 
 AB in R\ its point of contact; the determination of the point of 
 contact of GB is left as an exercise for the student.] 
 
 Problem 117. To describe an hyperbola^ four points A, B, C, D 
 of the curve and a tangent ad heiiig given (Fig. 111). 
 
 Let AB meet the given tangent in a, and BG, GD, DA meet 
 it in h, c, and d respectively. Find X the centre and F, P^ the 
 
 foci of the involution a, c and b, d; P or P^ will be the point of 
 contact of the given tangent, so that five points being known the 
 problem reduces to Prob. 115. 
 
188 
 
 CENTRE OF CURVATURE. 
 
 [In the figure Scj on hB = he, and adi on a parallel to bB = ad; 
 then c^d^ meets ad in X, the centre of the required involution, 
 and XP is a mean proportional between Xd and Xb; I\, the other 
 focus, of course lies outside the limits of the figure. 
 
 HAP and DC meet in U, and AD, CB in F, EF will meet 
 PB in ^, a point on the tangent at />.] 
 
 Problem 118. To find the centre and radius of curvature at 
 any 'point P of a given hyperbola (Fig. 112). 
 
 The construction is identical with that for the ellipse (Prob. 
 
 88). 
 
 Fig.II2 
 
 Draw PG the normal meeting the axis in G, GH perpendicular 
 to PG meeting the focal chord PF in H, and HO perpendicular to 
 PF meeting the normal in 0, the required centre of curvature. 
 
CHAPTER VL 
 
 THE RECTANGULAR HYPERBOLA. 
 
 If the axes of an hyperbola be equal, the angle between the 
 asymptotes is a right angle, and the curve is called equilateral or 
 rectangular. 
 
 If G is the centre, A a vertex, and F the corresponding focus, 
 it follows that CF' = 2AC\ for it has been shewn (p. 154) that 
 CF' = CA' + CB\ and in the rectangular hyperbola CB = CA. 
 
 Similarly FA'=2AX''j where X is the foot of the directrix, 
 i.e. the eccentricity is always ^2 : 1. 
 
 Conjugate diameters are equal to one another and are equally 
 inclined to either asymptote, for in any hyperbola 
 
 CP'-CD'=CA'-GB' and .-. GF^GB. 
 Also GPLD (fig. 113) is a rhombus and therefore CL bisects the 
 angle FGD. 
 
 Diameters at right angles to one another are equal, for if CF 
 be perpendicular to GP the angle BGF = the angle FGA = the 
 angle BGDy and therefore by symmetry GF = GD. 
 
 Corollary. The rectangles contained by the segments of chords 
 which intersect at right angles are equal since they are in the 
 ratio of the squares of the parallel diameters (p. 169). 
 
 Given three points on an equilateral hyperbola, a fourth is 
 also given, for if the curve pass through the three points AjB,G it 
 will also pass through the orthocentre of the triangle ABGy i.e. 
 
190 LOCUS OF CENTRE, THREE POINTS BEING GIVEN. 
 
 through the intersection of the perpendiculars from A, B, C on 
 the opposite sides. 
 
 This follows at once from tlie above corollary, for if ABC be 
 a triangle, and the orthocentre, and if CO meets ABC in 1), 
 the triangles DOA and DBC are similar, and 
 
 DO : DA y. DB . DC ', 
 .: DO.DC^DA.DB, 
 so that must be a point on the curve. 
 
 If P, Q, R are three points on the curve, the centre must lie 
 on the circle passing through the middle points of the sides of 
 the triangle PQR. For (fig. 113) let an asymptote meet the sides 
 
 
 
 ^\e, 
 
 .113. 
 
 
 / 
 
 y. 
 
 
 
 
 y^^^'^/ 
 
 / / 
 
 V N 
 
 Jy^ 
 
 1 
 
 /T^V 
 
 
 jh 
 
 J 
 
 <\ 
 
 —~iZ. 
 
 _l 
 
 f\ 
 
 
 ^i ^. 
 
 J 
 
 o^ 
 
 C 2£ 
 
 J 
 
 \a\ !P 
 
 
 
 / 
 
 \J^ 
 
 -v; 
 
 
 /l\^ 
 
 
 csT^ 
 
 \Ab 
 
 
 
 ^ 
 
 o 
 
 
 \ 
 
 \\ 
 
 >\ 
 
 PQ, PR in I and l^j and let dy e, f he the middle points of QR, 
 RP and PQ respectively. Let C be the centre of the hyperbola. 
 Then (7/* is conjugate to PQ and Ce to PR, 
 . •. the angle fCe =fCl + 1 fie 
 
 ^elfi^Glf 
 
 = PIJ. + Pll^ 
 
 ^fPe 
 
 = fde, 
 
THE KECTANGULAR HYPERBOLA. 191 
 
 since fdeF is a parallelogram, i. e. the circle through fde passes 
 also through C (Euclid iii. 21). 
 
 Four points are therefore in general sufficient to determine a 
 rectangular hyperbola, for the orthocentre of the triangle formed 
 by any three is necessarily a fifth point on the curve, which can 
 then be completed by the general method of Prob. 115, p. 185; 
 01* the centre can at once be determined as one of the points 
 of intersection of the two circles which can be described through 
 the centre points of the sides of the triangles formed by taking 
 any three of the four given points in succession. 
 
 Similarly a rectangular hyperbola can generally be determined 
 from four conditions, and the curve cannot in general be described 
 to satisfy a greater number. 
 
 If ^ r be an ordinate of a diameter PCP^ , QV^^FV .VP^. 
 For in any hyperbola (Prop. 5, p. 165) 
 
 (?r : FV . VF^ '.'.CD' : CF-, 
 but in the rectangular hyperbola CD — CF^ 
 .-. QV' = FV . VF^, 
 
 Hints for the solution of particular cases are given in the 
 following examples, but as they are usually simple it has not been 
 considered necessary to illustrate them by figures. 
 
 Given the following data, construct rectangular hyperbolas ful- 
 filling them. 
 
 a. An asymptote and focus. 
 
 [A line through the focus making 45" with the asymptote 
 meets it in the centre,] 
 
 h. An asymptote LC, a tangent FL and its point of contact F. 
 
 [Let the given tangent FL meet the asymptote in Z; on it 
 lake FL^ — FL, and draw Lfi perpendicular to the asymptote 
 lectin g it in C, the centre of the curve.] 
 
 c. The centre C, a tangent FT and its point of contact F. 
 [From C draw GY perpendicular to FT meeting it in Y. 
 
192 VARIOUS DATA. 
 
 The transverse axis bisects the angle PC F, and its semi-length CA 
 is a mean proportional between CP and CY.] 
 
 d. The centre C and two points P, Q on the curve. 
 [Produce PC to p and make Cp = CP, so that Pp is a diameter. 
 
 Describe a circle through the three points P, Q^ p. The tangent 
 to this circle at Q is parallel to the tangent to the curve at P, 
 which is therefore known.] 
 
 e. The centre C, a tangent PT and a point ^ of the curve. 
 [From Q drop a perpendicular ^iV on the given tangent, 
 
 meeting it in N : bisect QN in n, and draw nt parallel to PT. 
 A circle passing through C and N and touching nt will meet 
 P^ again in its point of contact.] 
 
 / The centre C and two tangents PP, QT. 
 
 [Produce TC to P, and make CT^^CT; through P^ draw 
 T^t parallel to QT and meeting PT in ^. C^' and Ct will be the 
 directions of a pair of conjugate diameters, which determine the 
 asymptotes.] 
 
 g. A focus F and two points P, Q. 
 
 [With centre P describe a circle, the radius of which 
 
 :PP :: 1 : J% 
 and with centre Q describe a circle, the radius of which 
 
 '.FQ ::\ : J2. 
 The directrix will be a common tangent to these two circles.] 
 h. A focus Fj a tangent PT and its point of contact P. 
 [With centre P describe a circle, the radius of which 
 : FP :: 1 : J~2', 
 draw FT perpendicular to PT meeting it in T. A tangent from 
 T to the circle will be the directrix.] 
 
 i. A focus F and two tangents PT, QT. 
 [From F draw GY perpendicular to PT meeting it in Y; pro- 
 duce FY to /, and make Y/= YF. Draw the circle which is the 
 
THE RECTANGULAR HYPERBOLA. 193 
 
 locus of the vertex of the triangle on base Ff^ and with the 
 sides terminating in F and f respectively in the ratio of ^2 : 1, 
 (Prob. 17, p. 21). This circle is a locus of the second focus. 
 Similarly the tangent QT will furnish a second locus, so that the 
 second focus must be at one of the points of intersection of the 
 two circular loci.] 
 
 h. A focus F^ a tangent QT^ and a point F. 
 
 [From F draw FY perpendicular to QT^ produce it to J\ 
 and make Yf= YF\ a circular locus of the second focus can be 
 determined from Ff as in the last example. With F and / as foci 
 and with distance between the vertices = i^P, describe an hyper- 
 bola, which will be a second locus of the second focus, which is 
 therefore at one of the intersections of the hyperbola and circle.] 
 
 I. Two tangents FT, QT and their points of contact P and Q. 
 
 [Bisect FQ in Y\ then YT is a locus of the centre. On FQ 
 describe a segment of a circle containing an angle equal to the 
 supplement of the angle FTQ \ the segment must be on the same 
 side of FQ as T, and is then a second locus of the centre, which 
 is therefore known.] 
 
 m. Given three points and a tangent. 
 
 [From the three points a fourth can be determined (p. 189), 
 and the curve can be drawn by the general method. Prob. 117, 
 p. 187.] 
 
 n. Given four tangents, AB^ BC, CD, DA. 
 
 [Draw the circle to which the triangle formed by some three 
 of the four tangents is self conjugate (Ex. 18, p. 55); it is a 
 locus of the centre. A second locus is the straight line joining 
 the points of bisection of the diagonals of the quadrilateral formed 
 by the tangents. The centre is therefore known.] 
 
 0. Given two points A, B and two tangents FT, QT. (Fig. 
 114.) 
 
 [Let AB meet FT in a and QT in h. Find X the centre, 
 and 0, Oj the foci of the involution A, B and a, h. The chord of 
 E. 13 
 
194 
 
 RECTANGULAR HYPERBOLA. 
 
 contact of PT and QT will pass through or Op and if it passes 
 through 0, TO^ will l^e the polar of and conversely. Through 
 
 ..y-Q 
 
 draw Opq perpendicular to AB. On AB make OA^-OA^ and 
 find OaS' an harmonic mean between OA^ and OB^ let Og' and ^'6>, 
 meet in R and join aS'^. A circle described with its centre on 
 OE perpendicular to SU and to pass through A^ and B will cut 
 Opq in points ^, q of the required curve.] 
 
 p. Given three tangents ^^, ^C, C-i and a point P. (Fig. 115.) 
 
 [The circle (centre /S', radius 80) to which the given triangle 
 ABC is self-conjugate (Ex. 18, p. 55) is one locus of the centre. 
 Bisect AB^ BG, CA in c, a, h respectively, and draw PA cutting 
 he in Z, PB cutting ca in J/, and PC cutting ah in N. The conic 
 described to touch the sides of the triangle ahc in the points 
 Xi/iV' (Probs. 81 or 111) is a second locus of the centre. In the 
 
EXAMPLES. 195 
 
 figure, the required conic is an ellipse and the circular locus cuts 
 
 it in the points and 0^ , either of which may be taken as the 
 centre. The points of contact of the given tangents are p, q, r.] 
 
 Examples on Chapters V. and VI. 
 
 1. Draw an hyperbola, the centre (C), one asymptote {CL) 
 and a directrix LL^ being given. 
 
 [Draw the axis GF perpendicular to LL^^ meeting it in X. 
 The vertex ^ is at a distance CL from 6'.] 
 
 2. Draw an hyperbola, the asymptotes (7Z, CL^ and the 
 distance CF of a focus F being given. 
 
 3. Given the base ^^ of a triangle and point of contact, F, 
 with base of the inscribed circle ; shew that the locus of vertex 
 of triangle is an hyperbola with foci A and B and vertex F. 
 
 4. Shew that the tangent at any point P of an hyperbola 
 bisects any straight line perpendicular to the axis AA^ and 
 terminated by liP^ A^P. 
 
 13—2 
 
196 EXAMPLES. 
 
 5. Draw the locus of the foci of parabolas passing through 
 two fixed points P and P^ and having their axes parallel to a 
 fixed line AB. 
 
 [The hyperbola described with P and P^ as foci and with 
 length of transverse axis = PJ/j the side parallel to AB of a 
 right-angled triangle on PP^ as hypotenuse.] 
 
 6. Given the centre C, vertex A, and a tangent PT meeting 
 CA in T, describe the hyperbola. 
 
 [The foot iV^ of the ordinate of the point of contact (P) may 
 be determined from GT : CA :: CA : CJV. P is then known. 
 The asymptotes cut ofi" equal distances on PT on each side of P 
 and make equal angles with CA (Prob. 19).] 
 
 7. Given the centre C, the axis C2\ a tangent PT and its 
 point of contact P, draw the hyperbola. (See last example.) 
 
 8. Determine the locus of the intersection of the bisectors of 
 the sides of the triangle formed by the asymptotes and any 
 tangent to a hyperbola. 
 
 [A similar and similarly placed hyperbola with axes reduced 
 in ratio 2:3.] 
 
 9. Given a focus F, tangent PT and point Q on an hyperbola, 
 draw the locus of the second focus. 
 
 [From F draw FY perpendicular to PT meeting it in Y: 
 produce FY to / and make Y/= FY. The required locus is the 
 hyperbola with foci /* and Q, and transverse axis = FQ.] 
 
 10. Given a line QT and two points P and F. From F draw 
 a perpendicular i^Zto ^7^ meeting it in Y, and produce FY to f^ 
 making Yf= YF, With P and / as foci and with PF as the 
 distance between the vertices describe an hyperbola. With F 
 and any point on this hyperbola as foci describe an ellipse to pass 
 through P, and shew that it will touch QT : 
 
 i.e. Given a focus, tangent and point of a conic, the locus of 
 the second focus is an hyperbola. 
 
i 
 
 EXAMPLES. 197 
 
 11. Given two tangents PT, QT to a rectangular hyperbola 
 and tlieir points of contact P, Q. Shew that if QR be drawn 
 perpendicular to PT, and PR to QP, R will be a point on the 
 curve. 
 
 12. In a given ellipse determine the pair of equal conjugate 
 diameters. 
 
 [They coincide with asymptotes of hyperbola having the same 
 
 13. Draw the loci of the points of trisection of a series of 
 circular arcs described on the straight line AB. 
 
 [Branches of two hyperbolas having their centres at the in- 
 ternal points of trisection of ^^ and asymptotes inclined 60" to 
 axis.] 
 
 14. Given the asymptotes and a point on a directrix, draw 
 the hyperbola. 
 
 15. From a given point P in an hyperbola draw a straight 
 line, such that the segment intercepted between the other inter- 
 section with the hyperbola and a given asymptote shall be equal 
 to a given line. 
 
 [With P as centre and the length of the given line as radius 
 describe a circle cutting the other asymptote. Either point of 
 intersection joined to P gives the line required.] 
 
 16. Given a focus F, and tangent PY to an hyperbola and 
 the length 2a of the transverse axis, shew that the locus of the 
 second focus is a circle. 
 
 [From F draw FY perpendicular to PY meeting it in Y ; 
 produce FY to / and make Yf= FY. f is the centre and 2a 
 the radius of the required circle.] 
 
 17. Shew that any point on the circle through the middle 
 points of the sides of a triangle ABG may be taken as the centre 
 of an equilateral hyperbola passing through A, B and C. 
 
198 EXAMPLES. 
 
 18. If four tangents to an equilateral hyperbola be given, 
 shew that either of the limiting points (p. 46) of the system of 
 circles described on the diagonals of the quadrilateral as diameters 
 may be taken as the centre of the hyperbola. 
 
 19. Given a focus F, a tangent FT, its point of contact P, 
 and the eccentricity, draw the conic. 
 
 [From F draw FT perpendicular to FP and meeting FT in T 
 
 which will be a point on the directrix. With P as centre and 
 
 FP 
 with radius r such that = the given eccentricity, describe a 
 
 circle. Tangents from T to this circle will be positions of the 
 directrix. Two solutions are generally possible.] 
 
 20. Draw normals to an ellipse, from a given point P. 
 
 [The normals pass through the intersections of the ellipse with 
 the rectangular hyperbola passing through F and the centre of 
 the ellipse, and having its asymptotes parallel to the major and 
 minor axes at distances respectively 
 
 ' and 
 
 where a and h are the semi-axes and a, y8 the co-ordinates of P.] 
 
 21. Draw normals to an ellipse from a point on the minor 
 
 axis. 
 
 [They will pass through the intersections of the ellipse with 
 the circle described through the foci and point.] 
 
i 
 
 CHAPTER VII. 
 
 EECIPROCAL POLARS AND THE PRINCIPLE OF DUALITY. 
 
 In page 31, it has been sliewn how to find the pole of a 
 given line and the polar of a given point with regard to a given 
 circle, and the principal properties of poles and polars have been 
 explained. 
 
 In pages 140 et seq. an extension has been made to the case of 
 an ellipse, and the properties there noticed are applicable to all 
 conic sections. 
 
 The pole of a line with regard to any conic being a point and 
 the polar of a point a line, it follows that any system of points 
 and lines can be transformed into a system of lines and points. 
 
 This process is called reciprocation, and it is clear that any 
 theorem relating to the original system will have its analogue in 
 the system formed by reciprocation. 
 
 Def. Being given a fixed conic section (2) and any curve (S), 
 we can generate another curve (s) as follows; draw any tangent 
 to *S^, and take its pole with regard to S;, the locus of this pole 
 will be a curve s, which is called the reciprocal polar of S with 
 regard to 2. The conic 5 with regard to which the pole is taken 
 is called the auxiliary conic. 
 
 A point (of the reciprocal polar curve s) is said to correspond 
 to a line (of the reciprocated figure S) when we mean that the 
 point is the pole of the line with regard to the auxiliary conic % ; 
 and since it appears from the definition that every point of s is 
 
200 EXPLANATION OF METHOD. 
 
 the pole with regard to 2 of some tangent to jS, this is briefly 
 expressed by saying that every point of s corresponds to some 
 tangent of S. 
 
 Theorem. The point of intersection of two tangents to S will 
 correspond to the line joining the corresponding points of s. 
 
 This follows from the property of the conic 2, that the point 
 of intersection of any two lines is the pole of the Hue joining the 
 poles of these two lines, (p. 141.) 
 
 Now if the two tangents to S be indefinitely near, then the 
 two corresponding points of s will also be indefinitely near, and 
 the line joining them will therefore be a tangent to s ; and since 
 any tangent to S intersects the consecutive tangent at its point 
 of contact, the above theorem becomes : If any tangent to S cor- 
 respond to a p>oint on s, the point of contact of that tangent to S 
 will correspond to the tangent through the point on s. 
 
 Hence we see that the relation between the curves is reci- 
 procal, that is to say, that the curve S might be generated from s 
 (through the auxiliary conic) in precisely the same manner that 
 s was generated from S. Hence the name "reciprocal polars*. " 
 
 Being given then any theorem of j)osition concerning any 
 curve S (i.e. one not involving the magnitudes of lines or angles), 
 another can be deduced concerning the curve s. For example, if 
 we know that a number of points connected with the figure S 
 lie on a right line, we know also that the corresponding lines 
 connected with the figure s meet in a point (the pole of the line 
 with regard to ^), and vice versd. 
 
 From any one such theorem another can be derived by suitably 
 interchanging the words "point" and "line," "inscribed" and 
 "circumscribed," "locus" and "envelope," &c., understanding by 
 the term envelope " the curve to which a series of lines drawn 
 according to any given rule are tangents." 
 
 The evolute of a curve, e.g. is the envelope of normals to the 
 curve. 
 
 * Salmon's Conic Sections, chap. xv. 
 
KECIPROCAL POLARS AND THE PRINCIPLE OF DUALITY. 201 
 
 Although the auxiliary conic 2 has hitherto been spoken of 
 as any conic whatever, it is most common to make this conic a 
 circle, considerable simplification being thereby introduced, and 
 generally unless the contrary is specially mentioned, reciprocal 
 polars may be understood to be polars with regard to a circle. 
 It has been shewn, p. 31, that the polar of any point with regard 
 to a circle is a line perpendicular to the line joining the point 
 to the centre, and conversely that the pole of any given line with 
 regard to a circle is on the line through the centre perpendicular 
 to the given line; in either case the product of the distances of 
 the pole and polar from the centre being equal to the square of 
 the radius, so that the polar of a given point or the pole of a 
 given line with regard to a given circle may always be found 
 by merely drawing tangents to that circle. The centre of the 
 auxiliary circle is frequently called the origin. 
 
 The advantage of using a circle for the auxiliary conic chiefly 
 arises from the two following theorems, which enable us to trans- 
 form by this method, not only theorems of position, but also 
 theorems involving the magnitude of lines and angles. 
 
 Theoeem. The distance of any point P from the origin 
 is the reci^jrocal of the distance Ot from the origin of the cor- 
 responding line pt ; 
 
 i.e. OP.Ot = r\ 
 where Ot is perpendicular to j)t and r is the radius of the auxiliary 
 circle. 
 
 Theorem. The angle TQT^ between any two lines TQ^ T^Q 
 is eqttal to the angle subtended at the origin by the corresponding 
 points 2^, P^ ; for Op is perpendicular to TQ and Op^ to T^Q. 
 
 Problem 119. To find the polar reciprocal of one circle, 
 centre C, radius GA^ with regard to another, centre 0, radius OJ/, 
 i.e. to find the locus of the pole p with regard to the circle (0) 
 of any tangent PT to the circle C. (Fig. 116.) 
 
 Find iT/ifj the polar with respect to the auxiliary circle 
 (centre 0) of C, the centre of the circle to be reciprocated ; i. e. if 
 
202 
 
 RECIPROCAL POLAR OF CIRCLE. 
 
 G is, as in the figure, outside the auxiliary circle, draw CM a 
 tangent to that circle and draw MAI^ perpendicular to OC, meet- 
 
 Fig.116. 
 
 
 
 A \ C,^'''^\_/''^^ X 
 
 A, 
 
 O n 
 
 \/x ^'"' /^ 
 
 la" 
 
 \ I 1 
 
 ^^ — >^ 
 
 N 
 
 ^ 
 
 ing it in X. Draw any tangent FT to (C); draw OT perpen- 
 dicular to FT, and find the pole p with respect to the auxiliary 
 circle of FT. Then by definition, 00 . 0X= r^^Op. OT, where r 
 is the radius of the auxiliary circle, 
 
 i.e. Op : 00 :: OX : OT 
 From p draw pN perpendicular to MM^ meeting it in N, and pn 
 perpendicular to 00 meeting it in n. Also draw OY perpen- 
 dicular to OF meeting it in Y, so that OT=FY. Then by 
 similar triangles Opn, 00 Y, 
 
 Op : 00 :: On : C7, 
 .-. Op :0C :: OX+On : FY+CY 
 :: nX : OF. 
 
 But nX=pXf 
 
 .-. Op : pN :: 00 : OF; 
 
 00 
 but the ratio ^ „ is constant, since both 00 and OF are fixed 
 
 distances. 
 
RECIPROCAL POLARS AND THE PRINCIPLE OF DUALITY. 203 
 
 Therefore the point p moves so that its distance (Op) from a 
 fixed point is in a constant ratio to its distance {pN) from 
 a fixed right line MM^; i.e. the locus of ^ is a conic section 
 of which is a focus, MM^ the corresponding directrix, and 
 
 OG 
 
 the eccentricity of which is ^^^ . The eccentricity is evidently 
 
 greater, less than, or equal to unity according as is outside, 
 inside, or on the circumference of the reciprocated circle. 
 
 Hence, the polar reciprocal of a circle is a conic section, of 
 which the origin is the focus, the line corresponding to the centre 
 is the directrix, and which is an hyperbola, an ellipse, or a para- 
 bola, according as the origin is outside, inside, or on the circle. 
 
 The tangents at A and A^, the extremities of the diameter 
 through 0, correspond to the vertices at a and a^ of the reciprocal 
 polar. [In the figure At touches the auxiliary circle and at is 
 perpendicular to 0C\ 
 
 The extremities of the latus rectum LL^ correspond to the 
 tangents parallel to OC. Therefore OL . CP — r^, where r is the 
 radius of the auxiliary circle. 
 
 The centre of the reciprocal conic is the pole with respect to 
 (0) of the polar of with respect to (C), i.e. if is outside ((7) 
 it is the pole of the chord of contact of tangents from to the 
 circle (C), and in that case the asymptotes are perpendicular to 
 these tangents. Of course if is inside (C) real tangents from 
 it to (C) cannot be drawn, and consequently the ellipse has no 
 real asymptotes. 
 
 Conversely of course the reciprocal of a conic section with 
 regard to a circle which has one of the foci for its centre is a 
 circle, with its centre at the pole of the corresponding directrix 
 and of radius {K) such that the ratio, R : distance between its 
 centre and the focus, is the eccentricity of the conic. 
 
 The above important property enables us to deduce from any 
 property of a circle, a corresponding property of a conic; and 
 since the proof of the existence of the relation in the circle will 
 usually be much simpler than a direct proof of the corresponding 
 
204 PROPERTIES OF CONICS FROM CIRCLE. 
 
 relation in the conic, the method is frequently of great value. It 
 will soon be found that the operation of forming the reciprocal 
 theorem will reduce itself to a mere mechanical process of inter- 
 changing the words "point" and "line," "inscribed" and "cir- 
 cumscribed," "locus" and "envelope," &c., as has been already 
 noticed ; but the method also furnishes admirable examples and 
 tests of draughtsmanship, and the actual construction of reciprocal 
 figures should I think be much more largely practised than it is. 
 
 Of course a little care is required in taking the original circles 
 so that the resulting conic may be of convenient proportions, but 
 a very little practice will enable this to be done and there is no 
 real difSculty in the construction itself. 
 
 A convenient ratio for the eccentricity of an ellipse is one not 
 very different from 3 : 4, which may therefore be taken as a guide 
 for the ratio of the radius of the circle to be reciprocated to the 
 distance of the origin from its centre ; and the auxiliary circle 
 should then be taken of such radius as to bring the length 
 between the poles of the tangents at the extremities of the dia- 
 meter through the origin, i.e. the length of the major axis, a 
 convenient one. The approximate position of these poles relatively 
 to any assumed radius is easily seen. The size of the reciprocal 
 conic depends entirely on the radius of the auxiliary circle. 
 
 As an example, fig. 117 gives the figure illustrating the fol- 
 lowing reciprocal theorems : 
 
 Theorem. Eeciprocal. 
 
 If a chord of a circle subtend a If two tangents to a conic move 
 
 constant angle at a fixed point on so that the intercepted portion of a 
 
 the curve, the chord always touches fixed tangent subtends a constant 
 
 a circle. angle at the focus, the locus of the 
 
 intersection of the moving tangents 
 is a conic having the same focus and 
 directrix. 
 
 C is the centre of the circle, M the fixed point on it, and PP, 
 the chord which moves so that the angle PMP^ is constant, and 
 which therefore always touches a circle described with centre C. 
 
RECIPKOCAL POLARS AND THE PRINCIPLE OF DUALITY. 205 
 
 F is the centre of the auxiliary circle, and since it is taken inside 
 the circle (C) the reciprocal polar of this circle will be an ellipse. 
 
 Find K the point corresponding to the line MP^ i. e. the pole 
 
 Fig.117. 
 
 of MP with respect to the auxiliary circle, (in other words draw 
 Fm perpendicular to MP meeting it in m, drav/ 'nxt a tangent to 
 the auxiliary circle touching it in t and draw tK perpendicular to 
 Fm, meeting it in K)\ find L the point corresponding to JfPj, 
 i. e. the pole of MP^ with respect to the auxiliary circle, (in other 
 words, since MP^ cuts the auxiliary circle, draw a tangent at one 
 of the points of intersection meeting FL drawn perpendicular to 
 J/Pj in L) ; then the line KL corresponds to the point M, and will 
 therefore be a fixed tangent to the reciprocal conic. 
 
 Find Q the point corresponding to the line PP^ , then the line 
 KQ corresponds to the moving point P, and the line LQ to the 
 moving point P^, and these lines are therefore moving tangents 
 to the reciprocal conic, intercepting on the fixed tangent a length 
 KL which subtends a constant angle at F, for since FK is per- 
 pendicular to MP and FL to MP^ , the angle KFL is equal to 
 the angle PMP^ which by supposition is constant. 
 
206 
 
 EXAMPLES, 
 
 Lastly, since Q corresponds to PP^ which is a tangent to a 
 circle centre G, the locus of Q must be the polar reciprocal of this 
 circle, and is therefore a conic with focus F and the polar of C 
 for directrix, i. e. a conic with the same focus and directrix as the 
 polar reciprocal of the circle MPP^ . 
 
 As in the figure F lies outside the circle to which PP^ is a 
 tangent, the locus of ^ is a hyperbola, the vertices of which are 
 the poles of the tangents at A and A^, the ends of the diameter 
 through the origin F, 
 
 Examples on Chapter YIT. 
 
 Below are given in parallel columns some examples of reciprocal 
 theorems : 
 
 1. The angles in the same seg- 
 ment of a circle are equal. 
 
 2. Two of the common tangents 
 of two equal circles are parallel. 
 
 3. If a circle be inscribed in a 
 triangle, the lines joining the vertices 
 with the points of contact meet in a 
 point. 
 
 4. If two chords be drawn from a 
 fixed point on a circle at right angles 
 to each other, the line joining their 
 ends passes through the centre. 
 
 5. Any two tangents to a circle 
 make equal angles with their chord 
 of contact. 
 
 6. If two chords at right angles 
 to each other be drawn through a 
 fixed point on a circle, the line join- 
 ing their extremities passes through 
 the centre. 
 
 If a moveable tangent of a conic 
 / meet two fixed tangents, the inter- 
 I cepted portion subtends a constant 
 angle at the focus. 
 
 If two conies have the same 
 
 focus, and equal latera recta, the 
 
 straight Une joining two of their 
 
 common points passes through the 
 
 \ focus. 
 
 ^ If a triangle be inscribed in a 
 \ conic, the tangents at the vertices 
 ineet the opposite sides in three 
 points lying in a straight line. 
 
 \ If two tangents of a conic move 
 so that the intercepted portion of a 
 fixdd tangent subtends a right angle 
 at the focus, the two moveable tan- 
 gents meet in the directrix. 
 
 The line drawn from the focus to 
 the intersection of two tangents bi- 
 sects the angle subtended at the focus 
 by their chord of contact. 
 
 The locus of the intersection of 
 tangents to a parabola which cut at 
 right angles is the directrix. 
 
RECIPKOCAL POLARS AND THE PRINCIPLE OF DUALITY. 207 
 
 [Take the fixed point on the circle as the centre of the auxiliary 
 circle, and the circle reciprocates into a parabola.] 
 
 7. The envelope of a chord of a The locus of the intersection of 
 circle which subtends a given angle tangents to a parabola, which cut 
 at a given point on the circle is a at a given angle, is a conic having 
 concentric circle. the same focus and the same directrix. 
 
 8. The rectangle under the seg- The rectangle under the perpen- 
 ments of any chord of a circle diculars let fall from the focus on 
 through a fixed point is constant. two parallel tangents is constant. 
 
 [Take the fixed point as centre of auxiliary circle.] 
 
 9. If lines be drawn from the The points of intersection of tan- 
 end of a diameter of a circle making gents to a parabola, which are equally 
 equal angles with a fixed straight inclined to a given straight line, lie 
 line in the plane of the circle, the on a fixed straight line passing 
 chords subtended by these lines are through the focus. 
 
 parallel. 
 
 10. The portion of any tangent The portion of the directrix inter- 
 to a circle intercepted between two cepted between chords drawn from 
 parallel tangents subtends a right the ends of any focal chord of a 
 angle at the centre. conic to any point of the curve sub- 
 tends a right angle at the focus. 
 
 11. Shew that the polar reciprocal of a parabola with respect 
 to a circle having any point (*S') of the directrix as centre is an 
 equilateral hyperbola. 
 
 [Draw the tangents to the parabola from the point S, which 
 will be at right angles to each other since S is on the directrix. 
 The reciprocals of their points of contact will be asymptotes to 
 the reciprocal curve, because their points of contact (the poles 
 of the tangents) are at an infinite distance. The tangents at the 
 vertices can easily be drawn, since they are the polars of the 
 points in which a line through S parallel to the bisector of the 
 angle between the asymptotes meets the parabola.] 
 
 12. Given three points A, B and C, on a parabola, and a 
 point L on the directrix, draw the curve. 
 
 [If the three points are reciprocated with respect to a circle 
 described with centre L, and a rectangular hyperbola described 
 
208 EXAMPLES. 
 
 passing through L and having the polars oi A, B and C for 
 tangents (Ex. 'p, p. 1 94), any point on the hyperbola when reci- 
 procated with respect to the same circle becomes a tangent to 
 the parabola.] 
 
 13. If a conic be inscribed in a quadrilateral, shew that the 
 angles subtended at a focus by the pairs of opposite sides are 
 together equal to two right angles. 
 
 [Reciprocate the well-known theorem : The opposite angles of 
 any quadrilateral inscribed in a circle are equal to two right 
 angles.] 
 
 14. With the centre of perpendiculars of a triangle as focus 
 are described two conies, one touching the sides and the other 
 passing through the feet of the perpendiculars ; prove that these 
 conies will touch each other, and that their point of contact will 
 lie on the conic which touches the sides of the triangle at the 
 feet of the perpendiculars. 
 
 15. An hyperbola is its own reciprocal with respect to either 
 circle which touches both branches of the hyperbola and inter- 
 cepts on the transverse axis a length equal to the conjugate 
 axis. 
 
\ 
 
 CHAPTER VIII. 
 
 ANHARMONIC RATIO AND ANHARMONIC PROPERTIES 
 OF CONICS. 
 
 AppUcatioyi of the signs + and — to determine the direction of 
 segments of a right line. 
 
 li A, B are two points in a straight line and it is necessary to 
 discriminate whether the length AB is to be measured from A 
 to B or from B to -4, it may be done by calling the one direction 
 positive and the other negative, the starting point in each case 
 being called the origin. 
 
 Hegard being paid to this convention we may evidently say 
 
 AB = -BA or AB + BA=0, 
 
 and an obvious interpretation of this equation is that if we go 
 
 from A to B and back again from ^ to ^ we are finally at zero 
 
 distance from the starting point. 
 
 The same thing is evidently true of any number of segments ; 
 for if we take three points A, B,G in any order in a straight line 
 and travel from A to B, then from B to C, and finally from C to A^ 
 we arrive at the point we started from, and really perform the 
 operation expressed by the equation 
 
 AB + BC + GA = 0. 
 Since — CA =AG^ this may also be written 
 AB + BC=^AC. 
 
 When the position of a point A is determined by its distance 
 from an origin 0, if we wish to refer it to another origin 0^ 
 anywhere on the line through and A, we can always take 
 OA:r.O^A-Ofi, 
 E. 14 
 
210 ANHARMONIC RATIO AND 
 
 for this is identical with the equation 
 
 OA-O^A + Ofi = 0, 
 and, since - O^A = AO^j with 
 
 OA+A0^ + Ofi = 0. 
 The difference of two segmeiits OA, OB of a straight line with 
 a common origin is always equal to JBA, whatever may be the 
 magnitudes and directions of the segrnents. 
 
 For the equation OA- OB = BA 
 is identical with OA + AB + BO = 0. 
 
 If a is the middle point of a segment Aa and M any point on 
 the line through Aa^ 
 
 ^^^^MA+Ma, ^^^ 3fA.3fa = Ma\'-'^\ 
 
 for between the three points M, a, A the relation holds 
 Ma + aA +A3I=0^ 
 Ma + aA = MA. 
 Similarly Ma + aa = Ma ; 
 
 therefore, adding these equations and remembering that aA=- aa, 
 since a is midway between A and a, 
 
 ,^ MA + Ma 
 
 Ma= — 2 — ; 
 
 also, multiplying together the right and left-hand members, we get 
 MA . Ma = ifah + Ma {aA + aa) + aA,aa 
 
 since aA = — aa. 
 
 Let Aj B, Cj J) he four points in a straight line, then the ratio 
 
 of the distances of one point A from two others B and D, divided 
 
 by the ratio of the distances of the remaining point C from the 
 
 same two (B and D), is called the Anharmonic Ratio of the range 
 
 A, B, C, D; i.e. the anharmonic ratio of the range ABC J) is the 
 
 AB CB 
 numerical value of the expression --^ ^ TTh^ which may also be 
 
 AU \j D 
 
ANHARMONIC PROPERTIES OF CONIC S. 211 
 
 written 4^ • ^ o^ ^5 . CD : CB . AD. The sign of the ratio 
 
 will de2:)end on the signs of the segments of which it is composed, 
 those which are measured in one direction being considered posi- 
 tive and those measured in the opposite direction negative. 
 
 Thus, if the four points are in the order from left to right 
 A BCD J the three terms AB, CD and AD in the above ratio are 
 positive and the term CB is negative; and the ratio itself is 
 negative. 
 
 Since four points in a straight line taken in pairs give six 
 segments, there are really six anharmonic ratios corresponding to 
 any range, three of which however are merely the inverse values 
 of the remaining three. Thus instead of taking the ratio of the 
 distances of A from B and D and dividing by the ratio of the 
 distances of C from B and D, we might take the ratio of the dis- 
 tances of A from C and D and divide by the ratio of the distances 
 of B from C and D, giving the expression 
 
 AC^_^BG_ AC^ BD 
 AD BD' ^'' AD' BC 
 and in this case if the points are in the order ABCD all the seg- 
 ments are of the same sign and the ratio is positive. 
 
 Again, we might take the ratio of the distances of A from B 
 and G and divide by the ratio of the distances of D from B and C, 
 giving the expression 
 
 AB DB AB DC 
 
 _: rjT* 
 
 AC DC AC DB' 
 where two of the segments are of the same sign and two of oppo- 
 site sign, so that the ratio is again positive. 
 
 In the above ratios the same point A has been associated suc- 
 cessively with the three remaining C, B, D. In the first A and C 
 may be said to be conjugate points, in the second A and B, and 
 in the third A and D, 
 
 Of the three fundamental ratios formed as above, two are 
 always positive and one negative, whatever the order of the points. 
 
 14—2 
 
212 BATIO OF A PENCIL IS CONSTANT. 
 
 Besides these there are the three inverse ratios 
 
 ^__^ :^__-?^ 4^_^^ 
 AB ' CB' AC'mC' AB'^DB' 
 
 It is of course necessary to retain throughout any investigation 
 the particular order adopted at its commencement. 
 
 The anharmonic ratio of the range A, B, C^ D is denoted by 
 {ABCD]. 
 
 If the anharmonic ratio of a range =-1, the segments are 
 in harmonic progression; for if the points occur in the order 
 
 ACBD and -ryr -f- — -y, = - 1, we have 
 AD BD 
 
 AG _ BC _CB 
 
 AD~ BD~ BD' 
 since BC and CB are measured in opposite directions, and there- 
 fore of course the three segments AG^ AB, AD are such that 
 the first : the third : : difierence between first and second : differ- 
 ence between second and third. If one of the points (D suppose) 
 
 AG BC 
 is at an infinite distance, the anharmonic ratio -n^-^-rrr. reduces 
 ' AD BD 
 
 to the simple ratio -jry,, for it may be written -^-p:, -^ >, ,, , and AD 
 ^ BC ^ BG BD' 
 
 is ultimately equal to BD. 
 
 Prop. 1. If four fixed straight lines which meet in he cut 
 hy any transversal in the 2>oints A, B, G, D (fig. 118), then will 
 {ABCD} be constant. 
 
 Draw the straight line aBc parallel to 0Z>, and meeting OA, 
 OC in a and c. 
 
 Then AB : AD :: aB : Oi) by similar triangles. 
 Similarly, CD : CB :: CD : cB 
 therefore AB .CD \ AD .CB y. aB : cB, 
 
 AB CB _ aB 
 ^^ AD ■ CD~ cB' 
 
 but — is a constant ratio for all positions of ac parallel to OD^ 
 
ANHARMONIC EATIO. 213 
 
 therefore — r- -f- t^-^ , wliich is the anharmonic ratio of the four 
 points A, B,C, I), is also constant. 
 
 Def. a bundle of lines drawn through one point is called 
 a pencil qfraySj or shortly a pencil. 
 
 The anharmonic ratio of a pencil of four rays is the anhar- 
 monic ratio of the range in which its rays are intersected by any 
 transversal. 
 
 Pencils and ranges are said to be equal when their anharmonic 
 ratios are equal, corresponding lines and points being taken for 
 the comparison. 
 
 Equiangular pencils are evidently equal. 
 
 The anharmonic ratio of a pencil is denoted by 0{AJBCI)]; 
 being the vertex and OA, OB, OC, OD the rays of the pencil. 
 
 Prop. 2. The transversal 7nay cut the rays of the pencil on 
 either side of the vertex. 
 
 For if, in the preceding article a transversal is drawn through 
 
 B, cutting OA in J^, 00 in 6',, and 01) in 2)^, where D^ lies in 
 
 DO produced, then, exactly as before, 
 
 A^B : A^D^ '.'. aB \ OD^, 
 
 and 6\Z>, '. C,B y. OD^ : cB; 
 
 , ^ A^B CB aB AB CB 
 
 therefore — . ^^^^ = ^ ^ Jj) ^^ CD ' 
 
214 PROPERTIES OF TRANSVERSALS. 
 
 If a transversal meet AO produced in a, BO produced in ^, 
 CO produced in y, and i>(9 in 8, 
 
 {ABCD] = {aPyh}. 
 
 If a transversal be drawn parallel to one of the rays of the 
 pencil (0(7) suppose, meeting the other rays in a, 6, d, we have as 
 before for the an harmonic ratio of the range ahcd, where c is 
 
 at an infinite distance, — ^ , which is therefore = -,->, H- ^c-^ . 
 ad' AD CD 
 
 If the pencil is harmonic — ^ = -1 ; therefore ah-- ad, or a is 
 the centre point of hd. 
 
 Problem 120. Given the anharmonic ratio \ of four points, 
 three of which are given in position, to determine the fourth 
 (Fig. 119). 
 
 Suppose that the three points A, (7, and D of the anharmonic 
 
 AC BC 
 ratio -T-^H- -j^^ = X are given and the point B required. Through 
 
 A draw any line and on it set off from A segments Aa, Aa whicli 
 
 FJg.ll9. 
 
 are to each other in the ratio \ : the segments must be taken on 
 
 the same side of -4 if A. is positive, and on opposite sides (as in 
 
 figure) if it is negative. Draw aC and a'D meeting in 6, and 
 
 draw hB parallel to Aa meeting the WxiqACD in B, which will be 
 
 the required fourth point of the range ; for by similar triangles we 
 
 have 
 
 AC Aa AD Aa' 
 
 W~Bb' ^'^^ BD ~Bb' 
 
 AC^^BG Aa ^ 
 
 •'• AD ' BD ~Aa'~ * 
 
AN HARMONIC RATIO. 215 
 
 the ratio being negative, since AC^ AB and BB are measured in 
 
 one direction and BC in the opposite. 
 
 The same construction determines the points C or Z) if the 
 
 AG BC 
 points A, B, B or A, B, C and the ratio y-r.-^ -jtj. = A. are given. 
 
 To determine C, e.g. draw Bb parallel to Aa meeting a' Z> in b, and 
 draw ab to meet AB in C. 
 
 If the points B, C, B are given, the given anharmonic ratio 
 
 may be written ^r-^ — — ^ = A, and the construction may be made 
 
 by substituting the point B for the point A ; by drawing, i.e. 
 through the point B any line and setting off on it from B seg- 
 ments Bb', Bb in the ratio \, and joining b' to B and b to C. 
 
 If X, instead of being a number positive or negative, is the 
 ratio of two lines of given length, these lengths may themselves 
 be set off from A to a and a\ on the same or on opposite sides of A 
 according as the ratio is positive or negative. 
 
 Prop. 3. If A,B,C,B are four points in a straight lins, 
 then AB . CB + AC . BB + AB . BC = 0, the general rule of signs 
 being observed (Fig. 120). 
 
 Divide by ^^ . CB, and the equation becomes 
 AC ^ BC ^_^CB 
 AB' BB'^ AB' CB~ 
 
 Draw tlie lines OA, OB, DC, OB, being any point, and 
 draw a transversal parallel to OB meeting OA, OB, OC in a,b,c ; 
 
21G 
 
 HOMOLOGOUS POINTS. 
 
 then 
 
 and 
 
 AB ' Dli~ ab' 
 
 ^'G ^I>G _hc _ AD^_^ CD 
 BA ' DA ~ ha ~ AJJ ' CB * 
 
 so that the above equation may be written 
 
 ac ch 
 
 ah ah ' 
 
 or ac + cb-\-ha = 0, 
 
 which is always true. 
 
 The above equation is formed by multiplying each term of the 
 identity BC + CD + DB == by the distance of A from the re- 
 maining point, the first term BG by AD, the second CD by AB, 
 and the third DB hy AC. 
 
 Prop. 4. If Hie anharmonic ratios of two systems of four 
 points A, B, C, D and a, h, c, d taken on two straight lines and 
 corresponding each to each are equal, and the lines are so placed 
 that two Jiomologous 2?oints A and a coincide, the three straight 
 lines joining the remaining pairs of hoinologous points will meet in 
 a point (Fig. 121). 
 
 For if not, let Bh and Gc meet in 0, and let OD meet the line 
 ac in d^ : the pencil 0, A BCD is met by two transversals AD 
 
 Fig.l2I. 
 
 and ad^, and therefore the anharmonic ratio 
 
 AB^DB _ah_^d^ 
 MJ ' DC ~ ac ' dTc ' 
 
ANHARMONIC RATIO. 217 
 
 but by hypothesis 
 
 AG'^DG~ac'^Jc' 
 
 therefore -7- = ^ , which is impossible unless d and d, coincide. 
 
 dc d^c ^ ^ 
 
 Two lines divided so that the anharmonic ratio of any four 
 points on the one is equal to the anharmonic ratio of the four 
 corresponding points on the other are said to be divided homo- 
 (jraphically. 
 
 Prop. 5. Jf the anharmonic ratios of two pencils of four rays, 
 corresponding each to each, are equal, and the pencils are so placed 
 that two corresponding lines coincide in direction, the three points 
 of intersection of the remaining homologous rays lie on a straight 
 line (Fig. 122). 
 
 Let and 0^ be the vertices of the pencils, the ray OA of the 
 one coinciding in direction with the ray Oa of the other, and let the 
 
 Fis.122. 
 
 homologous rays OB, Ofi meet in h, and OC, O^c in c ; the straight 
 line he will pass through the point d where OD intersects O^d, 
 for if not, let it meet OD in D and O^d in d^ ; then, since the 
 anharmonic ratios of the two pencils are equal, we have 
 
 ab ^ Db ab ^ dp 
 
 ac ' Do ac ' d^c ' 
 which is impossible unless d^ and D coincide. 
 
 Two pencils such that the anharmonic ratio of any four rays 
 of the one is equal to the anharmonic ratio of the corresponding 
 four rays of the other are said to be homographic. 
 
218 HOMOGRAPHIC DIVISION. 
 
 Pkoblem 121. Given any number of points A, B, C, D, E... 
 on a straight line, and any three corresponding points (as a, h, c) 
 on a second line, to complete the homographic division of the second 
 line (Fig. 121). 
 
 Place the lines at any angle with each other and with two 
 corresponding points (as A and a) coinciding. Let the lines 
 joining the remaining pairs of corresponding points {Bh and Gc) 
 meet in 0, then the lines OD, GE...&C. will meet the second 
 line in c?, e,, <fec., the required points of homographic division. 
 
 The construction is obvious from the known property of the 
 transversals of a pencil of rays. 
 
 It follows conversely that if a pencil of rays intersect any two 
 
 lines in points B,b; G,c the lines are divided homographically, 
 
 and that the point A in which the two lines intersect is its own 
 homologue in both divisions. 
 
 Problem 122. Given a pencil of rays . ABGDE... and 
 any three corresponding rays 0^ . abc of a second pencil, to com- 
 plete the second pencil so that the two shall be homographic 
 (Fig. 122). 
 
 Place the pencils so that two corresponding rays (OA, O^a sup- 
 pose) shall be coincident in direction, let the homologous rays in- 
 tersect in b and c ; the straight line be will intersect the remaining 
 rays of the given pencil in points on the required completing rays 
 of the second. The construction is obvious from the known pro- 
 perty of a transversal. 
 
 Conversely, when the corresponding rays of two pencils inter- 
 sect in points on a straight line, the pencils are homographic, and 
 the line 00, joining the centres is common to both pencils and is 
 coincident with its own homologue. 
 
 Prop. 6. If Ay B, G, D are four points on the circumference of 
 a circle, and from any point on the circumference the pencil 
 . ABGD is drawn, the anharmonic ratio of this pencil is constant. 
 
 For if Oi is any other point on the circumference the pencil 
 0, . ABGD is equiangular with the pencil . ABGD. 
 
ANHARMONIC PROPERTIES OF CONICS. 219 
 
 Prop. 7. If four fixed tangents to a circle are met hy any 
 variable tangent in A, B, C, D^ the anharmonic ratio of this range 
 is constaiit. 
 
 For the angles which the four points subtend at the centre are 
 constant, and therefore the ranges are transversals of equiangular 
 pencils. 
 
 If we reciprocate these tlieorems (Prob. 119), the four fixed 
 points in the first correspond to four fixed tangents to a conic, the 
 variable point corresponds to a variable tangent, the lines OA , 
 OB, &c. correspond to the points a, b, c, d in which the variable tan- 
 gent cuts the fixed tangents; and since points corresponding to 
 lines lie on lines through the centre of the auxiliary circle perpen- 
 dicular to the lines to which they correspond, the pencil formed by 
 joining a, &, c, d to the centre of the auxiliary circle is equiangular 
 with the pencil . A BCD, i.e. the anharmonic ratio of the range 
 abed is constant. 
 
 Prop. 8. The reciprocal theorem to Prop. 6 therefore is ^'•The 
 anharmonic ratio of the points in which four fixed taiigents to a 
 conic cut any variable tangent is constant." 
 
 Prop. 9. By exactly similar reasoning the reciprocal theorem, 
 to Prop. 7 is '''■The anharmonic ratio of the pencil formed by pining 
 any four fixed points on a conic to a variable fifth is constant." 
 
 If the recij^rocal figures be drawn, by observing the angles 
 which correspond to the constant angles in the circle, it will 
 be seen that tlie angles which the four points of the variable 
 tangent in the first theorem subtend at either focus are constant ; 
 and that the angles are constant which are subtended at the 
 focus by the four points in which the inscribed pencil meets the 
 directrix in the second theorem. 
 
 HOMOGRAPHIC RANGES IX THE SAME STRAIGHT LINE. DOUBLE 
 
 POINTS. 
 
 When two lines divided homographically are superposed, there 
 exist, in general, two points, each of which considered as be- 
 longing to the first division coincides with its homologue in the 
 
220 
 
 DOUBLE POINTS. 
 
 second division. They may be called double points, since each 
 represents two coincident homologous points. The double points 
 may become imaginary. 
 
 Let A, B, C, D... (fig. 123) be any points on a straight line, S 
 any point, and A J 3^ line through A making any angle with 
 AD^ and meeting *S'^, ISCy &c. in /?, y... respectively. 
 
 Fig.123. 
 
 The ranges ABCD... and A/3yS... are of course homographio, 
 and if the second range be rotated round A till it coincides in 
 direction witli AD, the point A, considered as belonging to the 
 first division, will evidently coincide with its own homologue in 
 the second division, as will also the point L determined by drawing 
 SL perpendicular to the line bisecting the angle DAJ. 
 
 Two homographic ranges in the same straight line formed as 
 above possess therefore two double points. 
 
 Instead however of the tv/o ranges being formed merely by 
 the rotation of the second about A, the second may in addition 
 be moved along AD into any position to the right or left, 
 bringing (as in the figure) the point corresponding to A to a, 
 /5 to 6, y to c... In this case also two double points in general 
 exist, which may be thus found : — 
 
 Problem 123. Given two homograpliic ramjes ABCD ..., abed. . . 
 
 in the same straight line, to determine the double points (Fig. 124). 
 
 Draw any circle whatever, and from any point 3£ on it draw 
 
ANHAEMONIC PROPERTIES OF CONICS. 
 
 221 
 
 MA, MB, MC cutting the circumference in A^, B^, C^, and draw 
 Ma, Mb, Mc cutting the circumference in a^, b^, c, . 
 
 -^--' 
 
 Draw Afi^, B^a^ intersecting in K, and A^c^ , C^a^ intersecting 
 in L, and draw IvL cutting the circumference in F and F^. The 
 lines IfF, MF^ will cut Aa in the required double points J 
 and J^ , 
 
 For {ABC J] 
 
 = M{A,Bfi,F] = a, [A.Bfi.P] (Prop. 6.) 
 
 '-= {NKLP}, 
 where N is the point of intersection of KL and A^a^ 
 and [abcJ] = M {a^b^c^F} = A^ [aJj^c^F] 
 
 ^{NKLF}, 
 i.e. the point J, considered as belonging to the first range, coin- 
 cides with its own homologue in the second and is therefore a 
 double point. Similarly for the point J^ . 
 
 It will be seen that the line FF^ is a Pascal line (Prob. 86) in 
 the circle, for CJ)^ and B^c^ intersect on it. 
 
 The double points may also be determined thus : — On Ab, 
 Ba as chords describe two circles passing also through any 
 arbitrary point G and intersecting again in G^, on Ac, Ca as 
 chords describe circles passing through G and intersecting again 
 
•)•>•), 
 
 INVOLUTION. 
 
 in G^. The circle through G, G^, G^ passes also through the required 
 double points of the ranges. 
 
 A third construction for the double points is shewn in fig. 123. 
 Through A, any point of the first range, draw A J, a line making 
 any angle with the given line. On it make Ap = ab, Ay = ac, 
 where a, b, c are the points of the second range corresponding to 
 the points A, £, C of. the first. Draw Bj3, Cy intersecting in S. 
 Through aS' draw SJ parallel to AG meeting A J in J, so that / is 
 the point on the range A^y ... corresponding to an infinitely dis- 
 tant point on the range ABG . . . ; and draw SI parallel to AJ, so that 
 / is the point on the range ABG... corresponding to an infinitely 
 distant point on the range A/3y.... Make aJ^ on the superposed 
 ranges = AJj so that the range A/Sy. .. Jis identical with the range 
 abc ... J^. Bisect IJ^ in 0, which will be equidistant from 
 the required double points. Find the point 0^ on the range abc... 
 corresponding to the point considered as belonging to the range 
 ABC... [i.e. join SO cutting AJ in o, and make aO^=Ao]. 
 Then the mean proportional between 00^ and OJ^ will be the 
 distance from of the required double points F and Q. 
 
 On page 17 a system of pairs of points on a straight line 
 such that XA.Xa^XB.Xb = XG.Xc=...=XP' = XQ' was 
 defined as a system in Involution, any two corresponding points 
 such as A^a being called conjugate points, the point X the centre, 
 and the points P and Q the foci of the involution. 
 
 Prop. 10. Wheii three pairs of conjugate points are in invo- 
 lution, the anharmonic ratio of any four of the points is equal to 
 the anharmonic ratio of their four conjugates, i. e. taking any four 
 A, B, C, a and their four conjugates a, b, c, A, 
 
 AB aB _ab Ab 
 
 AC aC ac Ac ' 
 
 for if is the centre of the involution 
 OA : OB :: Ob : Oa, 
 .-. OA-OB : OB :: Ob-Oa : Oa, 
 or AB :ab :: OB :0a, 
 
ANHARMONIC PROPERTIES OF CONICS. 223 
 
 and 0A+ Ob : Ob :: OB + Oa : Oa ; 
 
 .-. Ab :aB :: Ob : Oa, 
 .AB.Ab OB. Ob OA 
 
 Similarly of course 
 
 ab. 
 
 aB 
 
 ~ Oa' 
 
 
 Oa' 
 
 AC 
 
 .Ac 
 
 OA 
 
 AB. 
 
 Ab 
 
 ac. 
 
 aC 
 
 ~ Oa~ 
 
 ab. 
 
 aB' 
 
 AB. 
 
 Ab 
 
 ob.aB 
 
 
 
 AG. 
 
 Ac 
 
 ab . a(J 
 
 > 
 
 
 en 
 AB 
 
 f a 
 
 B ab 
 
 Ab 
 
 
 AC 
 
 ' ' aG~ac'' 
 
 'Ac 
 
 * 
 
 which may be written 
 
 which proves the proposition. 
 
 A series of points in involution consists of two homographic 
 ranges, the directions of which coincide, and in which to any 
 point whatever M of the line the same point m corresponds, 
 whether 31 be considered as belonging to the first or second system. 
 
 For consider M as belonging to the range ABC.y 
 and 7)1 „ „ abc.'f 
 
 then, since the ranges are homographic, 
 MA CA ma ca 
 MB'^TjB^mb'Vb' 
 
 If they are also in involution we must be able to interchange m 
 and 31, i.e. considering M as belonging to the range abc... and 
 771 as belonging to the range ABC.y we must have 
 
 Ma ca mA CA 
 Mb "c6 "^ ^CB' 
 Dividing each term of the first equation by the opposing term of 
 the second, 
 
 MA mA _ ma Ma 
 MB^^rnB^mb^Mb' 
 i.e. the anharmonic ratio of the four points M, A, B, m is equal 
 to the anharmonic ratio of their four conjugates 7;i, a, b, M, or 
 the points are in involution. 
 
224 INVOLUTION. 
 
 Prop. 11. It is always jyossihle to superimpose two liomographw 
 ranges, so that the two divisions shall be in involution. 
 
 For it has been shewn (p. 220) that two pairs of correspondiDg 
 points can be found equidistant from each other, by drawing viz. 
 through S (fig. 123) a line LI perpendicular to the line bisecting 
 the angle between the ranges when the second is placed at any 
 angle with the first and with two corresponding points A, a a.t the 
 intersection : the pairs of points A, L and ct, I are then equidistant. 
 If now the two ranges are superimposed with the point a coinciding 
 with the point Z, and the point I coinciding with the point A 
 (fig. 123a), the two ranges will be in involution. 
 
 The foci (p. 17) are points at which pairs of conjugate points 
 coincide, and their existence is only possible when the points of 
 any conjugate pair in the involution are both on the same side of 
 the centre. 
 
 Thus if the points are in the order ABah, the centre X 
 must fall between B and a in order that the products XA . Xa 
 and XB . Xh may have the same sign, and that sign will be 
 negative since the segments in each of the products are measured 
 in opposite directions ; but a square number is always positive, 
 and therefore no foci exist. 
 
 If three segments Aa, Bb, Cc are in involution and one ovei- 
 laps (as above) another, i.e. if the points are in the order ABab, 
 it will also overlap the third. This is evident if we consider that 
 if C lies to the left of X, and XC is greater than XA , c must be 
 on the opposite side of A", and Xc must be less than Xa and vice 
 versa, and similarly with regard to Bb. 
 
 Conversely, if the segment Aa does not overlap Bb, it cannot 
 overlap Cc, nor can Bb and Cc overlap. 
 
 The centre X forms with any two pairs of points A, a and B, h 
 
 an involution in which the conjugate to the centre is at an infinite 
 
 distance, for if x is conjugate to X, the anharmonic ratio of the 
 
 four points XABx^t\\Q anharmonic ratio of the four conjugate 
 
 points xdbXj 
 
 XA xA _xa Xa 
 
ANHAKMONIC RATIO AND PROPERTIES OF CONICS. 225 
 
 , ^ XA Xh . ^. . xB xa 
 
 but —r-T-, - -^F- ) ^"^^ thererore — -7 = -r , 
 
 XB Xa ' xA xb 
 
 or xA . xa = xB . xb (a). 
 
 Now X cannot lie between b and A because the segment Xx 
 must overlap both Aa and Bb, i. e. x must lie either to the right 
 of b or to the left of ^ ; if it lies to the right of b the segment xa 
 is greater than the segment xb and the segment xA is greater 
 than the segment xB, and therefore the equation (a) cannot hold 
 unless X is at an infinite distance, in which case the segments 
 xA, xa, xB, xb are ultimately equal. 
 
 Similarly x cannot lie to the left of A except at an infinite 
 distance, and similar reasoning applies to points in the position 
 shewn in figs. 12 and 13. 
 
 Prop. 12. If on two segments Aa, Bb of a right line as 
 chords, any two arcs of circles are described, their common chord 
 passes through the centre X of the involution A, a and B, b (Fig. 
 125). 
 
 For XA.Xa = XG.Xg = XB.Xb. 
 
 Fig.l25. 
 
 k 
 
 If the segments overlap as in the above figure, they may be 
 taken as diameters of the intersecting circles, and the chord Gg 
 will be perpendicular to the line AaBb. If the points are situated 
 as in figs. 12 and 13, circles described on Aa and Bb as diameters 
 will not intersect in real points, but the centre X will lie on the 
 radical axis of the two circles. 
 
 E. 
 
 15 
 
226 INVOLUTION. 
 
 If a third circle be described passing through the points Gg 
 and cutting Ah in the points M, m (fig. 125) it follows, since 
 
 XM. Xm = XG . Xg = XA . Xa = XB . Xb, 
 that 31 and m are another pair of conjugate points in the in- 
 volution. 
 
 The same is evidently true of amj line cutting the circumfer- 
 ences of all three circles, and we have the important proposition : — 
 
 If three circles pass throiogh two given points^ any straight line 
 meeting the circles does so in a 'series of points in involution^ the 
 two points on the same circle being conjugate. 
 
 When the three circles described on three segments in in- 
 volution as diameters intersect, straight lines drawn from either 
 of the points of intersection to the ends of each segment are 
 perpendicular to each other ; it follows, that when three segments 
 of a straight line are in involution, two 2)oints {real or imaginary) 
 exist, at each of which each segment subtends a right angle; and con- 
 versely, that if a right-angled triangle turns round that angular 
 point as centre, the segments which it intercepts on a fixed right 
 line in any three of its positions have their extremities in involution. 
 
 Problem 124. Given A, a and B, b, two pairs of conjugate 
 2)oints, andC a fifth point of the involution, to determifie c the point 
 conjugate to C. 
 
 Through any arbitrary point G describe segments of circles 
 having Aa, Bb as chords ; they will intersect in a second point g, 
 and a circle described through the three points G, g and C will 
 intersect AC in c, the required conjugate point. 
 
 Or tlius : — Take any arbitrary point G and draw GA, GB, GC; 
 di'aw any triangle with its vertices on these lines and two of its sides 
 passing through a, b. The remaining side will pass through c. 
 
 If the point C be at infinity, the same method will give us the 
 centre of the system. 
 
 The construction for this case is, "Through A, B draw any 
 pair of parallels Ah, Bk, and through a, b a. different pair of 
 parallels ah, bk; then hk will pass through the centre of the 
 svstem." 
 
RATIO AND PROPERTIES OF CONICS. 227 
 
 Prop. 13. Jf Aa, Bh, Cc are any three fixed segments of a 
 straight line, and a, /?, y their centre points, and if m is any point 
 on the line, the function mA . ma . (By + mB . mh .ya + mG . mc . a(i 
 is of constant value whatever the position of m, the general ride of 
 signs being observed. 
 
 Take M any other point on the line, then (p. 209) 
 
 mA = MA - Mm, ma = Ma - Mm ; 
 
 . •. 7nA . ma = MA . Ma - {MA + Ma) Mm + Mm' 
 
 = MA . Ma - 2iT/a . Mm + Mm^, 
 
 and mB . mb =^ MB . Mb -'IMfB . Mm + Mm% 
 
 and mC . mc = MO . Mc - 23fy . Mm + Mm^. 
 
 Multiply these equations by ^y, ya and ap respectively, then 
 since (p. 209) f3y + ya + a(3 = 0, 
 
 and (prop. 3) Ma . (3y + M^ . ya + My . a^ = 0, 
 we get mA . ma . (3y + 7nB . mb . ya + mC .mc. a(3 
 
 - 3fA . Ma . py + MB.Mb . ya + MG . Mc. a/S, 
 which proves the proposition. 
 
 Prop. 14. If three conjugate pairs of points A, a; B, b; C, c 
 are in involution, and a, /?, y the centre jyoints of the segments 
 Aa, Bb, Cc; if any poi^it m be taken on the same straight line, 
 
 mA . ma . /Sy + mB . mb . ya + mC . mc . ay8 = 0, 
 tJie general rule of signs being observed. 
 
 By the last proposition the value of the expression is constant 
 whether the points are in involution or no. When they are in 
 involution the value is zero when m coincides with the centre 
 of the involution, since then 
 
 mA . ma — 7nB . mb = mC . mc, 
 so that the equation may then be written 
 
 mA . ma ((By + ya + ap) = 0, 
 which is evidently true since (p. 209) ; 
 
 (3y + ya + a(3 
 is always zero ; and this proves the proposition. 
 
 15—2 
 
228 INVOLUTION. 
 
 Prop. 15. //* -4, a; B, h; C, c are three pairs of conjugate 
 points in involution, and a, p, y the centre points of the segments 
 Attf Bhj Cc, 
 
 AB .A h a^ aB.ah 
 
 AG . Ac ay aC . ac ' 
 with similarly formed equations for the remaining points. 
 
 In the equation proved in the last proposition, suppose that 
 the point m coincides with the point A. The first term then 
 becomes zero, since mA is zero, and the equation is 
 
 AB. Ah. ya + AC . Ac.ajS^O; 
 or, since ya = — ay, 
 
 AB.Ah _a^ aB.ah 
 AC . Ac ay aC . ac * 
 if we make m coincide with a. 
 
 Problem 125. Given two 2mi7's of points A, a and B, h in a 
 straight line, to find on the same line a fifth point such that the 
 2)roduct of its distances from one pair shall he to the product of its 
 distances from the other in a given ratio X, i.e. given A, a and 
 
 B, h, to determine a point M such that itTn^-jTri ^ ^ (Fig. 125). 
 
 Take any arbitrary point G and describe circles passing 
 through A, a, G and B, h, G and intersecting again in g. Their 
 centres will of course lie on lines perpendicular to Aa, Bh bisecting 
 these segments; let a and /3 be the points of bisection. Divide 
 
 aft in the point fi, so that ^ = X, i. e. on any parallel lines through 
 
 a and /? make a^ = the numerator, and (Sl' = the denominator of 
 the given ratio, the lengths al, /3l' being set off on the same side 
 of a/3 if \ is positive, and on opposite sides if it is negative. 
 
 Describe a third circle to pass through G and g, and with its 
 centre on a line through /w, perpendicular to Ah. This circle will 
 cut -46 in two points M and ?«, either of which fulfils the required 
 conditions of the problem. 
 
RATIO AND PROPERTIES OF CONICS. . 229 
 
 Proof. The points Mm as found are evidently in involution 
 with the points Aa and Bh (p. 226). 
 
 MA . Ma tia in A . ma , _ ^ , 
 
 but ^. = '^ by construction, 
 
 which proves the construction. 
 
 If the segments Aa^ Bh overlap the problem is always pos- 
 sible, but otherwise cases of impossibility may arise owing to the 
 position of the point //,. The problem becomes impossible if ^ 
 falls between F and Q^ the foci of the involution A^ a and B^ h. 
 
 1. If one of the segments falls entirely within the other (as 
 in fig. 12) their centre points a and /3 lie outside the segment FQ 
 and both on the same side of it. If /a falls within the segment 
 
 FQ the ratio ^ = X is positive, and its value lies between -yj— 
 ^p Jrp 
 
 . In order that the problem may be possible in this case A. 
 
 must be negative or must not be between these limits. 
 
 2. If the segments are entirely outside each other (as in 
 fig. 13) their centre points lie outside the segment FQ, but on 
 
 opposite sides of it. If /a falls on FQ the ratio —^ is negative, and 
 
 lip 
 
 its absolute value is between y-r and -^^. In order that the 
 
 1 p Qp 
 
 problem may be possible in this case X must be positive or must 
 
 not be between these limits. 
 
 Problem 126. Given two straight lines AL, BL^, and a fixed 
 point on each A and B. Through a given point F to drav) a 
 straight line meeting AL in a and BL^ in b, so that the segments 
 Aa and Bb shall be to each other in a given ratio X (Fig. 126). 
 
 Imagine two variable points a^, b^ to move along Aa and Bb 
 
 A a 
 respectively, so that in corresponding positions ^^ * = X. The points 
 
 in corresponding different positions would form homographic ranges 
 
230 
 
 ANHAEMONIC EATIO 
 
 on the two lines, for if a^, a^^ a^; h^, h^, b^ are corresponding posi- 
 tions of the moving points, 
 
 Aa^ 
 
 
 
 ~ Bb^^Bb^ ~ b.J),~ 
 
 Fig. 126. 
 
 Aa^ — Aa^ 
 Bb^-Bb^ 
 
 
 L, I, b 'B 
 
 and therefore the anharmonic ratio 
 Aa, ajd, Bb, 
 
 ^«3 ■ %% ~ ^h 
 
 M... 
 
 ¥/ 
 
 and the question therefore is to draw through B a line which will 
 meet the two ranges in homologous points. If the points bJ)J)^... 
 are joined to B, and the joining lines cut AL in a^ttga^..., the 
 ranges a^a^a^... and ttja^ag... are also homographic, since aja^ttg... 
 is a transversal of the pencil B . bfij)^... and the double points of 
 the ranges a^a^a.^... and aja^ag... are extremities of lines fulfilling 
 the conditions of the j^roblem. 
 
 Determine the positions of I and J^, the points which cor- 
 respond to infinity. [Through R draw BI^ parallel to AL meeting 
 BLi in /j, and make A I : BI^ = A; through B draw BJ^ parallel 
 to BL^ meeting AL in t/^.] 
 
 Bisect IJ^ in 0, and on BL^ take a point 0, such that 
 AO : BO^ = X. Draw BO^ meeting ^7^ in Q. Take a mean propor- 
 tional (Op) between OJ^ and OQy and on AL make Oa = -Oa^ — Op. 
 
 Either of the lines aB, ajl fulfils the required conditions. 
 
 In making use of the ratio -7-7- 
 
 segments, measured from the two fixed points A and B\ as other- 
 wise to a point a, we might have corresponding points b^ on 
 
 X, signs must be given to the 
 
AND PROPERTIES OF CONICS. 
 
 231 
 
 either side of the origin B\ so that if the directions in which 
 the segments are measured is not prescribed, the problem admits 
 of four solutions instead of two. 
 
 Problem 127. -Given two straight lines AL, BL^ and a fixed 
 point A^ B on each, to draw through a given point R a line meeting 
 AL in a and BL^ in h, so that the rectangle Aa . Bb shall have 
 a given value v (Fig. 127). 
 
 Exactly as in the last problem, if points a^a^a^... are taken on 
 AD, and points bfij)^... on BL^, connected by the relation 
 Aa^ . Bb^ = Aa^ . Bb^ = Aa^.Bb^...= v, 
 
 Fig. 127. 
 
 I 
 
 these points will form homographic ranges; and if the pencil 
 R.bbb... be drawn meeting AL in a^a^a^... respectively, the 
 double points of the homographic ranges a^a^a^..., a^a^ttg... will 
 be points on the lines required. 
 
 Corresponding to the point at infinity on the range a^a^a^... 
 we must evidently have the point B on the range bfij).^..., and 
 therefore the line BB meets AL in the required point J^. Cor- 
 responding to the point at infinity on the range a^aattjj... we have 
 the point I^ on BL^ found by drawing BI^ parallel to AL, and 
 then / is determined by making AI . BI^ = v. Bisect IJ^ in 
 and determine 0^ on BL^, so that AO . BO^ = v. Draw EO^ 
 meeting AL in Q, and take a mean proportional {Oj)) between 
 O'/j and OQ. Make Oa=^- Oa^ = Op, and Ba, Ra^ will be lines 
 fulfilling the conditions of the problem. 
 
232 ANHARMONIC RATIO 
 
 Problem 128. To draw a triangle having its vertices on three 
 given lines and its sides passing through three given 2^oints (Fig. 
 128). 
 
 Let JJ^, KK^, LL^ be the given lines, and A^ B, C tlie given 
 points. Through one of the points, as A, draw any three lines 
 
 Fig. 128. 
 
 J, 
 
 meeting two of the lines, as LL^ , KK^, in a, 6, c, ayb^c^ respectively. 
 Draw the pencils B . a6c, C . a^h^c^ cutting the third line JJ^ in 
 1, 2, 3, r, 2', 3' respectivelv, which will evidently be homographic 
 ranges. Find the double points JJ^ of the ranges (p. 221) and each 
 will be the vertex of a triangle fulfilling the required condition, for 
 
 {123J} = [ahcL] = {r2'3V} = {afi^c.K}, 
 and since the rays aa^, bb^, cc^ pass through^ so also must the 
 ray KL. 
 
 In the figure JKL, JJ^^L^ are the two triangles. 
 
 Prop. 16. If a quadrilateral ABGD be inscribed iii a conic, 
 and any transversal be drawn meeting the four sides in a, b, c, d 
 and the conic in e and g, then the three pairs of points ac, bd, eg 
 are in involution (Fig. 85.) 
 
 Let ABCD be the angles of the quadrilateral, AB, DC meeting 
 in E, and BC, BA in F. 
 
 Let a transversal cut AB in a, BC in b, CD in c, DA in d, 
 and the curve in e and g. 
 
 The rectangles de . dg, dA . dD are in the ratio of the squares 
 on parallel diameters, as also are the rectangles be . bg and bB .bC; 
 but the squares on the diameters parallel to AD and BC are in 
 the ratio 
 
 FA.FD: FC.FB; 
 
But 
 
 AND PROPERTIES OF CONICS. 
 dA .dD FB. FC 
 
 233 
 
 de.dg 
 he .bg 
 
 bB.bO ' FA.FD 
 Fig.85. 
 
 dA sin a 
 
 dD 
 
 dc ' 
 
 \ 
 
 sin c 
 
 da sin A ' 
 
 &mD ' 
 
 bB sin a 
 
 bC 
 
 sin c 
 
 ba sin B ' 
 
 be ~ 
 
 " sin (7 ' 
 
 dA.dD da. 
 
 .dc 
 
 sin B sin C 
 
 bB . bC ba. 
 
 , be ' 
 
 sm A sin D 
 
 da 
 
 .do 
 
 FA FD 
 
 ~ ba. 
 
 .be ' 
 
 FB • FC ' 
 
 de. dg 
 
 da 
 
 . dc 
 
 be . bg ba . be ^ 
 de ^ be bg ^ dg 
 da ' ba be ' de ' 
 i.e. the anliarmonic ratio of the four points d, e, a, b is equal to 
 that of their four conjugates 
 
 b, g, e, d, 
 or the three pairs of points are in involution. 
 
 Since the diagonals of the quadrilateral form a particular case 
 of a conic passing through tlie four points, it follows that the 
 points in which the transversal cuts the diagonals are another 
 pair in involution with ae, bd, &c. 
 
 Cor. If the transversal be a tangent to the curve (meets it, 
 that is, in two coincident points), it follows that the point of 
 
234 j^NHARMONIC RATIO 
 
 contact is a focus of the involution formed by the three pairs of 
 points in which the tangent cuts the opposite sides and diagonals. 
 
 MacLaurin' s nietliod of generating conic sections : — 
 Triangles are described, whose sides pass through three fixed 
 points A, Bj C , and whose base angles move on two fixed lines 
 Oa, Ob : the vertices will lie on a conic section (fig. 129). 
 
 Fig. 129. 
 
 Suppose four such triangles drawn, then since the pencils 
 through A and B are both homographic with the system through 
 C, they are homographic with each other; therefore A, B, F, 
 Fj, Fg, Fjj, lie on the same conic section (Prop. 9). Now if the 
 first three triangles be fixed, it is evident that the locus of V^ is 
 the conic section passing through 
 
 A,B,o, r„r,. 
 
 It follows of course that the locus of the intersection of 
 homologous rays in two homographic pencils is a conic section. 
 
 NewtonHs method of generating conic sections : — ■ 
 Two angles of constant magnitude move about fixed points 
 P, Q ; the intersection of two of their sides traverses the right 
 line AA^ ; then the locus of F, the intersection of their other twa 
 sides, will be a conic passing through P, Q (fig. 130). 
 Take four positions of the angles, then 
 
AND PROPERTIES OF CONICS. 235 
 
 but F.{AA^A^A^} = F.{VVJJ.^} 
 
 and Q.{AA^A^A^}^Q.{VV,VJ,}, 
 
 since the angles of the pencils are the same ; 
 
 .■.F.{vr,v,r^]^Q.{vvj,v,}, 
 
 and therefore, as before, the locus of V^ is a conic through 
 ^, (?, V, v„ V^. 
 
 Fig. 130. 
 
 Jf. Chasles' extension of Newton's method. 
 
 If the point A instead of moving on a right line moves on 
 any conic passing through the points P, Q, the locus of V is still 
 a conic section, since 
 
 Prop. 17. If there he any number of ^points a, b, c, d, &c. on a 
 right line, and a homographic system a^, b^, c^, d^, dec. on another 
 line, the lines joining corresponding 2^oints will envelope a conic. 
 
 For if we construct the conic touched bj the two given lines 
 and by three lines aa^, bb^, cc^, then, by the anharmonic property 
 of the tangents of a conic (Prop. 8), any other of the lines dd^ 
 must touch the same conic. 
 
 Problem 129. Given two points A and B and a line L ; given 
 also two lines Sa, Sb, and a point ; to find on L a point, Q, such 
 that ifOm, On be drawn parallel respectively to AQ andBQ, meeting 
 Sa in m and Sh in n, the line mn shall 
 
 (a) be parallel to a given direction R, 
 
 ifi) pass through a given point P (Fig. 131). 
 
236 
 
 ANHARMONIC RATIO 
 
 If two homographic pencils be drawn having as common 
 vertex, the envelope of the lines formed by joining the points in 
 which the rays of the one meet Sa to the points in which the 
 corresponding rays of the other meet Sh, is a conic section 
 (Prop. 17). 
 
 Fig. 131 
 
 Draw therefore any lines -41, A'2, -43, &c., meeting L in 
 Ij 2, 3,... and through draw parallels meeting Sa in c, d, e,... 
 and parallels to ^1, B2, £3,... meeting Sb in c^ d^ e^.... The 
 pencils 0{cde...), {c^d^e^...) are homographic because the pencils 
 A (1, 2, 3..,) and B(\, 2, 3...) are so, and therefore cCj, dd^, ee^ 
 are tangents to a conic touching Sa and Sh, and which can there- 
 fore be constructed (Probs. 84 and 114). 
 
 In the figure the point d^ is at an infinite distance (i.e. the tangent 
 ddi is parallel to Sb) because the line B2 is drawn parallel to Sb. 
 
 A line through parallel to the given line L is evidently a 
 tangent to the conic to be constructed, so that it is only necessary 
 to draw two pairs of lines Al, Bl, A2, B2, since five tangents to 
 the required conic are then known. 
 
AND PROPERTIES OF CONICS. 237 
 
 For the solution of the first part (a) draw mn, a tangent to 
 this conic, parallel to the given direction R, and meeting Sa in ni 
 and Sn in n. 
 
 Then AQ drawn parallel to Om and BQ parallel to On will 
 necessarily meet on the line L and determine the required point Q. 
 
 For the solution of the second part (^) it is evidently only 
 necessary to draw a tangent from P to the conic cutting Sa in ni^ 
 and Sh in n^^ and to draw through A and B parallels to Om^ and 
 On^ , which will meet in Q^ on the given line L. 
 
 This problem is sometimes of importance in questions of 
 Graphic Statics. 
 
 If A and B are on opposite sides of the line L, the conic is an 
 hyperbola if is situated in the acute angle formed by the lines 
 Sa and Sb ; and conversely if A and B are on the same side of L. 
 
 The conic will be a parabola when parallels to Sa, Sb, through 
 A and B, meet in the same point on L. 
 
 Examples on Chapter YIII. 
 
 1. Given on a conic three points A,B,C and three other points 
 a, b, c, determine on the conic a point P such that {ABCP]^{abGP]. 
 
 [Either of the points in which the Pascal line (Prob. 85) meets 
 the conic may be taken as the point P, i. e. draw Ac, Ca intersect- 
 ing in K, Be, Cb intersecting in L, and KL will cut the curve in 
 the required point. For if cC meets KL in 3f, the anharmonic 
 ratio of ABCP is that of the pencil c(ABCP), i.e. of the range 
 KLMP, and the anharmonic ratio of abcP is that of the pencil 
 C (abcP), i. e. again of the range XLMP.^ 
 
 2. Inscribe in a given conic a triangle with its sides passing 
 through three given points ABC. 
 
 [Draw any line through A meeting the conic in a and b, draw 
 bB meeting the curve in c, and draw Cc, which will not in general 
 pass through a but will meet the curve in d. Eepeat this twice, 
 giving a range aa^a^ and a second aa^ag. Find a point P such 
 that {aa^a^P} = {aa^a^P}. P will be a vertex of one such triangle.] 
 
 k 
 
238 EXAMPLES. 
 
 3. Given two straight lines AL, BL, draw a transversal 
 meeting them in F and /, so that Ff shall subtend given angles 
 a and ^ at two given points P and p. 
 
 [From the point A draw AP, Aj? and construct the angles 
 A Pa — a and Apb ^ /?, the points a and b being on £L. Imagine 
 A to slide along AL and a and b will form two homographic 
 divisions, and each of the double points gives a solution.] 
 
 4. Given two straight lines AL, PL, draw a transversal 
 meeting them in F and / passing through a given point P, and 
 such that Ff subtends a given angle ^ at a second given point p. 
 
 [Last example, the angle a being zero.] 
 
 5. Determine on a given straight line a segment which shall 
 subtend given angles at two given points. 
 
 [The two lines of Ex. 3 coincide in direction.] 
 
 6. Determine on a given straight line a segment of given 
 length which shall subtend a given angle at a given point. 
 
 7. Given two straight lines AL, PL and a point P, draw 
 through P transversals cutting AL in F, G and PL in /, g, so 
 that FG, fg are given lengths. 
 
 [Draw AP meeting BL in a. On AL make AA^-=FG, and 
 on BL make cia^=fg; draw Pa^ cutting AL in a; A^ and a will 
 form homographic ranges, the double points of which give solu- 
 tions.] 
 
 8. Given on two straight lines AL, BTj, two homographic 
 ranges; draw through two given points P and p, lines Pa, pa^ 
 passing through homologous points a, a^ of the ranges and con- 
 taining between them a given angle 6. 
 
 [Take any point A on the first line and its homologous point 
 B on the second. Draw pB and Pa meeting AL in a and making 
 an angle with pB. Imagine A to slide along AL to A^ and A^j 
 giving corresponding positions a^ a^, of a. 
 
 The range aajag is homographic with AA^A^ because the 
 pencils p [BB^B^ and P (aaja,) are equiangular. 
 
 The double points of these ranges give two solutions.] 
 
EXAMPLES. 239 
 
 9. Given two lines Aa, Ah intersecting in A, the point a 
 being fixed; given also a point S on the opposite side oi Ab 
 from a. Draw a line through S meeting Ah ^n P and Aa in p, 
 and so that AP^ap. 
 
 [Problem 126, the fixed points being A and a, and the ratio 
 one of equality.] 
 
 10. Two lines OAB, Oah meet in 0; A, B, a and h are fixed 
 points on the lines. If OAB remains fixed and Oah turns round 
 0, shew that the locus of the intersection (*S') oi Aa and Bh is a 
 circle having its centre (C) on AB, determined by drawing through 
 any one position oi S 2k parallel aS'C to the corresponding position 
 of Oah. 
 
 [The auharmonic ratio of A, B, 0, C is equal to the an- 
 hamionic ratio of ahO and an infinitely distant point, so that C 
 is fixed.] 
 
 11. Given two homographic ranges ABC... abc... on two 
 lines, determine two homologous segments KL, hi which shall 
 subtend given angles KPL = a, kpl = )8 at two given points P, p. 
 
 [Take any point, as A, on the first range and construct the 
 angle APF^a, and let a and / be the homologous points on the 
 second range to A and F on the first. Construct the angle 
 ap/^ = /3, where /i is on the second range. Suppose the point A to 
 slide along the first range and the points / and /*, will form on 
 the second range two homographic divisions, the double points of 
 which will evidently determine two solutions of the question. 
 Three angles, such as APP, have to be drawn to furnish three 
 pairs of points on the second range.] 
 
 12. Given two homographic ranges ABC... ahc... on two 
 lines, determine two homologous segments KL and M of given 
 lengths. 
 
 [The principle of the solution of the last example is evidently 
 applicable.] 
 
CHAPTER IX. 
 
 PLANE SECTIONS OF THE CONE AND CYLINDER. 
 
 Def. If any fixed point V be taken on a straight line passing 
 through the centre of a circle perpendicular to the plane of 
 the circle, and a straight line move so as always to pass through 
 the circumference of the circle and through the jDoint F, the 
 surface generated by the moving line is called a Bight Circular 
 Gone, and the line OV the axis of the cone. If any solid be 
 conceived as divided into any two parts by any plane passing 
 through the solid, the resulting plane surfaces of the solid in 
 contact with the cutting plane are termed sections of the solid. 
 
 The most convenient way of treating any question on the 
 sections of any solid figure, is by obtaining the projections of the 
 solid on two planes at right angles to each other, the projection 
 of a figure on any plane being, as already explained, the area 
 traced out on the plane by perpendiculars drawn from all points 
 of the figure to the plane. 
 
 The projection of any figure on a horizontal plane is called 
 its lylan, and on a vertical plane an elevation of the figure. In 
 any given position therefore a solid can have but one plan but 
 it may have any number of elevations, so that it is always possible 
 to take the vertical plane on which an elevation is projected 
 perpendicular to any plane of section of the solid. 
 
 For simplicity, the circular base of the cone will be supposed 
 to be horizontal, and the vertical plane of projection perpendicular 
 to the plane cutting the cone. 
 
PLANE SECTIONS OF THE CONE AND CYLINDER. 241 
 
 In figure 132 (p. 242) let o be the centre, and aoh a diameter of 
 a circle representing the base of a right circular cone resting on 
 the plane of the paper; draw any line xy parallel to aoh^ and 
 imagine the part of the paper above xy to be turned up along xy 
 so as to stand perpendicularly to the part in front of that line ; 
 draw aa', oo'v, and hb' all perpendicular to xy and meeting it in 
 a\ o' and h' respectively ; ol and V will be respectively the ele- 
 vations of a and h the extremities of the circular base, o' will be 
 the elevation of the centre, and oV will be the elevation of the 
 axis of the cone. The plan of the axis is obviously the point o. 
 Let the vertex of the cone be at the height o'v above the circular 
 base ah, then v' will be the elevation of the vertex; and if the 
 lines av\ h'v' be drawn and produced indefinitely the triangle 
 a'v'b' will be the elevation of the portion of the cone between the 
 vertex and the horizontal plane of projection. The angle a'v'h' 
 is called the vertical angle of the cone, and since the line vo 
 evidently bisects this angle either of the angles a'vo\ h'v'o' will 
 be the semi-vertical angle. 
 
 It is evident that any section by a plane perpendicular to the 
 axis, or parallel to the base of the cone, is a circle, the circle 
 becoming infinitely small (i.e. the section being a point) when 
 such plane passes through the vertex ; and that the section by 
 any plane through the vertex which cuts the cone in any other 
 point (i.e. which lies within the vertical angle of the cone) will be 
 two straight lines, the angle between which is greatest when the 
 plane passes through the axis, in which case the angle is equal to 
 the vertical angle, and the section is called a Principal Section. 
 
 Problem 130. To determine the section of a cone by a plane 
 which does not contain the axis, and does not pass through the 
 vertex. 
 
 Case I. Suppose the angle which the plane makes with the 
 horizontal plane to be equal to the angle va'o\ the base angle of 
 the cone, fig. 132. Let the plane intersect the base of the cone 
 in the line Im perpendicular to xy, the points I and m being on 
 the base of the cone, i.e. on the circle ab, and let the line Ir/i 
 E. 16 
 
242 
 
 PARABOLIC SECTION. 
 
 meet xy in I ; draw ZV parallel to av' meeting h'v' in 71. The 
 line Itti is the horizontal trace, and I'n' the vertical trace of the 
 
 Fig. 132. 
 
 section plane, those being the lines in which it cuts the planes of 
 projection respectively. The plan of the point n' will evidently 
 be on the line oh vertically below n\ i.e. if n'n be drawn perpen- 
 dicular to xy meeting oh in n, n will be the plan of n'. 
 
 Imagine a horizontal plane to cut the solid at any height 
 between the base and the point ti, as at pq' ; it will evidently 
 cut the cone in a circle of diameter p'q', and which would in 
 
PLANE SECTIONS OF THE CONE AND CYLINDEK. 243 
 
 plan have o for its centre, and it will cut the section plane in a 
 horizontal line the elevation of which is the point /, and the 
 plan of which is the line rry perpendicular to xy. The points of 
 intersection of the circle and line will evidently be points on the 
 desired curve of intersection, and therefore if r, r^ are the points in 
 which the plan of the line cuts the circle pq described with centre 
 o and radius equal to \p'q\ these will be points on the projected 
 curve of intersection. Similarly any number of additional points 
 can be found by taking a series of planes parallel to pq'. The 
 curve Ir^nrm will be of course the plan of the required curve. 
 
 Now imagine the plane of section to be rotated round its 
 horizontal trace hn until it coincides with the horizontal plane of 
 projection, carrying with it the various points of intersection as 
 found. In elevation they would of course travel over circular 
 arcs described with V as centre and with radii equal to the 
 distances between V and their respective elevations, while on 
 plan they would travel along lines through their respective plans 
 perpendicular to Im. 
 
 The point n would therefore reach N, the points r and r^ 
 would reach R and li^ and so on, and the curve IR^NRm would 
 be the true form of the section made by the given plane. 
 
 Inscribe in the cone a sphere which will also touch the given 
 plane of section. The elevation of such sphere will be the circle 
 touching v'a', v'h' and n!l'\ let it do so in the points g' , h' and/*', 
 and let the line g'h' meet I'n' in d'. 
 
 d' will be the elevation of the line of intersection of the given 
 section plane and of the plane through the circle of contact of 
 the cone and the inscribed sphere. On being turned down along 
 with the plane of section this line would therefore come into the 
 position DX, while the point f would come to F. 
 
 The required curve of intersection is a parabola having F for 
 focus and DX for directrix. 
 
 Proof. The line whose elevation is f'r, is a tangent to the 
 inscribed sphere, since it lies in a tangent plane to that sphere 
 (the given section plane) and passes through the point of contact 
 
 16—2 
 
244* ELLIPTIC SECTION. 
 
 of that plane. It is therefore equal in length to any other tangent 
 to the sphere drawn from the point whose elevation is r', and 
 since / is really on the surface of the cone, the length of the 
 tangents drawn from it to the sphere must be equal to the line 
 AY, which is evidently equal to d'r\ i.e. FR^^rd' -t\\e perpen- 
 dicular distance of R from DX ; therefore i? is a point on the 
 parabola described with focus F and directrix DX, and the same 
 of course holds for any other point of the curve. 
 
 Case II. Let the angle which the plane of section makes 
 with the horizontal plane be less than the angle v'a'o\ the base 
 angle of the cone (fig. 133). 
 
 Proceeding exactly as before, let the plane intersect the plane 
 of the base of the cone in the line Im perpendicular to xy, and 
 draw In'n" making any angle less than the base angle of the cone 
 with xy and meeting v'a\ vh' in n' and n". 
 
 Take any horizontal plane (as p'q') at any height between 
 n' and n", meeting v'a' in p\ v'h' in q', and In" in r', and draw 
 the plan of the circle in which this plane cuts the cone (the 
 circle described with centre o and radius op, or oq — \'p'q) and 
 of the line in which it cuts the section plane (through r per- 
 pendicular to xy). The points of intersection r and r^ of this 
 circle and line will be the plans of two points on the required 
 curve of section. Turning the plane round its horizontal trace 
 until it coincides with the horizontal plane the point n' reaches 
 JV, the point n" reaches N^ , and r and r^ come to R and R^ re- 
 spectively. Similarly any number of points can be found, and 
 it will be found that they lie on a closed curve. 
 
 Inscribe in the given cone spheres to touch also the given 
 plane of section (two such can be drawn, one above and the other 
 below the plane) ; let them touch the plane in^*' and/", and v'a, 
 vh' in g', g", and h', h" respectively : let gh' meet In in d', and 
 g"h" meet it in d ". 
 
 d' and d" will be the elevations of the lines of intersection 
 of the given plane of section with the planes of contact of the 
 cone and its inscribed spheres. 
 
PLANE SECTIONS OF THE CONE AND CYLINDER. 245 
 
 Suppose /' and f" and the lines througli d' and d" to be 
 turned down along with the plane of section so that /' comes to 
 F^ f" to F^^ and the lines to DX and Z^,Xj respectively, and the 
 curve of section will be an ellipse with foci F and F^ with 
 directrices DX and D^X^. 
 
 Fig. 133. 
 
 Proof. The line whose elevation is f'r' is a tangent to the 
 inscribed sphere, since it lies in a tangent plane to that sphere 
 (the given section plane) and passes through the point of contact 
 of that plane. It is therefore equal in length to any other 
 
246 HYPERBOLIC SECTION. 
 
 tangent to the sphere drawn from the point whose elevation is r'\ 
 and since / is really on the surface of the cone, the length of the 
 tangents drawn from it to the sphere must be equal to the line 
 }i!q\ which is always in a constant ratio to but is less than d'r, 
 since the angle d!r'(i is less than A'^''/, i. e. FR is always in 
 a constant ratio, smaller than unity, to the perpendicular dis- 
 tance of E from DK. 
 
 Therefore j5 is a point on the ellipse described with focus F 
 and directrix DX^ and the same of course holds for any other 
 point on the curve. 
 
 Case III. Let the angle which the plane of section makes 
 with the horizontal plane be greater than the angle vdo\ the base 
 angle of the cone (fig. 134). 
 
 The description of the last case applies exactly to the present, 
 and the figure is lettered to correspond. The plane will neces- 
 sarily cut both sheets of the cone, and the curve will consist of 
 two infinite branches. 
 
 It will be found to be an hyperbola with foci F and F^ and 
 with directrices DX and BX^ . ♦ 
 
 Proof. In this case the line Kq is always in a constant ratio 
 to but is greater than the line d'r. Hence the distance of any 
 point on the curve from the focus F is always in a constant ratio, 
 greater than unity, to its distance from the directrix DX. 
 
 The two straight lines in which a cone is intersected by a 
 plane through the vertex parallel to an hyperbolic section are 
 parallel to the asymptotes of the hyperbola. 
 
 The asymptotes may be thus found. They of course pass 
 through the point C midway between X and X,. Draw the 
 generators of the cone parallel to the given section plane, i.e. 
 draw 'v'w parallel to ZV, which will be the elevation of such 
 generators, and project w' to w and w^ on the base of the cone. 
 The tangent planes to the cone along the generators whose plans 
 are ow, ow^ and whose elevations are v'w' will intersect the given 
 plane of section in the required asymptotes. If therefore a 
 tangent at w to the circular base of the cone meet Im (the hori- 
 
PLANE SECTIONS OF THE CONE AND CYLINDER. 247 
 
 zontal trace of the given section plane) in W, W will be a point 
 on one asymptote, which will therefore be the line CW. Similarly 
 the second asymptote can be obtained from the point w^ . 
 
 Fig. 134. 
 
 Plan 
 
248 
 
 CONIC OF GIVEN ECCENTRICITY. 
 
 Problem 131. To cut a conic of given eccentricity from a 
 given cone (Fig. 135). 
 
 Let v'a!h' be the elevation of the given cone, and let the 
 m 
 eccentricity be — given by two lines m and n. From o', the foot 
 
 of the axis of the cone, set off along the axis a length o'd = n and 
 
 Fig. 135. 
 
 v' 
 
 
 
 1 ™ 1 
 
 
 
 
 
 \ ^ 
 
 
 
 1 " 1 
 
 
 
 
 
 
 
 
 
 \ 
 
 
 
 
 
 >rvl' 
 
 ; \ '' 
 
 
 / 
 
 \ '^ 
 
 y,-\q 
 
 
 o'e 
 
 Elevation 
 
 m; through d draw dh parallel to the base of the cone 
 
 meeting the slant side in h ; from e with radius hb' (the distance 
 between h and the foot of the slant side) describe an arc cutting 
 a'b' in g; the required section plane must be inclined to the hori- 
 zontal plane at the angle ego\ and all sections made by planes 
 inclined at this angle will have the same eccentricity. 
 
 Proof Produce ge to meet the slant side of the cone in q^ 
 and in the cone inscribe a sphere touching the plane of section 
 geq in the point f and the slant side v'b' in p : through p draw 
 px parallel to the base of the cone meeting geq in x. 
 
 In the triangle pqx, 
 
 pq : qx :: sin pxq : sin qpx ; 
 
 but pq =-fqy the angle pxq = the angle ego\ and the angle qpx = the 
 
 angle hb'g. 
 
 fa sin ego' eo' do m . , , , 
 
 •*• = ^—^iT = ~" -^ IT' = ~ ' s"^c® ^9 = '*o ; 
 qx &in hog eg ho n 
 
PLANE SECTIONS OF THE CONE AND CYLINDER. 249 
 
 but y is the focus, q the vertex, and x the trace of the directrix of 
 the section made by the plane geq. 
 
 If the conic is to be an hyperbola, i. e. if m > n, there is a limit 
 to the vertical angle of the cone in order that the problem may be 
 
 possible. It will be observed that the length eq is , where a 
 
 ° cos a 
 
 is the semi-vertical angle of the cone, and eg must evidently be 
 
 greater than eo or m. 
 
 Therefore — must be greater than cos a or a>cos~^ —.i.e. a 
 m ° 7/1 
 
 must be greater than the angle whose cosine is — , or in other 
 
 words the ratio of the heif];ht of the cone to length of slant side 
 
 must be less than — . 
 m 
 
 Problem 132. From a given cone to cut a conic of given 
 eccentricity and having a given distance FX between focus and 
 directrix (Fig. 135). 
 
 As in the last problem, draw some one plane of section of the 
 required eccentricity, as geq, and determine its focus f and the 
 trace [x) of its directrix. 
 
 Draw v'x, vf to the vertex of the cone ; on xf make xf = the 
 given distance FX, and through f draw fF parallel to xv' 
 meeting fv' in F. Through F draw a line parallel to gq meeting 
 the slant sides of the cone in A and A^ and xv' in X. This will 
 be the trace of the required plane of section, A and A^ being the 
 vertices, F a focus, and X the trace of one of the directrices. 
 
 Def. If a straight line move so as to pass through the 
 circumference of a given circle, and to be perpendicular to the 
 plane of the circle, it traces out a surface called a Right Circular 
 Cylinder. The straight line drawn through the centre of the 
 circle perpendicular to its plane is the axis of the cylinder. 
 
 The cylinder may evidently be regarded as a particular case of 
 the cone, the vertex being at an infinite distance from the base 
 so that the generators are ultimately parallel. 
 
250 
 
 ELLIPTIC SECTION. 
 
 As with the right circular cone, it is evident that a section of 
 the surface by any plane perpendicular to the axis is a circle, and 
 that a section by any plane parallel to the axis (i.e. passing 
 through the infinitely distant vertex) consists of two parallel 
 lines. 
 
 Problem 133. To determine the section of a right circular 
 cylinder hy a 'plane inclined at any given angle (0) to the axis 
 (Fig. 136). 
 
 Let Im be the line of intersection of the given plane of section 
 with the horizontal plane of the base of the cylinder, i.e. the 
 
 Fig. 136. 
 
 horizontal trace of the plane of section. Draw any ground line 
 xly perpendicular to Im, and through I draw Id' making the angle 
 
PLANE SECTIONS OF THE CONE AND CYLINDER. 251 
 
 d'ly equal to the complement of the given angle 0. Let o be the 
 plan of the axis of the cone, and through o draw oo'o" peipen- 
 dicular to xy, o'o" will be the elevation of the axis of the cone 
 on the vertical plane of projection, and Id' will be the trace on 
 the same plane of the given section plane. 
 
 With centre o and radius equal to that of the cylinder 
 describe a circle ah, and draw aoh perpendicular to Im; through 
 a and h draw aa'a", hh'h" perpendicular to xy, meeting it in 
 a and h' , and the rectangle a'ab'b" will be the elevation of the 
 cylinder. Let Id' cut a'a" in n and h'b" in n". 
 
 Imagine a horizontal plane to cut the solid at any height 
 between n and n", as at p'q; it will evidently cut the cylinder 
 ill a circle of diameter 2^'<i\ and which would in plan have o for 
 its centre, and it will cut the section plane in a horizontal line 
 the elevation of which is the point r, in which p'q cuts Id' and 
 the plan of which is the line rry perpendicular to xy. The 
 points of intersection of the circle and line will evidently be 
 points on the desired curve of intersection, and therefore if r, r^ 
 are the points in which the plan of the line cuts the circle ah 
 (which is of cour.-e the plan of the circle p'q') these will be the 
 plans of the points in which the horizontal plane at the height 
 a'jy' above the base of the cylinder cuts the required curve of 
 intersection. 
 
 Now imagine the plane of section to be rotated round its 
 horizontal trace Im until it coincides with the horizontal plane of 
 projection. In elevation the point r would travel over the cir- 
 cular arc r'R' struck with I as centre, meeting the ground line in 
 R', while on plan the points r and r^ would travel along lines 
 rR, r^R^ perpendicular to Im reaching the horizontal plane of 
 projection in the points R, R^ found by drawing R'R^R perpen- 
 dicular to xy. 
 
 Similarly any number of additional points can be found by 
 drawing a series of planes parallel to p'q, all of which will of 
 course cut the cylinder in circles, the plans of which are the 
 circle ah. 
 
252 FOCI AND DIRECTEIX. 
 
 The point ri travels in elevation over the arc 'r\!N\ and the 
 plan of N' is simultaneously on ah and on N'N perpendicular to 
 xy\ and the point n", the plan of which is 6, similarly reaches 
 the horizontal plane at N^, 
 
 The required curve of intersection is an ellipse having NN^ 
 for major axis, and for minor axis a length equal to the diameter 
 of the cylinder. 
 
 The minor axes of all ellipses which can be cut from the same 
 cylinder are consequently of equal length, but the length of the 
 major axis depends jointly on the diameter of the cylinder and 
 the inclination of the cutting plane to its axis, since 
 nri' = a'y cosec Q. 
 
 Just as in the case of the cone, if spheres be inscribed in the 
 cylinder touching the plane of section they will do so in the foci 
 of the curve of intersection. The elevations of these spheres are 
 the circles shewn in the figure touching the line M in the points 
 /' andy*/, and also touching ad\ h'h". f travels over the circular 
 arc f'F\ and F'F perpendicular to xy, meeting ah in F^ deter- 
 mines F, one of the foci. 
 
 The horizontal planes through the circles of contact of the 
 spheres and cylinder intersect the plane of section in the direc- 
 trices of the curve, d' is therefore the elevation of one of them, 
 which after rotation of the section plane round Im comes into the 
 position BX. 
 
 Proof. The line whose elevation is /// is a tangent to the 
 inscribed sphere, since it lies in a tangent plane to that sphere 
 (the given section plane) and passes through the point // in 
 which the sphere touches the plane. It is therefore equal in 
 length to any other tangent to the sphere drawn from the point 
 whose elevation is r', and since r' is really on the surface of the 
 cylinder, the length of the tangents drawn from it to the sphere 
 must be i''k', where r'k' is parallel to the axis of the cylinder, and 
 k' is on the circle of contact of sphere and cylinder. But r'k' : r'd' 
 in a constant ratio = cos 6, and r'k' = F^B; r'd' = EM, where JRM 
 is perpendicular to BX meeting it in J/, therefore i^^i? : HM 
 in a constant ratio, or the locus of i? is an ellipse. 
 
PLANE SECTIONS OF THE CONE AND CYLINDER. 253 
 THE OBLIQUE CYLINDER. 
 
 Def. If a straight line, whicli is not perpendicular to the 
 plane of a given circle, move parallel to itself, and always pass 
 through the circumference of the circle, the surface generated is 
 called an oblique cylinder. 
 
 The line through the centre of the circular base parallel to 
 the generating lines is the axis of the cylinder. 
 
 The section of the cylinder made by a plane containing the 
 axis and perpendicular to the base is called the principal section. 
 
 The section of the cylinder by a plane perpendicular to the 
 principal section, and inclined to the axis at the same angle as 
 the base, is called a suh-contrary section. 
 
 It is evident that any section by a plane parallel to the axis 
 consists of two parallel lines, and that any section by a plane 
 parallel to the base is a circle. 
 
 Problem 134. To determine the sub-contrary section of an 
 ohlique cylinder. 
 
 Let (fig. 137) be the centre of the circular base, and the 
 circle on ab as diameter the base of the cylinder ; let ob be the 
 plan of the axis. Draw xy parallel to ab, so that the elevation on 
 xy as ground line will be parallel to the principal section of the 
 cylinder; draw aa', oo\ bb' perpendicular to xy meeting it in 
 a\ o\ b', which will be the elevations of the corresponding points 
 of the base. Since the elevation is parallel to the principal section, 
 the angle which the elevation of the axis (i.e. the line o'c) makes 
 with the ground line will be the real angle which the axis itself 
 makes with the horizontal plane. Draw a'a^, b'b^ parallel to o'c, 
 these lines are the elevations of the bounding lines of the solid 
 projected on the vertical plane standing on xy. Draw any line 
 a^b^l making the same angle 6 with o'c' as o'c' makes with xy, 
 meeting xy in I, and draw Im perpendicular to xy. 
 
 Im will be the horizontal trace and la^ the vertical trace of a 
 plane of sub -contrary section ; and if this plane be rotated round 
 Im till it coincides with the horizontal plane, every point on the 
 
254 
 
 SUB-CONTRARY SECTION IS A CIRCLE. 
 
 surface of the cylinder between a/ and 5,' will evidently reach 
 a point on the circle on ah as diameter, i. e. the true form of the 
 sub-contrary section is a circle. 
 
 The horizontal projection of the sub-contrary section is the 
 ellipse having cc^ projected from c', the point in which a^h^ inter- 
 sects the axis of the cylinder as major axis, and afi^ the projec- 
 tion of rt/S/ as minor. 
 
 Problem 1 35. To determine the section of an oblique cylinder 
 hy a plane not parallel to the axis, to the base, or to a sub-contrary 
 section. 
 
 Case I. Let the plane of section be perpendicular to the 
 principal section (fig. 137). 
 
 Fig. 137. 
 
 The horizontal trace (de) of the plane of section must be drawn 
 perpendicular to ab, the plan of the axis. If the plane of section 
 makes an angle (<f>) with the horizontal plane, the vertical trace 
 must be drawn through d making this angle with xy. Let it 
 meet a'a^' in h' and b'b^' in k'. 
 
 Draw any circular section, as p'q, between h' and k' meeting 
 dk' in r/; the plan is of course the circle on pq as diameter pro- 
 
PLANE SECTIONS OF THE CONE AND CYLINDER. 255 
 
 jected from p' and ^' on ah^ and if the projection of the point r 
 cuts this circle in r and r^ these will be the plans of two points 
 on the required curve. If rr^ meet pq in ?z we have rn^ = pn . nq. 
 If now the plane of section be rotated round de till it coincides 
 with the horizontal plane, h' travels in elevation to H' and in 
 plan to H, k' travels in elevation to K' and in plan to K, and 
 r and r^ reach R and R^ respectively. Therefore if RR^ meet ab 
 in N, RN^ = pn .nq^p'r' . r'q' ; 
 
 but p'r' : h'r in a constant ratio, 
 
 and r'q' : r'k' in the same ratio, 
 
 . •. p'r' . r'q' : h'r' . r'k' in a constant ratio ; 
 
 but h'r' . r'k' = HJV . iVK, 
 
 . '. RN^ : HN . iV^ in a constant ratio, 
 
 or the locus of R is an ellipse (Prop. 4, p. 108). 
 
 Case II. Let the plane cut the cylinder in any manner 
 (fig. 138). 
 
 Let ah be the diameter of the base peri^)endicular to the hori- 
 zontal trace of proposed section plane. 
 
 The circle on ab is the plan of the base of the cylinder, ov the 
 plan, and o'v' the elevation of its axis, the elevation being projected 
 on a plane perpendicular to the proposed section plane. Lines 
 through a and h parallel to ov are of course the plans of the 
 generators through a and b, and if a and b are projected on to the 
 ground line at a' and b' lines through these points parallel to o'v' 
 will be the elevations of these generators and will be the bounding 
 lines of the solid as seen in the proposed elevation. 
 
 Im is the horizontal trace, and In^ the vertical trace of the 
 section plane; let a'n' parallel to o'v' meet In^ in n', and h'n^ 
 meet it in n^ ; n' and n^ are evidently points on the required 
 curve of intersection, and their plans n and n^ are found by pro- 
 jecting n' and n^ on to the plans of the generators through a and 
 b. Take any horizontal section of the cylinder between n' and n^, 
 as p'q ; the plan is of course a circle of diameter p'q', and its 
 position can be determined by projecting p and q' on to the plans 
 
256 
 
 ELLIPTIC SECTION. 
 
 of the generators through a and 6, as at p, q. This horizontal 
 section and the proposed section plane intersect in a line the 
 
 Fig.138. 
 
 elevation of which is r', the point in which p'q' cuts In^, and the 
 plan of this line cuts the circle pq in points r and r^ projected 
 from /, which are plans of points on the required curve of inter- 
 section. 
 
 Now rotate the section plane round Im, its horizontal trace, till 
 it coincides with the horizontal plane: in elevation the points 
 n', r\ nl travel over circular arcs to N\ R\ N^; in plan n, r, r^, w, 
 travel over lines perpendicular to Im to N, R, JR^, iV^, obtained 
 by projecting N\ R and iV/. 
 
 These are points situated on the true outline of the curve of 
 intersection, and any additional number of points can be obtained 
 in precisely the same manner. The curve is an ellipse having 
 NN^ as a diameter and RR^ as a corresponding double ordinate, 
 so that DD^, the diameter conjugate to NN^, can at once be drawn 
 
PLANE SECTIONS OF THE CONE AND CYLINDER. 257 
 
 by bisecting JVN^ in C, drawing through C a parallel to JRE^ or 
 to Iniy and making on it CD = CD^ = ao the radius of circular base 
 of cylinder. That the curve is an ellipse may be proved similarly 
 to Case I. 
 
 THE OBLIQUE CONE. 
 
 Def. If a straight line pass always through a fixed point 
 and the circumference of a fixed circle, and if the fixed point be 
 not in the straight line through the centre of the circle at right 
 angles to its plane, the surface generated is called an oblique cone. 
 
 The fixed point is called the vertex and the line joining the 
 vertex to the centre of the circle the axis of the cone. 
 
 The section of the cone made by a plane containing this axis 
 and perpendicular to the circular base, is called the principal 
 section. 
 
 The section made by a plane not parallel to the base, but 
 perpendicular to the principal section, and inclined to the gene- 
 rating lines in that section at the same angles as the base, is 
 called a sub-contrary section. 
 
 Problem 136. To determine the sub-contrary section of an 
 oblique cone (Fig. 139). 
 
 Let be the centre and oa the radius of the circular base, 
 and let ov be the plan of the axis. Draw a ground line xy 
 parallel to ov, and let v' be the elevation of the vertex on a 
 vertical plane standing on xy. Project o to o\ and the circular 
 base to a'b' , so that o'v' will be the elevation of the axis, and 
 a'v'b' the outline of the cone; a'v'b' is evidently also identical 
 with the principal section. 
 
 Draw any line e'd'l making the angle a'e7=the angle v'b'y, 
 and meeting v'b' in d' and xy in l] the angle e'd'v is evidently 
 equal to the angle va'b\ so that e'l may be taken as the vertical 
 trace of the plane of a sub-contrary section, the horizontal trace 
 of which must be the line Im perpendicular to xy. 
 
 E. 17 
 
258 
 
 SUB-CONTRARY SECTION. 
 
 Take any horizontal section as -p'ql between d! and e', the 
 plan of which will be a circle on 'pq as diameter, p and q being 
 
 Fig. 139 
 
 the projections of ^' and q^ on the plan of the axis or central 
 plane of the cone. The plane of this section intersects the plane 
 of sub-contrary section in a straight line, the elevation of which 
 is the point / in which 'p'c[ intersects le\ and the plan of which 
 is rr^ projected from r. If rr^ meet the circle on pq in r and 
 7*1, these will be plans of two points on the required curve of 
 sub-contrary section, and if rr, meet 'pq in n, 
 
 rri? = np .nq = r'q' . r'p' = r'd' . rV; 
 
 since a circle can be described round e'q'd'p'. 
 
 Rotate the plane of section round its horizontal trace till it 
 coincides with the horizontal plane of projection, and e', r' and 
 d' travel to E', R' and D' the corresponding positions in plan being 
 E, R and R^ and D their projections. These are of course points 
 on the real outline of the required curve, and if RR^^ meet ED 
 
 in xV, since 
 
 RN=rn, EN=e'r\ ND^r'd!, 
 
PLANE SECTIONS OF THE CONE AND CYLINDER. 259 
 
 we have RN'' = EF.ND, 
 
 or the locus of ^ is a circle on EI) as diameter, 
 
 i. e. the sub-contrary section of an oblique cone is a circle. 
 
 It is evident that all sections parallel to the base or to the 
 plane elm are also circles. 
 
 Planes parallel to the base, or to a sub-contrary section, are 
 called also Cyclic Planes. 
 
 Problem 137. To determine the section of an oblique cone hy 
 a 'plane not parallel to a cyclic plane and not passing through the 
 vertex (Fig. 140). 
 
 Case I. Let the plane be parallel to a tangent plane of the 
 
 cone, i. e. let it be parallel to a generator and perpendicular to 
 the plane containing that generator and the axis. 
 
 17—2 
 
260 PARABOLA FROM OBLIQUE CONE. 
 
 Let a'v'h' be the elevation of the cone, v the plan of the 
 vertex the elevation of which is v\ and ah the diameter of the 
 circular base parallel to the plane of the elevation. 
 
 It is convenient to take the plane of section perpendicular to 
 the plane of the elevation; so that its horizontal trace Im may be 
 drawn perpendicular to xy^ and its vertical trace must then be 
 drawn parallel either to a!v' or to h'v', since the plane itself must 
 be parallel to one or other of these generators — let Iv! parallel to 
 c^v' be its vertical trace. If Im cuts the circle on ah as diameter 
 in d and o?j, these will be points on the required curve of inter- 
 section, and if In! meets h'v in n', 71! will be the elevation of 
 another point, the plan of which will be w, the intersection of 
 hv and the projection of ti'. 
 
 Draw any horizontal plane as %>'(][ between I and 71, meeting 
 ci!v' in p', h'v' in q and hi! in r' ; this plane cuts the cone in a 
 circle the elevation of which is jo'g'', and the plan of which is a 
 circle on 'pq as diameter obtained by projecting ^' and c[ on av 
 and hv respectively. It meets the section plane in a line the 
 elevation of which is the point ?•', and the plan of which is the 
 line r^i projected from r' \ if this line meets the circle on pq in r 
 and r^, these are the plans of two points on the required curve 
 of intersection and similarly the plans of any additional number 
 of points can be obtained. 
 
 Rotate the section plane round its horizontal trace till it 
 coincides with the horizontal plane of projection; the point n' 
 travels in elevation to N' and the point / to R' ; in plan n, r, and 
 r^ travel along nN, rR and r^R^ perpendicular to Im till they meet 
 the projections of N' and R' respectively, and c?, R, N^ R^ and c?, 
 will be points on the real outline of the required curve of inter- 
 section. It is a parabola having the tangent at iV parallel to ^^1. 
 
 Proof. If K bisects RR,, KR^ = pV . r'q\ 
 
 Through n' draw Iin' parallel to j/q' meeting a'v' in A', then 
 h'n' = p'r'j 
 
 r'q' : rV :: h'n' : h'v'] 
 
PLANE SECTIONS OF THE CONE AND CYLINDER. 261 
 
 .*. p'r' .r'q' : Kn' . r'n' :: h'n' : h'v', 
 .'. KB? : h'n . r'n in a constant ratio, 
 
 but r'n = KX cos 6^ where 6 is the angle between KN and pq^ 
 and is constant ; 
 
 .-. KR^^KN multiplied by some constant, or the locus of K 
 is a parabola. 
 
 Case IT. Let the plane of section meet all the generating 
 lines on the same side of the vertex (Fig. 141). 
 
 Let a'v'U be the elevation of the cone, v the plan of v the 
 vertex, and ah the diameter of the circular base parallel to the 
 
 Fig. 141 
 
 ground line and therefore the plan of a'b'. Let the plane of 
 section be perpendicular to the vertical plane of projection, and 
 
262 ELLIPSE FROM OBLIQUE CONE. 
 
 draw its horizontal trace Im perpendicular to xy and its vertical 
 trace cutting a'v' m. h', and h'v' in k'. Project h' to h on ay and 
 h' to k on bv, then h and k are the p^cms of the points in which 
 the generators through a and b meet the section plane, i.e. are 
 the plans of two points on the required curve of intersection. 
 
 Imagine the cone cut by any horizontal plane as p'q' between 
 h' and k', the elevation of the curve of intersection will be the 
 line p'q', meeting a'v' in p' and b'v' in q' and Ik' in r' ; and the 
 plan will be the circle on pq as diameter, obtained by projecting 
 p' on av and q' on bv. The required plane of section cuts this 
 plane of circular section in a line the elevation of which is ?•', and 
 the plan of which is rr^ projected from r'. If rr^ meets the circle 
 on pq in the points r and r^, these are the plans of two points of 
 the required curve of intersection. Similarly the plans of any 
 additional number of points can be obtained. 
 
 Kotate the plane of section round its horizontal trace till it 
 coincides with the horizontal plane of projection; in elevation 
 h'j r' and k' travel to H', li', and K', and on plan h, r, 7\ and k 
 travel along hll, rB, r^ R^ , kK, perpendicular to bn till they meet 
 the projections of H', R' and K'. The points H, R, K, R^ are 
 points on the real outline of the required curve of intersection. It 
 is an ellipse having HK as a diameter, and RR^ as corresponding 
 double ordinate. • 
 
 Case III. Let the section plane cut both sheets of the cone 
 (Fig. 142). 
 
 Let a'v'b' be the elevation of the cone, v the plan of v' the 
 vertex, and ab the diameter of the circular base parallel to the 
 ground line, and therefore the j)lan of ab'. Let the plane of 
 section be perpendicular to the vertical plane of j^rojection, and 
 draw its horizontal trace hn perpendicular to xy, and its vertical 
 trace Ik' cutting b'v' in W, and a'v' in k'. Project K to h on bv, 
 and k' to k on av, then h and k are the plans of the points in 
 which the generators through a and b meet the section plane, 
 i.e. are the plans of two points on the required curve of intersec- 
 
PLANE SECTIONS OF THE CONE AND CYLINDER. 263 
 
 tion. Imagine the cone cut by any horizontal plane as ^tq ', the 
 elevation of the circle in which this plane meets the cone will be 
 
 Fig. 142 
 
 the line p'q meeting av in p', h'v in q ^ and Ik' in r', and the 
 plan will be the circle on 'pq as diameter obtained by projecting 
 ]} and q on av and hv respectively. The required plane of section 
 cuts this plane of circular section in a line, the elevation of which 
 is /, and the plan of which is rr^ projected from /. If rr^ meets 
 the circle on "pq in the points r and r^ , these are the plans of two 
 points of the required curve of intersection. Similarly the plans 
 of any additional number of points can be obtained. 
 
 Rotate the plane of section round its horizontal trace Im till 
 
264 EXAMPLES. 
 
 it coincides with the horizontal plane of projection ; in elevation 
 h\ r and U travel to H\ R and K\ and on plan A, r, r^ and k 
 travel along liH, rR^ r^R^ , kK perpendicular to Im till they meet 
 the projections of H\ K and K' . The points H, i?, R^ and K 
 are points on the real outline of the required curve of intersection. 
 It is an hyperbola having IlK as a diameter, and RR^ as corre- 
 sponding double ordinate of the branch through H. 
 
 The asymptotes are parallel to the generators of the cone 
 which are parallel to the plane of section. If therefore vw be 
 drawn parallel to Ik' meeting xy in io\ and w be projected to 
 meet the circular base ah in w and w^, the joZans of the asymptotes 
 will be parallel to vio, vv)^. Bisect hk in c, and draw cTT, cW^ 
 parallel respectively to vw and vw^, and meeting Im in TT, W^ 
 which will be points on the asymptotes, and they can therefore 
 be drawn through C the point of bisection of HK. 
 
 Examples on Chapter IX. 
 
 1. AVA^f an isosceles triangle, obtuse angled at F, is the 
 elevation of a cone. Shew that if VB be drawn meeting ^^^ in 
 R, and such that VBf-=AB.BA^ (Ex. 15, Chap, ii.) and any 
 plane be drawn having its vertical trace parallel to VB, and 
 horizontal trace perpendicular to ^^j, it will cut the cone in a 
 rectangular hyperbola. 
 
 2. Given a cone and a point inside it determine the conies 
 which have the given point as focus. 
 
 [Draw an elevation av'b' on a plane parallel to the plane 
 containing the axis of the cone and the given point, and lety" be 
 the elevation of the given point. The vertical traces of the re- 
 quired planes of section must be tangents at/' to the circles 
 touching av and 6V, and passing through /'. Two solutions are 
 generally possible.] 
 
EXAMPLES. 265 
 
 3. Shew that all sections of a right cone, made by planes 
 parallel to tangent planes of the cone, are parabolas, and that 
 the foci lie on a cone having with the first a comraou vertex and 
 axis. 
 
 [Shew that the foci of parallel sections lie on a straight line 
 through the vertex.] 
 
 4. Find the least angle of a cone from which it is possible to cut 
 an hyperbola, whose eccentricity shall be the ratio of two to one. 
 
 5. Cut from a right cylinder an ellipse whose eccentricity 
 shall be the ratio of the side of a square to its diagonal. 
 
 [In the cylinder inscribe a sphere, centre C ; determine a 
 point X in the horizontal plane through the centre such that 
 
 T 
 
 ■-^^^=the above ratio, where r is the radius of the sphere. The 
 
 required plane of section must be a tangent plane to the sphere 
 through the point X.'\ 
 
 6. Shew how to cut from a given cone a hyperbola whose 
 asymptotes shall contain the greatest possible angle. 
 
 [The plane of section must be parallel to the axis, pp. 246 and 
 241.] 
 
 7. Cut from a given cone the hyperbola of greatest eccen- 
 tricity. 
 
 [The plane of section must be parallel to the axis, p. 248.] 
 
 8. Different elliptic sections of a right cone are taken having 
 equal major axes ; shew that the locus of the centres of the 
 sections is a spheroid, oblate or prolate, according as the vertical 
 angle of the cone is greater or less than 90". 
 
 [Consider a series of sections perpendicular to a principal 
 section of the cone. The centre is a fixed point on a line of 
 constant length (the major axis), sliding between two fixed lines 
 (the two generators of that section). It therefore traces out an 
 ellipse which by revolution round the axis of the cone generates a 
 spheroid.] 
 
266 EXAMPLES. 
 
 9. Different elliptic sections of a right cone are taken such 
 that their minor axes are equal; shew that the locus of their 
 centres is the surface formed by the revolution of an hyperbola 
 about the axis of the cone. 
 
 [Consider a series of sections perpendicular to a principal 
 section of the cone. Take any section parallel to the base and 
 divide the diameter of that section, so that the product of the 
 two parts = h^ where h is the semi-length of the constant minor 
 axis; the corresponding elliptic section must pass through this 
 point of division, and all these points lie on a hyperbola, the 
 asymptotes of which are the generators of the principal section 
 taken (Prop. 1, p. 160).] 
 
 10. Shew how to cut a right cone so that the section may be 
 an ellipse whose axes are of given lengths. 
 
 [The centre of the section made by the plane perpendicular to 
 any principal section must be the intersection of the ellipse and 
 hyperbola in which such principal section cuts the surfaces re- 
 ferred to in examples 8 and 9.] 
 
 11. Shew how to cut from a right cone a section of given 
 latus rectum. 
 
 [Any point i^ on a hyperbola described as in Ex. 9 may be 
 taken as focus, and the plane of section must be a tangent 
 plane at F to the sphere inscribed in the cone, and passing 
 through F.'\ 
 
CHAPTER X. 
 
 CYCLOIDAL CURVES. 
 
 When one curve rolls without sliding upon another, any point 
 invariably connected with the rolling curve describes another 
 curve, called a roulette. The curve which rolls is called the 
 generating curve, and the curve on which it rolls is called the 
 directing curve, or the base. 
 
 Only a few of the simpler examples of roulettes are here 
 given, the first being the most simple of all, viz. the cycloid. 
 
 B^EF. The cycloid is the path described by a point on the 
 circumference of a circle, rolling upon a fixed right line, in one 
 plane passing through the line. 
 
 In the construction this plane coincides with the plane of the 
 
 paper. 
 
 r 
 Problem 138. To describe a cycloid^ the diameter of the circle 
 
 being given (Pig. 143). 
 
 Let ^i> be the diameter of the given circle, C its centre, and 
 suppose that the tracing point is the point B, and that at the 
 momenb A is the point of contact of the circle with the directing 
 line. Draw the directing line XA Y a tangent at A to the circle. 
 The tracing point B will evidently reach the guiding line at points 
 X and Y on opposite sides of A such that AX—AY=i\\Q semi- 
 circumference ABf since each point of the semi-circumference 
 comes down successively on a corresponding point of the line. 
 
 The following geometrical construction gives an exceedingly 
 close approximation to the length of the circumference of a 
 circle : — From C, the centre, draw a radius Gil making an angle of 
 30" with the radius CB, and draw HK perpendicular to AB meeting 
 it in K. At A, the extremity of the diameter through B, draw a 
 
268 
 
 THE CYCLOID. 
 
 tangent to the circle and on it make AL= 3 . AB. KL will be 
 
 Fig. 143. 
 
 very nearly the circumference of the circle and its semi-length 
 may be taken for the length AX ov A Y. 
 
 [In the figure L does not fall within the limits of the paper, 
 but if J ^ is bisected in h and hk on a parallel to the tangent at A 
 be made = 3 times the radius of the circle, Kk may be taken as the 
 semi-circumference. ] 
 
 Divide up ^X into any number of equal parts (say 8) as at 
 a', h', c',... and divide the semi-circumference AB into the same 
 number as at a, b, c,... Draw a line through C parallel to XAY, 
 which will evidently be the path of the centre of the circle, i. e. as 
 the circle rolls along AX the centre will always be on this line; 
 and draw a'l, b'2, c'3... perpendicular to AX, the points 1, 2, 3, &c., 
 being on the path of the centre. The point a will evidently come 
 down to a', b to b\ and so on; and when a has come to a', the 
 centre of the circle will be at 1 and the tracing point will be on a 
 line making an angle with a'l equal to the angle aCB, which is of 
 course equal to ACff, since Aa = gB. Draw IG parallel to Cg and 
 make lG = Cg, the radius of the rolling circle. G will be a point 
 on the required curve. 
 
 Similarly, when b has rolled down to b', the centre of the circle 
 will be at 2 vertically above b', the tracing point will be on a line 
 making with b'2 an angle = the angle bCB, i. e. = the angle AC/, or 
 
CYGLOIDAL CURVES. 269 
 
 it will be on a line 27^ parallel to C/" and at a distance from 2 equal 
 to tlie radius of the circle. 
 
 Similarly for the remaining points c', d\ &c. 
 
 It will be noticed that the lengths 16^, 2^, &c., may be deter- 
 mined without actual measurement by drawing through g^f, &c., 
 parallels to AX meeting the corresponding lines through 1, 2, &c., 
 in the points (r, F, &c., the figures ICgG, 2C/F are parallelo- 
 grams and therefore in each case lG = Cg, 2F= Cf, and so on. 
 
 The curve should be drawn free-hand through the series of 
 points thus found, and the half loop corresponding to the circle 
 rolling on ^ Z may be found by the same construction or may be 
 put in by symmetry. The line XS is a tangent to the curve at 
 the point X. 
 
 The length AX may be determined arithmetically by multi- 
 plying the length of the radius AC by 3. 14... and may then be 
 laid down by scale : the diagonal scales usually supplied with 
 cases of mathematical instruments can conveniently be used for 
 the purpose. In many works on geometry the length ^X is 
 determined by dividing up the semi-circle into any number of 
 equal parts (say n) and laying oif along AX the length of the 
 chord of one of the parts repeated n times. This method is radi- 
 cally bad and should never be adopted : if the number of equal 
 parts into which the semi-circle is divided is small it gives only a 
 very rough approximation to the truth, while if the number is 
 increased it is almost impossible to measure the length of the 
 chord so accurately but that in repeating it n times an appreciable 
 error will be introduced. A long length should in fact never 
 ■ be determined as the sum of a series of short ones. 
 ""^ To draw the normal at any 'point of a cycloid. 
 
 In all roulettes the normal at any point passes through the 
 corresponding point of contact of the rolling and guiding curves. 
 This point is called the Instantaneous Centre. The direction of 
 motion of the tracing point will evidently at any moment be 
 perpendicular to the line between it and the point about which 
 the rolling curve is turning, i.e. the corresponding instantaneous 
 
270 THE TROCHOID. 
 
 centre, and since the direction of motion at any point must co- 
 incide with the tangent at that point, the normal must pass 
 through the instantaneous centre. 
 
 In the figure, when the tracing point is at E the centre is at 3 
 and c' is the instantaneous centre, so that Ec is the normal at E ; 
 this is evidently parallel to eA, e being the point in which a 
 parallel to AX through E meets the circle on AB as diameter, so 
 that the normal at any point P may be thus constructed : — 
 
 Through P draw a parallel to the directing line -4 F meeting 
 the circle on AB in the point Q. The normal at P will be parallel 
 to AQ, and since the angle AQB is a right angle the tangent at P 
 will be parallel to QB. 
 
 If the normal at P meet the directing line in M and PM be 
 ^ produced to S so that Pas' = '2PM, S will be the centre of curvature 
 at the point P. The e volute of the cycloid is two equal semi- 
 cycloids, the vertices being at X and Y and the cusp on BA 
 produced at a distance from A=AB. 
 
 Let the tangent at P meet the tangent at the vertex in T, 
 then the length of the arc BP of the cycloid is double the intercept 
 TP of the tangent, i. e. double the chord BQ of the circle. Hence 
 the whole length of the cycloid is 4 times the diameter of the 
 generating circle. 
 
 ^ ^Bra^ If, as in the cycloid, a circle rolls along a straight line, 
 any point in the plane of the circle but not on its circumference 
 traces out a curve called a Trochoid. 
 
 PROBLBai/139. To describe a trochoid, the diameter of the 
 circle and the distance of the tracing point from its centre being 
 given (Fig. 144). 
 
 Let ABhQ the diameter of the given circle, C its centre, and 
 CP the distance of the tracing point from the centre. 
 
 Draw XA Y a tangent to the circle, and as in the last problem 
 determine the length AX or AY equal to the semi-circumference 
 of the circle AB. Draw C8, the path of the centre, through C 
 parallel to XAY, and through X draw XS perpendicular to XA. 
 
CYCLOIDAL CURVES. 
 
 271 
 
 Divide CS into any number of equal parts (8 in the fig.), and with 
 centre C and radius CP draw a circle. The point F in which this 
 
 circle cuts AB produced will be the vertex of the required curve. 
 Divide the semi-circumference of the circle into the same number 
 of equal parts Fg, gf, &c., as has been chosen for the division of 
 the path of the centre. 
 
 Draw IG parallel to Cg and gG parallel to ^X : their intersec- 
 tion G will be a point on the required curve. Similarly 2F paral- 
 lel to Cf and /F parallel to AX will intersect in a point on the 
 curve, and so on in succession. When B has come down to X the 
 tracing point will evidently be at F^ vertically below X on 8X 
 produced so that 8F = CF ; the tangent at F^ is parallel to AX. 
 
 The construction is obvious from that of the cycloid. 
 
 In the figure a second trochoid is drawn generated by a point 
 Q inside the rolling circle, to which the foregoing description 
 applies exactly by the substitution of Q for F. 
 
 To draw the normal at any point of a trochoid. 
 
 Consider for a moment the point F. When the tracing point 
 is at F the centre of the rolling circle will be at 2 and the point 
 of contact of the rolling circle and directing line will be H on AX 
 vertically below 2 ; i.q. H will be the instantaneous centre, and 
 therefore FH will be the normal at F, since the direction of motion 
 of F must be perpendicular to FH. But FH is parallel to/A, 
 
272 THE EPI-CYCLOID. 
 
 since the triangles F2H,fCA are in all respects equal and are 
 similarly situated, and therefore the normal at any point R may 
 be thus constructed : — 
 
 Through R draw a parallel to the directing line meeting the 
 circle described with C as centre and CP as radius in the point r, 
 and the normal RM will be parallel to the line joining r to A^ the 
 lowest point of the rolling circle when its centre is C. 
 
 To find the centre of curvature at any point R*. 
 
 Find Ky the position of the centre of the rolling circle corre- 
 sponding to R. (/f will of course be vertically above M.) Join 
 RK and draw 31 JV perpendicular to RM meeting RK in iT. Draw 
 IfS perpendicular to the guiding line meeting R3I in S. S will 
 be the required centre of curvature. 
 
 -" Def. The Epicycloid is the path described by a fixed point 
 on the circumference of a circle rolling on the convex side of a 
 fixed circle, both circles lying in the same jjlane. 
 
 Problem 1X0. To describe an epicycloid, the radii of Wie 
 rolling and directing circles being given (Fig. 145). 
 
 Let be the centre of the directing circle, OA its radius, 
 AC the radius of the rolling circle, C, on OA produced, its centre, 
 and let B be the other extremity of the diameter through A. 
 Suppose B to be one position of the tracing point. As the one 
 circle rolls round the other let the point B come down to X on 
 the one side of A and to Y on the other, X and F being on the 
 directing circle. The arc AX will necessarily be equal to the 
 arc AYf and equal to the semi-circumference of the rolling 
 circle. 
 
 These points may be thus determined : — 
 Let the length of the semi-circumference AB be S, then 
 JS = 7r.AC, 
 IT being the circular measure of two right angles. 
 
 * The construction for the centre of curvature of this and the following 
 roulettes was given by M. Savary in his Legons des Machines a VEcole 
 Poly technique, and is quoted by Williamson, Differential Calculus, 3rd ed,, 
 p. 345, where its proof is given. 
 
CYCLOIDAL CURVES. 273 
 
 Let 6 be the circular measure of the angle subtended by the 
 
 Fig. 145. 
 
 arc AX (the length of which is S), at the centre of the directing 
 circle ; then 
 
 S = e.A0 = 7r.AC; 
 
 .'. 6 : TT '.'.AC lAO, 
 or if n is the number of degrees in the angle A OX, 
 71 : 180" :: AC : AG, 
 
 or «=180»^, 
 which determines n. 
 
 [In the figure AG ^ SAC so that the angle AGX contains 
 60°.] 
 
 Draw the path of the centre of the rolling circle, i.e. an arc 
 with centre G, and radius OC, and let GX produced meet it in 8. 
 Divide up the arc (78 into any convenient number of equal parts 
 (8 in the fig.) and draw the radii 01, G2... cutting the directing 
 circle in a' h' .... Divide up the semi-circumference of the rolling 
 circle into the same number of equal parts Aa, ah... . 
 
 K 18 
 
274 NOKMAL AND CENTRE OF CUEVATUREJ. 
 
 As the one circle rolls on the other, the point a will evidently 
 come down to the point a\ h to h' and so on : when a has come to 
 a\ the centre of the rolling circle will be at the point 1, and 
 the tracing point will evidently be on a line making with a\ 
 an angle equal to the angle aCB which is equal to the angle ACg. 
 Hence an arc described with centre 1, and radius CB, will inter- 
 sect an arc described with centre 0, and radius Og, in a point 
 G of the required curve, for the triangles 6^10 and gCO are 
 equal in all respects : — i.e. G is the position of the tracing point 
 corresponding to a\ being the point of contact of the rolling and 
 directing circles. 
 
 Similarly an arc described with centre 2, and radius CB will 
 intersect an arc described with centre 0, and radius 0/ in a point 
 F of the required curve, and so on in succession for the points 
 3, 4, &c. 
 
 The arcs gG,fF, &c. will cut the corresponding arcs described 
 with the successive centres 1, 2, &c. in two points, but it is 
 evident by inspection which of the points must be taken, viz. 
 that on the side of the corresponding radius 01, 02, &c. remote 
 from OA. 
 
 The radius 0X8 is a tangent to the curve at the point X. 
 
 To draw the normal at any point P of an epi-cycloid. From 
 P with the radius AC oi the rolling circle describe an arc cutting 
 the path of the centre in K. [It will do so in two points but 
 the one lying within the angle POB must be taken.] This will 
 be the position of the centre of the rolling circle when the tracing 
 point is at P. Draw KO cutting the directing circle in J/, the 
 point of contact between the circles when .the tracing point is at 
 P: i.e. if is the instantaneous centre corresponding to P. 
 
 Therefore PM is the normal at P. 
 
 To find the centre and radius of curvature at any point P. 
 From M the instantaneous centre draw MN perpendicular to 
 PM meeting PK, the radius of the rolling circle when the tracing 
 point is at P, in iV. Then NO (0 being the centre of the guiding 
 
i 
 
 CYCLOID AL CURVES. 275 
 
 circle) will cut FM produced in S the required centre of cur- 
 vature. 
 
 Def. The Hypo-cycloid is the path described by a fixed 
 point on the circumference of a circle rolling on the concave side 
 of a fixed circle, both circles lying in the same plane. 
 
 Problem 141. To describe a hijpo-cycloid the radii of the 
 rolling and directing circles being given (Fig. 145). 
 
 OA is the radius of the directing circle, and its centre, AC 
 is the radius of the rolling circle, and B' the tracing point when 
 the centre is at C. The construction is identical with that for 
 the epi-cycloid. In the figure the radius AC is equal to AC the 
 radius of the epi-cycloid, and B' of course reaches the directing 
 line at X and Y — the points F' and D' are the positions of the 
 tracing point when the points b^ and d^ are the points of contact 
 of the rolling and directing circles. 
 
 Def. When, as in the epi-cycloid, a circle rolls on the convex 
 side of another, any point in the plane of the rolling circle, 
 but not on its circumference traces out a curve called an Fpi- 
 trochoid. 
 
 Problem 142. To describe an epi-trochoid^ the rolling and 
 guiding circles , and the position of the tracing point being given 
 (Fig. 146). 
 
 [In the figure the tracing point is assumed outside the rolling 
 circle; it might be inside it.] 
 
 Let be the centre of the directing circle, OA its radius, 
 ^Cthe radius of the rolling circle; G, on OA produced, its centre; 
 let B be the other extremity of the diameter through^, and F on 
 CB produced be one position of the tracing point. As in the 
 epi-cycloid determine an arc ^X or ^7 of the guiding circle equal 
 in length to the semi-circumference of the rolling circle, so that 
 B comes down to X and Y as the circle rolls round: i.e. construct 
 angles AOX and AOY each containing n degrees Where 
 
 [In the figure AO = ZAC bo that n = 60.] 
 
 18—2 
 
276 
 
 THE EPI-TROCHOID. 
 
 Draw the path of the centre of the rolling circle, i.e. the 
 circular arc with centre 0, and radius OC, and produce the 
 
 
 Fig 
 
 .146. 
 
 P 
 
 
 5^ 
 
 P^ 
 
 ^'"'''^ 
 
 ■^-^.^^ 
 
 
 \ 
 
 B~"^\ 
 
 \ 
 
 d 
 
 _.ji 
 
 
 - K 
 
 n'' 
 
 >) 
 
 h 
 
 ^^S? 
 
 m>^ 
 
 X / 
 
 ^^ -^ — 
 
 ^>s 
 
 \ 
 
 :.-.X 
 
 *■ 
 
 ■c,f--. 
 
 I 
 
 
 *>.- 
 
 
 
 radius OX to meet it in 8. Divide up C8 into any convenient 
 number of equal parts CI, 12, '&c. — (8 in the figure), and divide 
 up the semi-circle drawn through F, with centre C, into the same 
 number Pg, gf, &c. "With centra 1, and radius equal to CP, 
 describe an arc, and with centre 0, and radius Og, describe a 
 second arc cutting it in G. G will be a point on the curve. 
 Similarly with centre 2, and radius equal to CP, describe an arc, 
 and with centre 0, and radius Of, describe a second arc cutting 
 it in F. F will be a point on the curve, and so on in succession 
 for the points 3, 4, &c. 
 
 The arcs gG, /F, &c. will cut the corresponding arcs described 
 with the successive centres 1, 2, &c. in two points, but it is 
 evident by inspection which of the points must be taken, viz. 
 that on the side of the corresponding radius 01, 02, &c. remote 
 from OA. 
 
 The radius 0X8 is a normal to the curve at the point P^. 
 
CYCLOIDAL CURVES. 277 
 
 To draw the normal at any point R of an epi-trochoid. 
 
 Find Z'the corresponding position of the centre of tlie rolling- 
 circle, i.e. with centre R, and radius equal to OP, describe an arc 
 cutting the path of the centre in K. [It will do so in two points, 
 but the one must be taken lying within the angle ROB.'\ 
 
 Draw KO cutting the directing circle in M. M will be the 
 instantaneous centre corresponding to R. Therefore RM is the 
 normal at R. 
 
 To find the centre and radius of curvature at any point R. 
 
 From M the instantaneous centre draw MN perpendicular to 
 RM meeting RK (K being as above) in J^. Then, if is the 
 centre of the directing circle, OJV will cut the normal RM pro- 
 duced in Sf the required centre of curvature. 
 
 Def. The Hyjyo-trochoid is the curve traced out by any point 
 in the plane, but not on the circumference of a circle, rolling 
 on the concave side of a fixed circle, both circles lying in the 
 same plane. 
 
 Problem 143. To describe a hypo-trochold, tJie directing and 
 rolling circles^ and the position of the tracing point being given 
 (Fig. 146). 
 
 [In the figure the tracing point is inside the rolling circle, 
 but by the above definition this is not a necessary condition.] 
 
 OA is the radius of the directing circle, and its centre, AC 
 is the radius of the rolling circle, and Q the tracing point when 
 the centre is at C". The construction is identical with that for 
 the epi-trochoid. 
 
 Companion to the cycloid. 
 
 Def. If a line liE (Fig. 147) be drawn perpendicular to a 
 fixed diameter ^^ of a circle, meeting it in JV, and the circle 
 itself in e, and if JSfE be made equal to the arc Be, the locus of 
 the point B is called the Companion to the Cycloid. 
 
 Problem 144. To describe the companion to the cycloid, the 
 generating circle being given (Fig. 147). 
 
 C is the centre, and ABdi. diameter of the given circle. Through 
 
278 COMPANION TO CYCLOID. 
 
 A draw XAT a tangent to the given circle, and on it make 
 AX = AY = the semi-circumference. (Prob. 138.) Divide AY 
 
 Fig. 147. 
 
 into any convenient number of equal parts as at a', h\c' .., and 
 divide the semi-circumference AB into the same number of equal 
 parts as at «, 5, c... 
 
 It will be observed that the lettering proceeds from A in the 
 one case, and from ^ in the other. 
 
 Through a', h\ c ... rule perpendiculars to AY^ and through 
 «, 6, c... rule parallels to AY. The intersections of corresponding 
 lines ^^ D, E^ F ... are points on the required curve. 
 
 The construction is obvious. 
 
 To draw the tangent at any point P. 
 
 Draw PM parallel to AX meeting the circle in /), and the 
 diameter ^^ in M. Make Cm on CB = Mp, and join m to K 
 the extremity of the diameter perpendicular to A C. The tangent 
 at P is parallel to mK. The curve has parallel tangents at 
 points equi-distant from CK. 
 
 To find the radius of curvature at any point P. 
 
 It is easily proved analytically that p = —frn » where p is the 
 
 radius of curvature, m and M are points corresponding to P as 
 above, and a is the radius of the generating circle. 
 
CYCLOIDAL CURVES. 279 
 
 Make Km^ on KC = Km, and draw m^R perpendicular to KC 
 meetiug Km in R, also make Kk on KG = CM. Through m^ 
 draw m^s parallel to kR meeting KR in s, and Ks will be the 
 length of the required radius of curvature. Make PS on the 
 normal at P = Ks, and S will be the centre of curvature at P. 
 
 Evidently Ks \ KR v. Km^ : Kk, 
 
 KR . mK 
 
 or Ks 
 
 CM ' 
 
 7nK' 
 
 but KR : Km :: Km^ : CK, or Zi^ 
 
 J. mK\ 
 
 a. CM ^ 
 
 Examples on Chapter X. 
 
 1. Shew that if the diameter of the rolling circle be half 
 that of the directing circle, the hypo-cycloid becomes a straight 
 line. 
 
 2. Shew that if the diameter of the rolling circle be half that 
 of the directing circle any hypo-trochoid becomes an ellipse. 
 
 3. Shew that if AOB be a diameter of the guiding circle, and 
 P any point on it, the hypo-cycloids described by the circles 
 having AP and BP as diameters, and P as tracing point, are 
 identical. 
 
 4. -4 is a fixed point on the circumference of a circle of 
 radius R. The points L and M are taken on the same side of 
 A such that arc AL^m. arc AM, where m is a constant. Shew 
 that LM will always touch the epi-cycloid described with a circle 
 
 of radius r (= ) rolling on a circle of radius p^R-2r, the 
 
 \ m+ 1/ ° r 5 
 
 point A being the centre of the loop, and the centre of the 
 guiding circle coinciding with that of the given one. 
 
 [As a numerical example take R = 3|, m = 4.] 
 
280 EXAMPLES. 
 
 5. ^ is a fixed point on the circumference of a circle of 
 radius R. The points L and M are taken on opposite sides of A, 
 such that arc AL = m . arc AM, where w is a constant. Shew that 
 LM will always touch the hypo-cycloid described with a circle of 
 
 radius r = j rolling under a circle of radius p = B+2r, the 
 
 m — 1 r ' 
 
 point A being the centre of the loop and the centre of the guiding 
 
 circle coinciding with that of the given one. 
 
 6. Shew that the radius of curvature of an epi-cycloid at 
 any point varies as the perpendicular on the tangent at the point, 
 from the centre of the fixed circle. 
 
 7. Shew that the evolute of the epi-cycloid described with 
 guiding circle of radius a and rolling circle of radius 6 is a similar 
 
 2 
 
 figure, the radii of the fixed and generating circles being — —7 and 
 ^^ respectively. 
 
 8. Shew that the evolute of the hypo-cycloid is a similar 
 figure, the radii of the fixed and generating circles being — — 
 
 and TTT respectively. 
 
 a — lb 
 
 [To make a practicable figure h must be much smaller 
 than a.] 
 
 9. If a parabola rolls on another equal parabola shew that 
 lo^ 
 other. 
 
 the locus of the focus of the rolling one is the directrix of the 
 
CHAPTER XI. 
 
 SPIKALS. 
 
 When a line rotates in a plane about a fixed point of its 
 length, and a point travels continuously in the same direction 
 along the line according to some fixed law, the path of the moving 
 point is called a spiral. The fixed point is called the pole; a fixed 
 line in the plane passing through the pole from which the position 
 angle of the moving line may be measured is called the initial 
 line, and the line drawn from the pole to any point of the curve is 
 called the radius vector of that point. 
 
 After rotating through four right angles the revolving line 
 comes back to the position it occupied at starting, but there is of 
 course a different value for the length of the radius vector, and since 
 the position angle may increase without limit, so too does the value 
 of the radius vector. Spirals consequently extend to an infinite 
 distance from the pole, and consist of a series of convolutions 
 round it. 
 
 Cases of mathematical instruments usually contain a diagonal 
 scale, the unit of which is half-an-inch, and on which lengths can 
 be read to two places of decimals. In the numerical examples 
 which follow, this scale is intended to be used. 
 
 Def. In the Spiral of Archimedes the length of the radius 
 vector is directly proportional to its position angle. 
 
 Let r be the length of the radius vector of any point, 6 the 
 angle which it makes with the initial line ; the above definition is 
 expressed symbolically by the equation r=.aB, where a is any 
 numerical constant. 
 
282 
 
 SPIRAL OF ARCHIMEDES. 
 
 In this equation 6 is the circular measure of the position angle, 
 and therefore r — a when is unity, i.e. when the number of 
 degrees in the position angle is 57*2957... i.e. corresponding to this 
 angle measured from the initial line, the tracing point is at a 
 distance of a imits (inch, or any other that may be chosen) from 
 the pole; when r = 0, ^ = 0, or the initial line is the position of the 
 revolving line when the travelling point is at the pole. 
 
 Problem 145. To describe the spiral of ArcJdmedes, the pole, 
 two points on the curve, and the unit of the curve being given 
 (Fig. 148). 
 
 Let be the pole, P and Q the two points on the curve which 
 we will suppose to be on the same convolution; and let OQ be 
 
 Fig. 148 
 
 greater than 0P\ let 6 be the angle between OP and the initial 
 
 line, and the length L the given unit. 
 
 0P = a6 
 
 0Q = a{6 + Q0P); 
 
SPIKALS. 283 
 
 therefore OQ-OP = ax circ. meas. of QOP 
 
 OQ-OP 
 circ. meas. of QOP ' 
 OQ-OP can be measured by scale, the number of degrees in 
 the angle QOP can be measured by a protractor and its circular 
 measure can be obtained from a table of the circular measures of 
 
 OP 
 angles, and the numerical value of a thus calculated: then^ = — , 
 
 the length OP being of course measured on the same scale as that 
 used for determining OQ — OP^ which gives the circular measure 
 of the angle between OP and the initial line, and the correspond- 
 ing number of degrees can be obtained from the table. 
 
 To take a numerical example : 
 
 Let the unit of length be \ an inch. Suppose 
 0^ = 2, 
 OP =1-5, 
 and the angle QOP - 60", the circular measure of which is 
 3-14159.. 
 
 3 
 2-1-5 -5 
 
 1-0472... 
 
 ^ = •477. 
 
 1-0472 1-0472 
 then ^^_J^^3-14 
 
 the number of degrees corresponding to which may be taken 180". 
 
 The initial line will therefore be the line OA. If the tracing 
 point after one complete revolution of the generating line cuts OP 
 again in F we have 
 
 OP = ae 
 
 and 
 
 0P' = 
 
 = a(0+27r), 
 
 
 therefore 
 
 OF -0P = 
 
 = 27ra. 
 
 
 Successive points on the cur^^e may at once be found thus : — 
 Construct an angle QOP = angle QOP; with centre and radius 
 OP describe an arc cutting 0^ in^; on OQ produced make 
 
284 TANGENT AND EADIUS OF CURVATURE. 
 
 Qr = Qp and with centre and radius Or describe an arc cutting 
 OR in i? a point of the curve. 
 
 A A 
 
 Similarly if ROS = FOQ, and Qs on OQ produced = 2Qp, an arc 
 described with centre and radius Os will cut OS in S a. point of 
 the curve. (In the figure S coincides with A on the initial line.) 
 
 In like manner points can be found nearer the pole than F by 
 constructing angles on the side of OP remote from Q equal respec- 
 tively to FOQ, 2F0Q, 3F0Q, &c., and diminishing the radii 
 A^ectores by the constant difference pQ. 
 
 To draw the tangent at any point of the curve. 
 
 A known expression for the angle which the tangent at any 
 
 r 
 point makes with the radius vector is ^ = tan~^ — , i.e. the tangent 
 
 a 
 
 of the angle is the radius vector divided by the given constant of 
 
 the curve. 
 
 Therefore to draw the normal at any point Q, on the radius 
 
 OG at right angles to OQ measure a length OG = a, the constant 
 
 of the curve, and QG will be the normal at Q, for evidently 
 
 tan OG^^ - ^ = - = tan (jf). 
 
 Ubr a 
 
 Hence if a circle be drawn with centre 0, and radius = a, nor- 
 mals at all the points on the curve can at once be drawn by merely 
 joining them to the corresponding points in which such circle cuts 
 the perpendicular radii. 
 
 The initial line is a tangent at the pole. 
 
 If p is the radius of curvature at any point 
 
 p : J^^T? ;: a' + r" : 2a' +r', 
 so that p can be calculated without much difficulty. 
 
 Peoblem 146. To describe the spiral of Archimedes^ tlie pole, 
 the initial line and the constant of the curve being given (Fig. 149). 
 
 Here a is given in the equation r = aO. Let be the pole, 
 and OA the initial line. In the figure, the unit being the length 
 L, a =-239. 
 
SPIRALS. 
 
 285 
 
 Determine some convenient length of radius corresponding to 
 a multiple (n) of 4 right angles ; say the greatest distance to which 
 
 Fig. 149. 
 
 it is proposed to draw the curve. In the figure e.g. A is taken at 
 angular distances of 8 right angles from the initial line (i.e. n = 2), 
 so that OA = -239 x i-rr 
 
 = •239x4x3-14159... 
 
 =: 3-60 units. 
 
 Draw OD at right angles to OA and divide up the quadrants 
 formed at into any number (m) of equal parts (in the figure m=S) 
 and draw the radii OB, OC, &c. through the points of division. 
 Divide OA into 4 . m . w equal parts. In the figure therefore 
 OA is divided into 24 equal parts. Then arcs drawn through the 
 successive points on OA with centre will intersect the corres- 
 ponding radii in points on the curve. The point P in the figure of 
 course bisects OA, and after one complete convolution has been 
 found the curve can be completed by measuring from B, C, &c. 
 on the successive radii a constant distance BQ, CR, &c. =AP. 
 
 I THE RECIPROCAL OR HYPERBOLIC SPIRAL. 
 
 5 J ^- 
 
 -^ Def. In this curve the length of the radius vector is inversely 
 proportional to its position angle. 
 
286 
 
 RECIPEOCAL SPIRAL. 
 
 The equation to tlie curve may therefore be written 
 
 1 
 
 where r is the length of any radius vector, 6 the circular measure 
 of the angle it makes with the initial line, and a a numerical 
 constant. 
 
 When ^ = 0, r is therefore infinite, and r diminishes as 6 in- 
 creases, but the curve does not reach the pole for any finite value 
 
 of 6. Corresponding to the value 6 = 1, r = -; i.e. the radius 
 
 vector makinp; 57 '2957... deorrees with the initial line is - units 
 long. 
 
 A line parallel to the initial line and - units distant from it, is 
 an asymftote to the curve. 
 
 Problem 147. To draw the reciiwocal spiral, the pole, the 
 initial line and the unit and constant of the curve being given 
 (Fig. 150). 
 
 Let be the pole and OA the initial line. In the figure « = ^ 
 the unit being the length L. 
 
 Fig. 150. 
 
 4 Asymptote 
 
 10\ 
 
 >.; 
 
 -^ia 
 
 12 
 
 N 
 
SPIRALS. 287 
 
 Draw 04: perpendicular to OA and with as centre, and - 
 
 as radius describe a circle 4, 8, 12,... and divide it up into any 
 number of equal parts, as at 1, 2, 3... 
 
 Draw the line OB making 57*2957... degrees with OA and 
 cutting the circle in B ; B will be a point on the curve. 
 
 Determine the length of radius vector corresponding to any 
 convenient division of the circle — say the radius making 45" with 
 the initial line — i.e. determine 
 
 1 4 24 
 
 Draw the line 02, and produce it to C making (9(7 = 7*63 units. 
 C will be a point on the curve. As the angle doubles the radius 
 diminishes one half; so that if 00 is bisected in d, the length Od 
 will be the length of radius vector making a right angle with the 
 initial line, i.e. D on the line 04, OD being equal to Od, is another 
 point on the curve. 
 
 Bisect OD in e and make OB on 08 = Oe ; B will be a point 
 on the curve. 
 
 OB is also of course — ^OC. 
 
 Similarly OB the radius corresponding to ^ = 2;r is ^OB or 
 
 G the point on the curve corresponding to ^ = § • j is at a 
 
 distance § of OC from 0. OH the radius corresponding to ^ = 3 . - 
 is of course ^OG or i of OC. OK the radius corresponding to 
 
 77 . 
 
 ^ = 6 . ^ is ^OR. 
 
 OM the radius corresponding to 6 = ^ .- is |0(7, and OiV the 
 
 TT 
 
 radius corresponding to ^ = 5 . - is ^OM or ^OC. 
 
288 THE LITUUS. 
 
 In the second convolution 
 
 OP on OCj i.e. corresponding to ^ = 9 . ^ is \0C^ 
 
 OQ on OD „ 
 
 IT 
 
 OR OH OH ,, „ „ :=11.^ is3-\0C, 
 
 OS on OE „ „ „ = 12 . ^ is J^^^. 
 
 atid so on, and similarly any additional number of points can be 
 obtained. 
 
 In the figure OF bisects the angle AOG and therefore 
 OV^^.OG, 
 
 W bisects the angle AOC and W= 2 . 00. 
 
 To draw the tangent at any point p. 
 
 Draw the radius Oq of the circle described with centre and 
 
 radius - perpendicular to Op. 2^9 ^^^^^ ^® ^^^ tangent at p. 
 a 
 
 To determine the centre and radius of curvature at any point p. 
 Draw the normal j^;m perpendicular to the tangent pq and 
 
 meeting qO in m. On pq make pr = Oq = ~, and pn = mq. Then 
 
 a 
 
 ns drawn through n parallel to rm, meeting pm in 5, determines s 
 the required centre. 
 
 THE LITUUS. 
 
 In this curve the radius is inversely proportional to the square- 
 root of the angle through which it has revolved. Its equation 
 is therefore 
 
 r 
 or as it may also be written 
 
I 
 
 SPIRALS. 289 
 
 The radius therefore diminislies as the angle increases and 
 is of infinite length when ^ = : it never vanishes however large 
 may be, so that the spiral never reaches the pole, but makes an 
 injSnite series of convolutions round it. 
 
 Problem 148. To draw the Lituus, the pole, the initial line 
 and the unit and constant of the curve being given (Fig. 151). 
 
 Let be the pole, and OA the initial line. In the figure « = 1 
 
 the unit being the length Z. Draw 00 perpendicular to OA, 
 
 and determine the value of r corresponding to 6 = ^j i-®* ^^ ^ 
 
 being the circular measure of a right angle. 
 In the figure 
 
 1 _ 1 ^ 
 
 Make Oc on OA equal to this length, and make OB on AO 
 produced equal to unity on the scale adopted. A mean pro- 
 portional between OB and Oc will evidently be the required 
 length 00, i.e. a semi-circle on Be will cut 00 in 0, a point on 
 the curve. 
 
 Draw radii OG, OH bisecting the quadrants COB, BOB. 
 
 Trisect Oc in e and g, and take two parts measured from 
 as Og. A mean proportional between OB and Og will be equal 
 to the length OG at which the curve cuts the bisector OG of the 
 right angle COB. 
 
 Bisect Oc in d. A mean proportional between OB and Od 
 will give the length of the radius vector 01) corresponding to 
 
 Divide Oc into five equal parts, and take two of them from 
 as Oh. A mean proportional between OB and Oh will give 
 the length Off oi the radius vector corresponding to 
 
 ^ = T- 
 
 19 
 
290 
 
 THE LITUUS. 
 
 A mean proportional between OB and Oe (^rd of Oc) gives 
 
 q 
 
 OE the length corresponding to ^ = — . 
 
 Similarly a mean proportional between OB and fOc would 
 give OK the radius corresponding to ^= Jtt, and a mean pro- 
 
SPIRALS. 2.91 
 
 portional between OB and ^Oc would give OF corresponding to 
 6 =-- 27r, but this is more easily determined by making it equal 
 \0C, for since the square of the radius is inversely proportional 
 to the angle, the radius diminishes J as the angle increases four 
 times. 
 
 For the same reason the length OJ on HO produced will be 
 20D since the angle AOJ—^ of two right angles: 
 
 Draw the angle AOP to contain 57. 29... degrees; the arc 
 subtending this angle is equal to the radius, i.e. corresponding 
 to it, = 1, and therefore OP the corresponding radius must 
 
 contain - units (in the fiojure OP ^ 3). 
 
 2 
 Bisect the angle AOP by OQ, and make OQ^ = —^ (in the figure 
 
 ci 
 
 0^ = Vl8 = 4-24...). 
 
 (2 is a point of contrary flexure in the curve, i.e. at that point 
 it becomes convex towards the initial line, the radius of curvature 
 being infinite. 
 
 Bisect the angle AOQ by OB, and make OP = twice OP ', R 
 will be a point on the curve. 
 
 Bisect ^0^ by OS, and make 0>S' = twice OQ; S will be a 
 point on the curve. 
 
 In the second convolution the following table gives the values 
 of r corresponding to successive values of 6 difiering by 45", and 
 similarly for the third convolution. 
 
 T 
 
 If 2 be taken as the numerator of all the fractions the suc- 
 cessive denominators evidently differ by unity. 
 
 The values of r may of course all be calculated arithmetically, 
 instead of being obtained geometrically from the calculated value 
 of one of them. 
 
 19—2 
 
SO- 
 
 TANGENT TO LITUUS. 
 
 e 
 
 ,.2 
 
 
 ...I 
 
 > 
 
 i.e. r must be a mean proportional between 05 and - Oc 
 
 ...I 
 
 lo^ 
 
 loc 
 
 
 
 ^^^^i 
 
 ^.oc 
 
 i"" 
 
 Sir 
 
 > 
 
 loc 
 
 -4' 
 
 .V» 
 
 .. ^0. 
 
 2.4' 
 
 > 
 
 ^Oc 
 
 --^T 
 
 S- 
 
 B* 
 
 4rr 
 
 1- 
 
 loc 
 
 To draw the tangent at any point. 
 
 A known expression for the angle which the tangent at any 
 point makes with the radius vector is 
 
 ,^ = tan-(--y; 
 
 .-. tan<^ = --^, =-2^. 
 T'a 
 
 The value of tan <^ for any point can therefore easily be 
 calculated numerically, and the corresponding number of degrees 
 obtained from a set of tables; the angle then being plotted by 
 means of a protractor. The minus sign in tlie above expressions 
 denotes that ^ is always greater than a right angle when measured 
 on the 6 side of the radius. It becomes more and more nearly a 
 right angle as the angle increases. At the poiut Q corresponding to 
 
 The tangent may be constructed geometrically, though not 
 very conveniently, thus : — 
 
 20P' 
 
 we have 
 
 *^°*=-?^' = 
 
SPIRALS. 293 
 
 where OP is tlie radius corresponding to unit angle. Determine 
 a length I such that 
 
 OB : OP :: 20P : I, 
 
 so that l = 20P^, since 0£ is unity on the scale adopted ; 
 
 I 
 
 .'. tan d) = :, . 
 
 r 
 
 The value of r^ is known, because it is some definite fraction of 
 Ac. At G on the curve for example it is the length Og. From 
 the point at which the tangent is required measure any convenient 
 fraction of the length r^ along the radius vector, from the ex- 
 tremity draw a line perpendicular to the radius, and measure on 
 it the same fraction of the length I, and the required tangent will 
 pass through the point thus obtained. 
 
 In the figure Gm is ^Og, and m?i is ^l, then Gn is the 
 tangent. 
 
 Owing to the rapid diminution of r^ as the angle increases 
 the method very soon becomes impracticable. 
 
 If p is the radius of curvature at any point, and r the corre- 
 sponding radius vector, 
 
 p : J 4: + a'r' :: r (4 -f a'r') : 2 (4 - a'r'). 
 
 The Logarithmic or Equiangular Spiral. 
 
 In this spiral the radius increases in a geometric while the 
 angle increases in an arithmetic ratio. The angle of revolution 
 is therefore proportional to the logarithm of the length of the 
 radius vector, v/hence it derives its first name ; it is called equi- 
 angular because in it the tangent at any point makes a constant 
 angle with the radius vector. 
 
 This constant angle is called the angle of the spiral. 
 
 The equation to the curve is generally expressed in the form 
 ^ r = a\ 
 
 ^■b^here a is some constant on which the form of the curve de- 
 ^Hpends. From it evidently 
 ^F log r^eioga-, 
 
 m 
 
294 LOGARITHMIC SPIRAL. 
 
 and since the logarithm of 1 is 0, r must evidently be of unit 
 length when ^ = 0, i.e. the curve must cut the initial line at unit 
 distance from the origin. 
 
 If this condition is not fulfilled the equation to the curve is 
 of the form r — ha where h is another constant, and in this form 
 the initial line must be taken so that it cuts the curve at a distance 
 h from the origin. 
 
 The known constant value ^ of the angle which the tangent 
 at any point makes with the radius vector is 
 
 <h — tan~^ -, , 
 
 log, a 
 
 where e is the base of Napierian logarithms, i.e. ^ is an angle 
 such that 
 
 tan d> =^ 
 
 log, a 
 
 The value of log,„e is 0-43429448. 
 
 From the definition of the curve it follows that any radius 
 vector is a mean proportional between the two at equal an- 
 gular distances from it on opposite sides. This property gives 
 the best method of constructing the curve geometrically when 
 the pole and two points are given or determined. 
 
 Problem 149. To draw a logarithmic spiral, the value of the 
 constant in tJie equation, and the unit of the curve being given 
 (Fig. 152). 
 
 Let the equation be r= 1 . 15| , the unit being the length L. 
 
 Take the pole, and 0^ the initial line — ^the curve will cut 
 this line in the point M at unit distance from 0. 
 
 Suppose the revolving line to have made one complete revo- 
 lution, so that it again coincides with OA ; the corresponding 
 value of will be the circular measure of four right angles ; — 
 
 i.e. 27r or 2 (3-14159...) = 6-28318. 
 
SPIRALS. 
 
 The corresponding value of r is given by 
 
 logr=6-283181og(M5) 
 = 6-28318 X -0606978 
 = -381376 ; 
 . -. r = 2'4:l very nearly — = OiT, 
 and iY is a second point on the curve. 
 
 Fig. 152. 
 
 295 
 
 1 /V"\ 
 
 ^f^ 
 
 
 ^___^p / V 
 
 .V 
 
 
 ^>^l^ 
 
 ;\ • 
 
 A 
 
 
 yf 
 
 
 \ ' y^ > 
 
 / 
 
 
 
 / 
 
 
 Make OP on MO produced a mean proportional between OM 
 and ON, and P will be a tliird point on the curve. Through 
 draw QOp at right angles to OJ/, make OQ a mean proportional 
 between OP and ON, and Q will be a point on the curve. 
 
 Similarly if the curve cuts QO again ini?, OR : ON :: ON : OQ 
 which determines P. To do so evidently all that is necessary is 
 to draw NP parallel to PQ or perpendicular to QN, and thus a 
 series of points lying on two lines perpendicular to each other, 
 and passing through the pole can be determined. 
 
 It is of course easy to' interpolate points between those of 
 
 kthe original series ; for bisect the angle TOP by the line OS, and 
 make OS a mean proportional between OT and OP, and on SO 
 
296 TANGENT, &C. 
 
 >S^ and V will be points on the curve. 
 
 Draw OTTat right angles to OS, and make OW a. mean pro- 
 portional between OS and OV (i.e. on SV describe a semi-circle 
 cutting OF" in W), and W will be a point on the curve. Then 
 a series of points on the lines OS and W can be obtained by- 
 drawing, as shewn by the dotted lines, parallels to SW and WV 
 alternately. 
 
 The angle between any tangent and its radius vector is given 
 by the equation 
 
 _, -43429448 
 
 <^ = tan 
 i.e. tan <^ = 
 
 iog(M5) ; 
 
 43429448 
 
 •0606978 
 = 7-155, 
 whence (f>= 82" nearly (more exactly Sl^.SS'). 
 
 The tangents can therefore be drawn at all the points found 
 by drawing lines through them making this angle with the radii. 
 
 The dotted part of the curve arises from negative values of 
 the angle of rotation; it never reaches the pole. 
 
 Centre of Curvature. 
 
 The centre of curvature at any point S can easily be deter- 
 mined when the angle between the radius and tangent is known. 
 Draw the normal SC, and from the pole draw OC perpendicular 
 to OS the radius vector ; G will be the required centre. 
 
 Problem 1 50. To describe an equiangular spiral, the pole 
 and two points S and K on the curve being given (Fig. 152). 
 
 Let OS=r^, OK = r,, and the angle KOS= a. (In the figure 
 aS^=3-3, OK =2-78, smd KOS= 1-22173... = the cm. of 70".) 
 
 The angle of the spiral may be determined from the following 
 equation — 
 
 ^_ alog,„c _ 1-22173 X -43 429 _...-, g. 
 ^ ^ ~ log r, - log r," -5135139 --4440448"' ' 
 
 whence <^ = 82" nearly. 
 
SPIRALS. 297 
 
 The constant a of tlie curve is then given by 
 
 log,„„J-2?i^^=:^^.0609, 
 ^''^ tan<^ 7-124 
 
 . • . a = 1 -15 very approximately. 
 
 Taking OK as the initial line the equation to the curve may 
 
 be written r = r^a . 
 
 Draw OJ at right angles to OK and on it take a length OJ 
 It 
 
 equal to r^a^, i.e. determined from the equation 
 
 log (9/= log 7-^+^ log a 
 
 = '4440448 + 1-5707 x -0609 = 3-47. 
 We have now two points on radii at right angles to each other, 
 and other points can at once be found by the preceding problem. 
 
 Any number of j)oints on the curve can be found without 
 determining either £» or <^ by making each radius a mean propor- 
 tional between the two at equal angular distances from it. Thus 
 the radius bisecting the angle KOS must be a mean proportional 
 between OK and OS^ and the radius making an angle 2a with OK 
 must be a third proportional to OK and OS. 
 
 Points at equal angular distances can easily be found by 
 Problem 8, when the lengths of two radii separated by that angu- 
 lar distance are known. 
 
 In practice cfi should always be determined, and tangents 
 drawn at all the points found, because these tangents are of great 
 assistance in tracing the curve through the points. 
 
 Problem lol. To inscribe a Logarithmic Spiral in a given 
 parallelogram (Fig. 153). 
 
 Let A BCD be the given parallelogram, a the circular measure 
 of its acute angle. [In the figure AB = 3, AD = 4, the unit being 
 the length L, and the angle BAD contains 75", so that its circular 
 measure is 1-309...] 
 
 Let p and q be the perpendicular distances between the oppo- 
 site pairs of sides, p being greater than q. 
 
298 LOG. SPIRAL IN GIVEN PARALLELOGRAM. 
 
 In the fig. 'p = 3-86, and ^ = 2-89. 
 
 Fig.l53. 
 
 ~^-^^ 
 
 If <^ be the angle of the spiral, it can be determined from the 
 equation 
 
 tan^ 
 
 logjo-logg 
 
 or with the above dimensions 
 
 1-309 X -43429 
 
 *^^*^~ -5865873 --4608978 
 
 •5385 
 
 = 4-284, 
 
 •1257 
 .-. <^ contains 77" very nearly. 
 
 Next determine the number (iV suppose), the log. of which 
 Trloge 
 
 i.e. in the present case log JV 
 
 tan (f> ' 
 3-1416 X -43429 
 
 4-284 
 = -3185, 
 .'. from a table of logs iV-2-08. 
 
 Divide the perpendiculars p and q so that one portion shall be 
 to the other :: 1 : iV^, and lines drawn through these points of 
 
SPIRALS. 
 
 299 
 
 division parallel to the sides of the given parallelogram will inter- 
 sect in the pole of the required spiral. In the figure the per- 
 pendicular Gc is divided by making Cd on CB — unity on any 
 convenient scale, and (ie = 2*08 on the same scale, then dmn parallel 
 to ec divides Cc at the point ni in the required ratio. Similarly 
 Aa is divided in n, and nO and mO perpendicular to Aa and Cc 
 respectively intersect in the required pole. 
 
 The Involute of the Circle. The Evolute of a Curve has already 
 (p. 91) been defined as the locus of the centres of curvature, and 
 considered with respect to its Evolute the curve is called the 
 Involute of its Evolute. If an inextensible string be imagined to 
 lie in contact with the evolute and to be kept stretched while 
 gradually unwound from it, a certain fixed point on the string will 
 describe the corresponding involute. The free portion of the 
 string will be a tangent to the evolute at the point it quits it, and 
 a normal to the involute at the point reached at the moment by 
 the tracing point. 
 
 Problem 152. To draw the Involute of a given circle to pass 
 through a given point (Fig. 154). 
 
 1st. Let the given point be on the circle. Let C be the 
 centre and AB ?i diameter of the given circle, and let J. be a point 
 
 Fig. 154. 
 
 4 5 6 7 D 
 
300 INVOLUTE OF CIRCLE. 
 
 on the involute. Draw the tangent at A, and on it determine a 
 length AD equal to the circumference (see p. 267). Divide AD 
 into any convenient number of equal parts ^1, 1.2,... (fee, and the 
 circumference into the same number AV, V'2'.... Draw tangents 
 to the circle at 1', 2'... 
 
 If we imagine a string unwound from the circle starting from 
 A, — when its point of contact is 1', i.e. when the free portion of the 
 string is a tangent to the circle at 1', the length of the free portion 
 will of course be equal to the arc AV^ or to the length ^1 of the 
 straight line AD. Make V£J on the tangent at 1' equal to ^1, 
 and U will therefore be a point on the curve. Similarly make 2'F 
 on the tangent at 2' equal to A2, and i^ will be a point on the 
 curve, and so on in succession. 
 
 2nd. Let the given point be P. Through P draw a tangent 
 to the given circle meeting it in p. If A is the point where the 
 required involute through F would meet the circle and be the 
 circular measure of the angle subtended at the centre by the arc 
 
 Ap we have 6 = — j-^j— ; but the length of the arc Ap must be the 
 
 line Fp so that if the lengths Fp and AC he measured on any 
 scale the numerical value of 6 can be calculated, and the corres- 
 ponding number of degrees obtained from a table. This of course 
 determines A and the construction reduces to the first case. 
 
 As the distance from the pole increases and the points found on 
 the curve get further and further apart, others can be determined 
 between those of the original series by bisecting the corresponding 
 arcs of the circle and divisions of the straight line ADj as shewn 
 at M. 
 
 Tangents to the circle are of course normals to the involute, 
 and the centre of curvature at any point is the point of contact of. 
 the tangent drawn from that point to the circle. 
 
 The involute of the circle is the locus of the intersection of 
 tangents drawn at the points where any ordinate meets a circle 
 and the corresponding cycloid. 
 
EXAMPLES. 301 
 
 Examples on Chapter XI. 
 
 1. Draw a spiral of Archimedes to touch a given line, the 
 pole and the constant {a), and unit of the curve being given. 
 
 [If r is the length of rad. vector to the point of contact of the 
 given tangent, and p the length of the perpendicular on it from 
 the pole 
 
 -f^.x/a^^f. 
 
 Construct therefore a rectangle equal to the sum of the two 
 rectangles 
 
 ;7x|, andpx y^a-+| 
 
 (Prob. 18.) 
 
 The last expression is of course the length of the hypotenuse of a 
 right-angled triangle, the sides of which are a and -: , and is con- 
 
 sequentlv always greater than ^. The negative sign in the above 
 
 equation therefore gives an imaginary result. A mean propor- 
 tional between the sides of the rectangle constructed as above is 
 the required length n] 
 
 2. Draw a spiral of Archimedes to touch a given line PT at 
 a given point P, and to have a given pole 0. 
 
 [Through F draw Pa perpendicular to PT, and through 
 draw Oa perpendicular to OP meeting Pa in a. The length Oa is 
 the unit of, and is proportional to, the constant of the curve, and 
 the initial line is at an angle POA from OP given by 
 
 OP 
 
 circular measure of POA — ^— .1 
 
 Oa -* 
 
 3. Draw a reciprocal spiral, the pole 0, and two points P, Q 
 on the curve being given. 
 
 [Compare problem 145. Let OP = r, OQ = r^ of which let r be 
 the greater ; the angle POQ = a, and the angle between the initial 
 
302 
 
 
 
 
 
 EXAMPLES. 
 
 line 
 
 and 0P = 
 
 -e, 
 
 then 
 
 1 
 r 
 
 1 
 
 
 11 1 r-r, 
 
 = aa, or a = - . ' . 
 
 7'j r a rr^ 
 
 The value of a can be obtained from a table of the circular 
 measures of angles, and if a fourth proportional I be determined to 
 r — r^j T and r^ 1 ^ 
 
 a 
 
 which determines a. Any convenient scale can be used for measur- 
 ing I and the unit of that scale will then be the unit of the curve ; 
 
 then Q-^~ x - , the length of r beinf]f measured on the same scale. 
 a T 
 
 The initial line can then be drawn.] 
 
 4. Draw a reciprocal spiral, . the pole 0, a point P on the 
 curve and the tangent at that point being given. 
 
 [Draw OT perpendicular to OP meeting the tangent at P in 
 
 T. OT =: - , so that the constant of the curve is known. If the 
 a 
 
 circular measure of the angle between the initial line and OP is 6 
 
 \__ 1 OT 
 
 0P~ OT ' ""^ OP' 
 
 and the initial line can be drawn.] 
 
 5. Draw the Lituus, the pole and two points P and Q on 
 the curve being given. 
 
 [Let OP = r, OQ = i\^ r being greater than r^ ; the angle 
 POQ = a, and the angle between the initial line and 0P= 6. 
 
 Then - = aje, 
 
 T 
 
 ~ = a J 6 + a, 
 

 
 EXAMPLES. 
 
 
 
 
 r'e = r^'{6 + a\ 
 
 
 
 
 f\ — r, 1 
 
 
 
 r - r* ' 
 
 
 Take a fourth 
 
 proportional I to 
 
 
 
 
 rj, r + r^ and r- 
 
 ■'•i 
 
 Q 
 
 
 e^J-f, 
 
 
 303 
 
 L 
 
 then 
 
 and can be calculated, the lengths r^ and I being measured on any 
 convenient scale.] 
 
 6. Draw an equiangular spiral to touch three given lines AB, 
 BC, CA in three given points F, Q, R respectively. 
 
 [On PR as chord describe a segment of a circle containing an 
 angle equal to the external angle between the tangents AB and 
 GA. This is a locus of the pole. Similarly on PQ as chord 
 describe a segment of a circle containing an angle equal to the 
 external angle between the tangents AB and BG which will be a 
 second locus. The pole is thus determined.] 
 
 7. Draw an equiangular spiral of given angle (<^) to touch 
 three given lines AB, BG, GA. 
 
 [Suppose the spiral is to touch BA and BG produced. 
 Through B draw a line dividing the angle ABG so that the 
 perpendicular (^j>J dropped from any point on it on AB is to the 
 perpendicular {p^ dropped from the same point on BG as 1 : a* 
 where a is the constant of the required curve, and a is the circular 
 
 measure of the supplement of the angle ABG, i.e. — = a*. 
 
 P\ 
 
 a is of course the number whose logarithm is 
 
 0-43429448 x cot <^, 
 
 and can therefore be obtained from a table of logarithms. The 
 line so drawn is a locus of the pole. Similarly draw a line through 
 A dividing the angle between BA produced and AG, so that 
 
304 EXAMPLES. 
 
 n B 
 
 — = « , where q^ and q^ are perpendiculars on AB, AC respectively, 
 
 and 7> is tlie circular measure of the angle BAG. This line will 
 be a second locus of the pole which is therefore known.] 
 
 8. Draw an equiangular spiral, the pole 0, and two tangents 
 TP, TQ being given. 
 
 [Draw perpendiculars f^ , p^ on TP^ TQ from of which let p^ 
 be the greater ; then 
 
 Avhere a is the constant of the curve, a the circular measure of the 
 angle between the tangents alternate with that in which lies, 
 and </) the constant angle between the tangent and radius vector. 
 </> can therefore be determined from a table of logarithms.] 
 
CHAPTER XII. 
 
 MISCELLANEOUS CURVES. 
 
 The Harmonic Curve or Curve of Sines. 
 
 In this curve the ordinates are proportional to the sines of 
 angles which are the same fractions of four right angles as the 
 corresponding abscissae are of some given length. It is the curve 
 in which a musical string vibrates when sounded. 
 
 Problem 153. To draw the Harmonic Curve, the length and 
 amplitude of a vibration being given (Fig. 155). 
 
 Let AB hQ the given length, AO the given amplitude. With 
 centre on BA produced describe a semi-circle 4^4', and divide 
 it up into any convenient number of equal parts. Bisect -4^ in 
 (7, and divide u^ AG and CB into the same number of equal parts 
 chosen for the semi- circle. Draw the successive ordinates la, 
 lb, &c,, and from the corresponding points on the semi-circle draw 
 parallels to AB meeting the ordinates in a, b,... &c., which will 
 be points on the curve. The length from J. to (7 is half a wave 
 length which will be repeated from C to ^ on the other side of 
 AB. (7 is a point of inflection on the curve, the radius of curva- 
 ture there becoming infinite. 
 
 To draw the tangent at any point P. 
 
 Through P draw pFM parallel to AB, cutting the semi- circle^ 
 in p'j and make FM=AC. Draw pm perpendicular to OA 
 cutting it in m, and make Jfrn' on J/P = Om^ Through M draw 
 MN perpendicular to BM or AB, and on it make Mg— 3.14... 
 on any convenient scale. On J/P make Mk = unity on the same 
 E. 20 
 
306 
 
 HARMONIC CURVE. 
 
 scale, and draw mN parallel to kg cutting MN in N'. iV will be a 
 point on the tangent at P. 
 
 ^ ,- 
 
 The lines corresponding to m'N will of course be parallel for 
 all points on the curve, so that the points k and g need only be 
 found once. 
 
 A parallel to kg through the point 6 (the quadrisection of CA ) 
 cutting 4^ in T determines AT and CT, the tangents at A 
 and (7. 
 
MISCELLANEOUS CURVES. 
 
 807 
 
 Ovals of Cassini. 
 
 When a point moves in a plane so that the product of its 
 distances from two fixed points in the plane is constant, it traces 
 out one of Cassini's ovals. The fixed points are called the foci. 
 The equation of the curve is therefore rr^ = k^, where r and r^ 
 are the distances of any point on the curve from the foci and 
 ^ is a constant. 
 
 Corresponding to any given foci an infinite number of ovals 
 may of course be drawn by varying k. 
 
 Problem 154. To describe an oval of Cassini^ the foci F and 
 F^ and the constant k of the curve being given (Fig. 156). 
 
 Draw a line . through F and F^ and bisect FF^ in C : through 
 C draw BCB^ perpendicular to FF^, and with F as centre and 
 
 radius = k describe an arc cutting BCB^ in B and B^ . B and B 
 will evidently be points on the curve. 
 
 Draw FK perpendicular to FF^ and make FK=k, and on 
 CF make CA and CA^ each= CK. A and A^ will be points on the 
 
 20—2 
 
o08 CASSINl'S OVALS. 
 
 curve, for CA' = CK' = CF' + FK\ 
 
 .'. CA' - CF' = k' = (CA + CF)(CA - CF); 
 but CA + CF= F^A and CA-CF = FA, 
 
 .-. FA.F^A^k'. 
 
 With centre F and any radius greater than FA and less than 
 FA^ describe an arc dJ) cutting FA in d. Through K draw Kd^ 
 perpendicular to dK and cutting FF^ in d^. A circle described 
 with centre F^ and radius Fd^ will cut the arc dD in D, a point 
 on the curve. 
 
 Evidently by symmetry D^, the intersection of arcs of the 
 same radii as the above but struck from the opposite foci as 
 centres, will also be on the curve, and so also will be the inter- 
 sections on the other side of AA^. Similarly any number of 
 points may be found. 
 
 An alternative method may be adopted as soon as two points 
 such as A and Z>, not very far apart, and the two corresponding 
 points A^ and D^ are found. If two series of terms in geo- 
 metrical progression are found, FA and FD being successive terms 
 of the one and FA^ and FD^ successive terms of the other 
 (Problem 8), circles struck with the corresponding terms of 
 each as radii and with the opposite foci as centres intersect in 
 points of the curve, the radii increasing from the one focus and 
 diminishing from the other. This is shewn in the figure, and this 
 construction moreover enables at once any number of ovals to be 
 drawn, the intersection of any two circles of opposite series 
 being taken as a starting point, and the successive intersections 
 giving succeeding points. The second curve drawn in the figure 
 is an example of this. 
 
 It may be noticed that a circular arc with centre at tho 
 focus coincides very closely with the oval at the vertices A 
 and A^. 
 
 To draw the tariff ent at any j^oint P. 
 
 The angle FPG which the normal at any point P makes with 
 the focal chord FP is equal to the angle which the other focal 
 
MISCELLANEOUS CURVES. 309 
 
 chord F^P makes with the chord CP drawn from P to the 
 centre. 
 
 TJie Cissoid of Diodes. 
 
 This curve, named after Diodes, a Greek mathematician, 
 who is supposed to have lived about the sixth century of our 
 era, was invented by him for the purpose of constructing the 
 solution of the problem of finding two mean proportionals. The 
 curve is generated in the following manner : — 
 
 kin the diameter ACB of the circle AD BE (fig. 157) make 
 AN=BM, and draw MQ and NE perpendicular to AB, and let 
 MQ meet the circle in Q^ then AQ and NR intersect in a point on 
 the curve, i.e. the locus of this intersection is the Cissoid. 
 ^ ..,,., liN QM J AM. MB . , ..^ . 
 By similar triangles -^ = -^ = ^^ , since AQB is a 
 
 right angle; or if we call RN'=y, AJ}^ = x, and the radius of the 
 circle a, 
 
 y_ J{2a-x)x 
 
 X '2a — X 
 
 V 2a -x' 
 
 which is the equation to the curve referred to rectangular axes 
 with A as origin and AB as axis of x. 
 
 "t3* 
 
 Problem 155. To describe the Cissoid corresponding to a 
 circle of given diameter (Fig. 157). 
 
 Of course the above description is really a construction for 
 the curve, since by it any number of points can be determined. 
 The curve may also be described by continuous motion thus : 
 
 Draw a diameter AB of the circle, and the tangent at B. 
 If ^ is a point on the curve, this tangent will be an asymptote. 
 Through C, the centre of the circle, draw a parallel to the tangent 
 at B of indefinite length, and make ^0 on CA produced equal to 
 A C. Cut a piece of paper to a right angle as abc, and on one side 
 of it mark ofi* from the angle the points d, c, making bd = dc=AC\ 
 the radius of the given circle. If the paper be now placed 
 so that the edge ba passes through 0, and the point c is always 
 on BCD, the point d will be on the curve, and by moving it 
 
810 
 
 THE CISSOID. 
 
 the positions of any number of points can easily be marked off 
 on the paper. The curve is evidently symmetrical about AB, 
 
 Fig. 157. 
 
 there is a cusp at A, and D and E, the extremities of the diameter 
 perpendicular to AB, are points on the curve. 
 
 To draw the tangent at any point P. 
 
 From P, with radius AC, mark off L on the diameter ECD. 
 Through L draw LG parallel to AB, and through draw OG 
 parallel to PL, meeting LG in G, G will be a point on the 
 normal at P, and the tangent is therefore perpendicular to PG. 
 
 It may be noted that the area included between the curve 
 and the asymptote is three times the area of the generating 
 circle. 
 
 The problem of finding two mean proportionals between two 
 given quantities a and h is, to find two quantities m and n 
 such that 
 
 m^ = an and n^ = mb, 
 
 or that m^ = a^b and n'' = ah^. 
 
 By means of the cissoid corresponding to the circle, the radius 
 of which is equal to a, the smaller of the given quantities a and h, 
 the first term m can easily be found thus : 
 
MISCELLANEOUS CURVES. 811 
 
 Make CS on the diameter DCE=h. By hypothesis S will 
 
 always fall beyond E. Draw BS cutting the cissoid in K. Then 
 
 AK will cut CS in a point T at sl distance from C equal to the 
 
 required quantity m, i.e. CT^ = m^ = a%. For draw the ordinate 
 
 Kn. By similar triangles 
 
 CT AG ^„3 Kn^ 3 
 ^- = -^ , or CT^= -r—^ a% 
 Kn An An^ 
 
 CS BC BC 
 and 
 
 Kn Bn 2a-An' 
 
 ... CS=^ a; 
 2a— An 
 
 but An is the x and Kn is the y of the point Z, and it has been 
 therefore already proved that 
 
 ... ^r3.-^.^3«' 
 
 'la -An An\ 
 ^'"^ a^^CSa^ 
 
 2a -An 
 = a'h. 
 When m is found the second mean proportional n can be found 
 by similar triangles, for 
 
 a '. m :: n : h. 
 
 If CS or h be made equal to 2a, m will be the length of the 
 side of a cube^ the volume of which is twice that of a cube of side 
 a, since in this case m^= 2a^. 
 
 The ConcJioid of Nicomedes. 
 
 If through a fixed point a straight line POp be drawn 
 meeting a fixed right line LM in R, and RP^ Rp be taken each 
 of the same constant length, the locus of P and p is called the 
 conchoid. 
 
 If CD be drawn perpendicular to LM meeting it in A, and 
 OA = a, RP=b, and AOR^O, 
 
 OP=OR+RP=-^+h. 
 cos 
 
312 
 
 THE CONCHOID. 
 
 Also Op = OR-Bp, since we go in the positive direction from 
 to R, and in the negative from R to p; 
 
 .: Op 
 
 i, 
 
 SO that the polar equation of the curve, being the pole and AD 
 the initial line, will be 
 
 (r ± h) cos = a. 
 
 Problem 156. To draw the GoncJioid, the constants a and h 
 being given (Fig. 158.) 
 
 Draw the line OD^ and make OA on it =a, and AD, Ad each 
 = ^• 
 
 FIg.lBS. 
 
 D 
 
 Through A draw LAM perpendicular to OA ; LM will be an 
 asymptote of the curve. Draw any line OP through meeting 
 LM in Rf and on it make RP = Rp = b. 
 
 By definition P and p will be points on the curve, and 
 similarly any additional number of points may be determined. 
 
 The curve is evidently symmetrical about OD. 
 
 If b is less than a, the form of the curve is that shewn by the 
 dotted lines. 
 
 When b- a the point is a cusp on the curve. 
 
 To draw the nor7nal at any point Q, 
 
 Let OQ meet LM in r; draw rG perpendicular to LM and 
 OG perpendicular to OQ intersecting in G, which will be a point 
 on the required normal ; for the line OQ is moving so that it 
 always passes through while a fixed point on it is travelling along 
 
MISCELLANEOUS CUKVES. 31-5 
 
 LM j i.e. at the moment the line is moving along OQ (or turning 
 about some point on OG), and also along LM (or turning about 
 some point in tG\ i.e. G is the centre of instantaneous rotation. 
 
 The Witch of Agnesi. 
 
 Let AB (fig. 159) be a diameter of a circle, iO/a line per- 
 pendicular to AB meeting it in iV and the circle in 3f. If F be 
 taken on iOf produced so that 
 
 An ~ AN' 
 
 the locus of the point P is tlie curve called the Witch. 
 If a be the radius of the circle we have from the above 
 PN^ _MN' BF 2a -AN 
 4a' ~~AN' ~ AN-~ AN ' 
 
 or putting AN=x, and PN=y^ 
 
 xy^ = ia^ (2a - x), 
 
 which is the equation to the curve referred to rectangular axes 
 with A as origin and AB axis of x. 
 
 Problem 157. To describe the Witch of Agnesi corresponding 
 to a circle of given diameter (Fig. 159). 
 
 Let AB be the given diameter, C its centre ; draw the tangent 
 at B, and through A draw any number of lines AE, AF,...&.c., 
 cutting the circle in F, F, &c., and the tangent at B in 6,/,... 
 &c. Lines drawn through F and e respectively parallel and per- 
 pendicular to the tangent will intersect in Q, a point on the curve ; 
 similarly lines through F and /intersect in Pj and so any number 
 of points can be determined. 
 
 The construction is obvious from the definition of the curve. 
 
 The curve is symmetrical about ^^ aud cuts the diameter per- 
 pendicular to AB at distances from the centre equal to the diame- 
 ter ; the tangents at these points pass through B. 
 
 If CB be bisected in B and BK be drawn perpendicular to ^^ 
 meeting the curve in X, K is a, point of inflection on the curve. 
 The tangent to the circle at A is an asymptote to the curve. 
 
314 
 
 WITCH OF AGNESI. 
 
 To draio the tangent at any point T. 
 
 Through T draw tTv parallel \>o AB meeting the tangent at B 
 in t and the asymptote in v. Draw Aw perpendicular to At meet- 
 
 ing the ordinate tlirough C, the centre of the circle in w. The 
 tangent at T is parallel to vw. 
 
MISCELLANEOUS CURVES. 315 
 
 THE CATENARY. 
 
 The curve in which a heavy inextensible string, freely sus- 
 pended from two points, hangs under the action of gravity, is 
 called the Catenary. If the mass of a unit length of the string is 
 everywhere constant, i.e. if the string is of uniform density and 
 thickness, the curve in which the string hangs is called the 
 Common Catenary. 
 
 Investigation of the conditions of the statical equilibrium of 
 the string gives for the curve of the common catenary the well- 
 known equation 
 
 X a-. 
 
 2' = |{«' + ^"1' 
 
 tlie axis of y being a vertical line through the lowest point of the 
 curve, and the axis of a; a horizontal line in the plane of the string 
 at a distance c below the lowest point, c is the length of string, 
 the weight of which measures the tension at the lowest point, and 
 e is the base of Napierian logarithms. 
 
 At a distance c from the origin measured along the axis of a;, 
 the corresponding value of 2/ is 
 
 at a distance 2 c it is 
 
 Ik+«-'i. 
 
 tK+^-^ 
 
 and so on ; and if we make c the unit of length the corresponding 
 values of y are \ {e^ + 6~^}, 
 
 and so on. 
 
 The third column of the following table gives the value of - 
 
 at the corresponding points along the axis of x as shewn by the 
 first column 
 
 e = 2-718281828... lo^JO = -43429448... 
 
 I. 
 
316 
 
 THE CATENARY. 
 
 Abscissae J{e'^+e ^} 
 
 x=j 1(1-28405 + -77880) 
 
 y 
 
 c 
 1-03142 
 
 x=^- 
 
 1(1-6487 + -60653) 
 
 1-1276 
 
 3c 
 
 i(2-J17 + -47144) 
 
 1-294422 
 
 x=c 
 
 1(2-71828 + -36788) 
 
 1-54308 
 
 x^lc 
 
 1(7-389 + -13534) 
 
 3-76217 
 
 x=-.Zc 
 
 1(20-0855 + -049787) 
 
 10-0676... 
 
 a;= 4c 
 
 1(54-598 + -018316) 
 
 27-308... 
 
 Problem 158. To draw the common catenary, the unit c being 
 given. 
 
 Example 1. {c=OA) fig. 160. 
 
 Draw the horizontal line Ox and the vertical line Oy. On Oy 
 measure OA =c. A will be the lowest point of the curve. Set off 
 from along Ox lengths Oa = ah — hd = c, and draw the ordinates 
 through a, h, d... parallel to Oy. 
 
 On the ordinate through a measure from a a length ap^ ~ (the 
 number in third column of above table opposite aj = c) x c, i.e. 
 1-54308 X c (e.g. if c is Y ^^ ^^ ^^Y necessary to measure off on a 
 diagonal scale of half inches a length 1-54). p^ will be a point on 
 the curve. Similarly on the ordinate through h measure hp^ = 
 (number in column 3 opposite a; = 2c) x c, ie. 3-76217 x c. p^ will 
 be a point on the curve. Similarly for ordinate through d. 
 
 Points can of course be found between A and p^ by using the 
 fractions of c given in the table. 
 
 Example 2. {c=OA) fig. 161. 
 
 The points p^^ p^, p^, p^ on the ordinates through a, h, d, e, 
 where Oa = ab = bd=de = |c, 
 
 are given by the table : the next point furnished by the table 
 would be on the ordinate through /, where e/= Oe. Points on 
 
MISCELLANEOUS CURVES. 
 
 817 
 
 ordinates between e and f may be found without calculation as 
 follows : 
 
 d 3C 
 
 Any three equidistant ordinates (2/„_j, 2/„, Vn^^ ^re connected 
 by the relation 
 
 where h is some constant, i.e. if eg^^de 
 
 e^' + F 
 
 9P, 
 
 dpz 
 
318 THE CATENARY. 
 
 Construct the right-angled triangle AOm, with hypotenuse 
 Am ^ api, the ordinate at distance Oa — de = eg from origin : the 
 length Om is the value of the constant h. 
 
 li p^q^ be drawn parallel to Ox and meeting 0^ in j^, 
 
 so that the required length gp^ can be determined by taking a 
 third proportional to dp^ and mq^. 
 
 Similarly, if gh = eg= Oa, 
 
 ep^ '. mq^w mq^ : Jip^i 
 or, since eh = he, hp^ may be determined from 
 hp^ : mj^ :: m,q^ : hp^, 
 where m^ is a point on Ox such that Am^ = bp^. 
 
 To draw the tangent at any point (p^ say). 
 
 With centre and radius OA describe a circle; through p. 
 draw p^q^ parallel to Ox and meeting Oy in q^. The tangent at ])^ 
 will be parallel to one of the tangents which can be drawn from q^ 
 to the above circle. 
 
 From g^ the foot of the ordinate at p^, draw gt perpendicular to 
 the tangent at p^ meeting it in t. gt= OA, the c of the curve, and 
 p^t is the length of the arc of the curve between p^ and the lowest 
 point, i.e. p^t = arc Ap^. 
 
 To determine the centre and radius of curvature at any point 
 (asp,). 
 
 Draw the normal at p^ meeting the horizontal axis Ox in G. 
 On the normal make p^S=pfi. /S' will be the required centre, 
 and Sp^ the radius of curvature. 
 
 Problem 159. To draw a catenary, the vertex A, the axis Ay 
 and a point Q being given (Fig. 161). 
 
 The following method is approximate only, but gives tolerably 
 close results provided the depth of A below Q does not exceed 
 two-thirds of the distance of Q from Ay. 
 
 Find on Ay the centre (F) of the circle passing through A and 
 Q, and determine the length of the circular arc A Qj i.e. from a 
 
MISCELLANEOUS CURVES. 
 
 319 
 
 table of the circular measure of angles get the circular measure 
 corresponding to the number of degrees in the angle AFQ and 
 
 multiply this number by the length FA measured on any con- 
 venient scale. [In the figure AFQ contains 64", the circular mea- 
 sure of which is 1*117, and FA =5, the unit being J inch; the 
 length of the circular arc ^^ is therefore 5*585 units.] From Q 
 set off downwards on a parallel to Ai/ drawn through Q the length 
 QL = the circular arc AQ as above determined, and let the horizon- 
 tal through A meet QL in k, and make XiV on Q produced through 
 L a third proportional to twice Qk and kL ; i.e. take 
 
 Zi\^ : JcL :; kL : 2 . Qk. 
 
 ]V will be a point on the axis Ox of the required catenary, i.e. 
 c is determined for the required curve. 
 
 [iV is easily determined by inflecting from L to Ak produced a 
 length Lk^ = twice Qk; produce k^L to n making Ln = kL. 
 
S20 THE CATENARY. 
 
 n will be a point on the required axis of Ox; for by the 
 similar triangles Lkk^ , LNn, 
 
 LN : Lk :: Ln : Lk^ :: Lk : 2 . Qk] 
 
 The construction is based on the assumption that the length of 
 the arc of a catenary near the vertex does not sensibly differ from 
 the circular arc passing through its centre and extremities j and 
 the point N is determined so that the tangent from it to a circle 
 with centre Q and radius QL shall be equal to Nk. 
 
 Problem 1 60. To draw a catenary, a point of suspension P, 
 the tangent FT at that pointy and the depth PK of the loop being 
 given (Fig. 161). 
 
 Draw the horizontal through K meeting PT in R. 
 
 On PR produced make RT = RK, and draw TI^^ perpendicular 
 to PT meeting PK in N^. 
 
 KN^ = the unit c for the required curve. PT is the length of 
 the arc between P and the lowest point, and a known expression 
 for its length is 
 
 where x=^AI{. Also 
 
 P^,=-^{ehe% 
 PN.+PT ^ 
 
 or - log e = log PJ^^ + PT- log c, 
 
 1.8. X 
 
 which determines the vertex^. 
 
 '( loge j' 
 
 Problem 161. To draw a catenary, tJie axis Oy, a point P 
 on the curve, and the tangent PT being given (Fig. 161). 
 
 Through P draw PN^ parallel to Oy, and PM perpendicular 
 to Oy meeting it in M. Let the angle TPN^ = 6, and if PT is 
 
MISCELLANEOUS CURVES. 321 
 
 = length of arc between P and A the vertex, we have if TN^ is 
 perpendicular to PT, 
 
 PT-PiT,COS^=||€''-€~''|, 
 
 and PiV, = ||€<=+€ « |, 
 
 PAT 
 
 .'. PiYj (1 + COS ^) = c.cS 
 
 but c = :riV^ = PiV>in^, 
 
 , e ?^ ^^ J 
 .'. cos^=sin^€'' or € ** =cot^, 
 
 PM. , J 
 
 ••• -y-l0g€ = l0gC0t^, 
 
 r_iog€_ 
 
 jlogcot- 
 
 By means of a table of logarithms, the value of c can be 
 calculated, and when the length N^T is known, the points N^ and 
 T are of course easily determined. 
 
 THE TRACTORY OR ANTI-FRICTION CURVK 
 
 The involute of the Catenary is called the Tractrix or Tractory. 
 Since in the catenary (fig. 161) gt drawn from the foot of the 
 ordinate at any point P, perpendicular to the tangent at P, meets 
 it in a point t such that Pt = arc of catenary measured from the 
 lowest point, t is evidently a point on the involute of the catenary 
 and tg is a tangent to the involute. Also tg is constant (p. 318) 
 and equal to OA, and therefore the Tractory is a curve such that 
 the intercept on its tangent between the point of contact and a 
 fixed right line is constant. This fixed length is called the 
 constant of the curve. 
 
 E, 21 
 
822 THE TRACTOEY. 
 
 The equation of the tractory may be written 
 
 + X + Jf -'if =^0, 
 
 tlog 
 
 y 
 
 where OA (fig. 162) is the axis of y^ ON the axis of x and OA = t 
 the constant of the curve. 
 
 Peoblem 162. To draw a Tractory the constant t being given 
 (Fig. 162). 
 
 Describe the catenary corresponding to the unit t == OA 
 (Problem 158). 
 
 -^«is 
 
 Flg.l62. 
 
 / 
 
 .^/ 
 
 In the figure since ON = OA, QN the ordinate of the catenary 
 r543... X OAj and so for other points. 
 
MISCELLANEOUS CURVES. 323 
 
 Draw QP the tangent at Q (p. 318) and NP perpendicular to 
 QP and therefore parallel to Op. P is a point on the tractrix as 
 akeady shewn, and similarly other points can be determined. 
 
 The centre of curvature at P is of course the point Q. 
 
 The line ON is an asymptote to the curve, and by the revolu- 
 tion of the curve round ON a solid is generated, the form of 
 which has been adopted for the foot of a vertical shaft working 
 in a socket or step. This pivot is known as Schiele's Anti-Friction 
 Pivot. The theoretical advantage of the adoption of the form in 
 this case is that the vertical wear of the pivot and step is every- 
 where equal. 
 
 INVERSE CURVES. 
 
 Def. If on any radius vector OP drawn from a fixed origin 
 0, a point P" be taken such that the rectangle OP . OP' is con- 
 stant, the point F is called the inverse of the point P; and if P 
 describe any curve, P describes another curve called the inverse 
 of the former, with respect to the pole 0. 
 
 Let be the pole and P, Q two points on any curve, and let 
 Pj, Q^ be the inverse points, then by definition 
 
 OP . OP^ = OQ . OQ^ = ¥ suppose. 
 
 A circle can therefore be described round PQQ^P^ and hence 
 the triangles OQP and OP^Q^ are equiangular. (Euc. in. 22.) 
 PQ ^OP ^ OP .OQ ^ OP.OQ 
 ''' 'P,Q\~OQ,~OQ.OQ,~ F • 
 
 Since the angle OQ^P^ = the angle OPQ, it follows that when 
 Q moves up to and coincides with P so that PQ becomes the 
 tangent at P, ^i moves up to and coincides with P^ and Q^P^ 
 becomes the tangent at P^, and the angle OP^T^ between OP^ and 
 Q^P^ produced is equal to the angle OPQ^ so that the tangents to 
 a curve and its inverse at corresponding points make equal angles 
 with the radius vector but on opposite sides of it. 
 
 21—2 
 
324 THE LIMAgON. 
 
 The Limagon, The inverse of an ellipse or hyperbola with 
 respect to a focus is called a limagon. The polar equation to an 
 ellipse or hyperbola, the focus being the pole and the major axis 
 
 the initial line, is r = — . — — where a and h are the maior and 
 
 a 1 + ecos^' '^ 
 
 minor axes of the ellipse or the transverse and conjugate axes of the 
 
 hyperbola, and e is the eccentricity of the curve (pp.99 and 154). 
 
 If r be produced to a length r' such that rr' = k^ (Def. p. 323), 
 the above equation becomes 
 
 -T = - T-. a or r = -y^ (1 + e COS 6), 
 
 which is of the form / = ^ cos ^ + ^ the equation to the Lima^on. 
 
 
 the positive sign being taken for an hyperbola, negative for an 
 
 ak" A 
 
 ellipse, and B=-j^ so that -^ = e the eccentricity of the conic. 
 
 Hence the constant for the Inverse being given, the values of 
 A and B for the limagon corresponding to any particular conic 
 can be calculated — and conversely the equation to the Limagon 
 being given, and also the constant kj the particular conic of which 
 it is the inverse may be determined by solving the above two 
 equations for a and b. 
 
 Evidently A is less than B in the inverse of the ellipse, and 
 greater in the inverse of the hyperbola. 
 
 Problem 163. To describe a Limaron, the equation to tJie 
 curve being given (Fig. 163). 
 
 Let the given equation he r = A cos 6 + B. 
 
 Draw a circle of diameter OD = A^ and on DO set off from D 
 on each side of D lengths DM^ Din each equal to B. M and m 
 are evidently the points corresponding to the values of ^, zero and 
 180", being the pole; i.e. OD must be the initial line. 
 
MISCELLANEOUS CURVES. 325 
 
 Through draw any line whatever cutting the circle in Q. 
 
 Flg.l63 
 
 I 
 
 On it from Q on each side of Q set off lengths QPj Q}^ each equal 
 to B. P and p will be points on the curve ; 
 
 for OP = OQ + QF = OI)cobDOQ + QP 
 
 = A cos 6 + B, 
 
 and Op=QP-OQ = QP- OD cos DOQ 
 
 = QP+ OB con (180 + DOQ) 
 = A cos + £, 
 jhe 6 in this case of course corresponding to the radius 0/?. 
 
326 NORMAL TO LIMAgON. 
 
 Similarly, by drawing a series of lines through and setting 
 ojff on them from the points where they cut the circle, the constant 
 length B any number of points can be determined. 
 
 In the figure the outer curve with plain letters is the inverse 
 of an ellipse, and the inner one with suffixed letters the inverse 
 of an hyperbola. 
 
 The values of the constants are ^ = 2-1, 
 
 B for the outer curve = 2*4, 
 
 B „ inner „ = -84, the unit being the length I. 
 
 Hence corresponding to the value F = 1-7 we have 
 
 and 2-4 = -Y2-«^, 
 
 whence a=3 and 6 = 1-46 the semi-axes of the ellipse of which 
 the figure is the inverse ; and corresponding to the value ^^ = 9 
 we have 
 
 2-i = |7ci^TF, 
 
 whence a = 2*0 3, 
 
 6 = 4-66, 
 the semi-axes of the hyperbola of which the inner curve is the 
 inverse. 
 
 To draw the normal at any 'point B of a Limagon. 
 
 Through D draw DG parallel to OP, meeting the circle on 0Z> 
 as diameter again in G^ which will be a point on the required 
 normal. 
 
 To find the centre of curvature at any point P. 
 On OP as diameter describe a semicircle, and draw Q V perpen- 
 dicular to OP meeting it in V. On PGj the normal at P, make 
 
MISCELLANEOUS CURVES. 327 
 
 Pv = PV and draw vX parallel to GV meeting PT in X. On 
 P V make Pg = 2 . PG and draw Gx through G parallel to Pg and 
 = PX. gx will intersect PG in s, the other extremity of the 
 diameter of curvature at P, so that S the required centre is the 
 point of bisection of Ps. 
 
 Proof. It is easily shewn analytically that if p is the radius 
 of curvature at P, 
 
 ^~ 2A' + SAB cos e + B' ' 
 
 where A and B are the constants of the curve and 6 is the 
 angle DOP. 
 
 But A' + 2AB cos 6 + B' = PG', 
 
 and . •. A'+ AB cos e = PG'- (B' + AB cos 6), 
 
 ^ PG^ 
 
 **• f'~2.PG'-(B' + ABcose)' 
 
 But QV'^OQ . QP^ABcos 6, 
 
 and PV = PQ' + QV' = B' + AB cos 0, 
 
 PG^ 
 *'• ^~2PG'-Pr' 
 
 By construction PX : PV :: Pv : PG, 
 
 and Pv^PVy .'. PX.PG = PV\ 
 
 •*• ^~2PG-PX' 
 
 i.e. 2p : PG :: 2PG : 2PG-PX, 
 
 or 2p:2p-PG::2PG:PX, 
 
 but Ps :Gs::Pg: GX, 
 
 le. Ps:Ps-PG::2PG:PX, 
 
 .'. Ps=2p. 
 
 The limagon is an epi-trochoid, the diameters of the directing 
 and rolling circles being equal. 
 
328 THE CARDIOID. 
 
 The Inverse of a Parabola is called a Cardioid, i.e. a Cardioid 
 is a Limagon in the equation of which the constants A and B are 
 equal. 
 
 Its equation is therefore r = ,4 (1 + cos ^). 
 
 The inner loop disappears in this case, and the origin is a 
 cusp on the curve. 
 
 Problem 164. To describe a Cardioid, the equation to the 
 curve being given (Fig. 164). 
 
 Let the given equation be 
 
 r = ^ (1 + cos 6). 
 
 Fig.i64. 
 
 Draw a circle of diameter OD = A and on OD produced set off 
 D3f=A. 
 
 M is evidently the point on the curve corresponding to zero 
 value of 6, being the pole; i.e. OD must be the initial line. 
 
 Through draw any line whatever cutting the circle in (?, 
 and on OQ produced make QP = OD = A. P will be a point on 
 
L 
 
 MISCELLANEOUS CURVES. 829 
 
 the curve, for 
 
 0F=0Q + QP = ODcosD0Q+QP = A (l+cosO). 
 
 Similarly, any number of points on the curve can be ob- 
 tained. 
 
 To draw the noi'mal at any point P. 
 
 Through D draw DG parallel to OP meeting the circle again 
 in G. (? is a point on the required normal. 
 
 THE LEMNISCATE OF BERNOULLI. 
 
 The inverse curve of the Rectangular Hyperbola with respect 
 to its centre is called a Lemniscate. 
 
 The polar equation to the rectangular hyperbola, the centre 
 being the pole, and one of the axes the initial line, is 
 7^cos2e = a\ 
 
 If any radius vector OP, being the centre, is produced to 
 P so that OP . OP' = k^, where k is any constant, P will by 
 definition be a point on the inverse. 
 
 If OP = r, OP' =. r', this may be written 
 
 TV' = k' or r'^ = -3COs2^; 
 a 
 
 the polar equation to the lemniscate may therefore be written 
 
 The lemniscate is a particular case of the ovals of Cassini, the 
 distance between the foci being J2K and the product of the focal 
 
 distances of any point of the curve being -^ , 
 
 Problem 165. To describe a lemniscate, the constant of the 
 curve being given (Fig. 165). 
 
 Draw any two lines OB, Ob at right angles to each other. 
 On OB make OA - 0^, = the constant K of the curve. A and A^ 
 
330 THE LEMNISCATE. 
 
 are evidently points on the curve corresponding to tlio values of 
 
 6, zero and 180". On Ob make Oa= OA, and with as centre, 
 and ^a as radius, describe a quadrant of a circle Bh. 
 
 Draw any line OD through meeting the circle in D, and 
 draw DN perpendicular to OA meeting it in N. With A as 
 centre, and ON as radius describe an arc cutting Oh in p, and 
 make OP, OP^ on OD each = Op. P and P^ will be points on the 
 curve. 
 
 Similarly any additional number can be determined. 
 
 The curve passes through the origin for r = when 26 = 90*', 
 and lines drawn through making 45" with OA (the initial line) 
 are tangents to the curve at 0. 
 
 Proof, The equation to the curve may be written 
 r' =K' (2 cos^ ^ - 1) or "^f = cos^ 0, 
 
 but 
 
 cos' DON: 
 
 oM 
 
 2K' 
 
 o^w 
 
 0D\ 
 
 .-. t''=6n~\-K\ 
 
 which by construction it does since 
 
 AP^ON and OA = K, 
 
MISCELLANEOUS CURVES. 
 
 831 
 
 Between the values 90" and 270" for 20^ cos 20 is negative, 
 and consequently no real values for r exist. 
 
 The length OQ corresponding to an angle AOQ = 30'^ is ^OB, 
 and the tangent at ^ is parallel to OB. 
 
 To draw the tangent and normal at any point. 
 
 The angle OFG between the radius vector OP and the normal 
 PG is twice the angle POA. Considered as one of Cassini's ovals 
 the foci are at F and F^ where OF = OF^ = \0B^ and the normal 
 may of course be dra\^n in the manner given for those curves, 
 i.e. by making the angle F^PG ^ angle OPF. 
 
 Problem 1G6. Given two points A and 0, and a line OB 
 through one of them, to determine the locus of a point P moving so 
 that the angles which OP makes with PA and with a parallel to 
 OB through P, shall he equal (Fig. 166). 
 
 On OA as diameter describe a circle, and through draw a 
 perpendicular to OB. 
 
332 INCIDENT AND EEFLECTED RAYS. 
 
 With as centre and any radius less than OA, describe a 
 circle cutting the circle on OA in a and a^ , and the perpendicular 
 through in 6 and b^. 
 
 Draw Aa meeting parallels to OB through h and Sj in F 
 and Pj, and draw Aa^ meeting the same parallels in Q and Q^ . 
 F, Q, F^ and Q^ will be points on the required locus, for the 
 triangles ObQ^ Oa^Q, e.g. are equal in all respects. 
 
 Similarly any additional number of points can be determined 
 as shewn. 
 
 The curve extends to an infinite distance on both sides of 0, 
 and has an asymptote parallel to OB on the opposite side to A 
 and at the same distance from OB as -4 ; or if AI^ be drawn 
 perpendicular to OB and NX on it be made equal to AN, the 
 asymptote passes through X. 
 
 The internal and external bisectors of the angle AOB are 
 tangents at to the two branches of the curve passing through 
 that point. The tangent at A is inclined to OA at an angle OAT 
 = angle AOBy and parallels to OB at distances from it = 0A are 
 tangents to the curve. The points of contact L and M of these 
 last are determined by drawing LAM perpendicular to OA, 
 At some point beyond 3£ the curve becomes convex to the 
 asymptote. 
 
 This problem is a solution of the question : — to find the point 
 on a spherical mirror, on which a ray from any point A must 
 impinge in order that it may be reflected parallel to a given 
 direction. 
 
 For if be the centre of the mirror, the circular arc repre- 
 senting the section of the mirror by the plane passing through 
 Aj 0, and the line OB through parallel to the given direction, 
 will of course cut the curve in points such that the incident and 
 reflected rays make equal angles with the normals at those points. 
 In other words the problem is to find the point F on a given circle 
 at which the lines AF, FB, A being a given point and FB being 
 parallel to a given line make equal angles with the normal at F, 
 
 The whole curve in such a case need not be drawn, since it 
 
MISCELLANEOUS CURVES. 
 
 333 
 
 is easy to find points on the curve in the neighbourhood of the 
 part of the mirror required and to draw an arc of the curve 
 through them. 
 
 Problem 167. Given three points A, i>, (7, to determine the 
 locus of a poi7it P moving so that the angles which PC makes ivith 
 PA and PB are always equal (Fig. 167). 
 
 Let ^C be greater than BG. On AC and BC &h diameters 
 describe circles, and with centre C and any radius not greater than 
 
 J^ Fig. 167. 
 
 BC describe an arc cutting the circle on AG in a and a^, and the 
 circle on BG in b and h^. The lines Aa, Aa^ will intersect both 
 the lines Bb and Bb^ in points on the required locus. Only three 
 of the intersections are shewn in the figure, viz. the points /*, Q 
 and Pj the fourth not falling within the limits of the paper. 
 Similarly any additional number of points can be determined as 
 shewn. 
 
 The curve extends to an infinite distance on both sides of the 
 line AB, and has an asymptote parallel to the line joining C to the 
 centre point of AB, and which cuts AB between A and D the 
 foot of the perpendicular from C on AB at a distance DE from 
 D, which may be thus determined. 
 
334 INCIDENT AND REFLECTED RAYS. 
 
 Let BC = a, AC=h, AD = m, BD = n and CD^h. 
 It can be shewn analytically that the length 
 d^ + h^ 
 
 'm-n\'-\-2h\' 
 On DA m2i'keI)F=DJ3, therefore AF=m-n. 
 Draw FG perpendicular to A£ meeting BC in 6^, so that 
 FG = 2I)G = 2h; 
 
 .-. AGi' = AF\' + FG\' = 7n-ni' + 2h\\ 
 
 I Draw GK perpendicular to ^ 6^ meeting AB in K, so that by 
 similar triangles 
 
 AF : AG :: AG : AK ; 
 
 .-. AG[' = AF.AK. 
 
 In the figure K is beyond the limits of the paper, but if ^ 6^ is 
 bisected in g, and gk is drawn perpendicular to AG meeting AB 
 ink, Ak = lAK and therefore AG\'' = 2.AF. Ak 
 
 The above expression for DF therefore becomes 
 
 Draw CL perpendicular to AC and make CL = CB so that 
 Ar = a' + b\ 
 
 On AB make Al=AL, and through I draw O/ parallel to 
 KL meeting AL in J/. (In the figure AL is bisected in L so 
 that ^Zj is parallel to KL.) By similar triangles 
 
 Al.AL AL\' 
 
 AM : Al :: AL : AK or ^if 
 
 AK 
 
 i.e. -4illf will be the required length DE. The asymptote can 
 then be drawn through E parallel to the line joining G to the 
 middle point oi AB. 
 
MISCELLANEOUS CURVES. 335 
 
 The internal and external bisectors of tlie angle ACB are 
 tangents at C to the two branches of the curve passing through 
 that point. 
 
 The tangents AT, BT^ at A and B make angles CAT, CBT, 
 with CA and CB equal respectively to the angles CAB, CBA. 
 
 This problem is a solution of the question : — to find the point 
 on a spherical mirror on which a ray from A must impinge in 
 order that it may be reflected to B ; — for if C be the centre of the 
 mirror, the circular arc representing the section of the mirror by 
 the plane passing through A, B and C will of course cut the curve 
 in points such that the rays from A and B make equal angles 
 with the normals at the points. In other words the problem is 
 to find the point P on a given circle at which the lines AP, BP, 
 A and B being given points make equal angles with the normal 
 at P. 
 
 The whole curve in such a case need not be drawn, since it 
 is easy to find points on the curve in the neighbourhood of the 
 point required and to draw an arc of the curve through them. 
 
 Magnetic curves. 
 
 The locus of the vertex of a triangle described on a given 
 base and having the sum of the cosines of the base angles constant, 
 is called a magnetic curve. 
 
 If ^j5 be the given base, and P a point on the locus, we must 
 therefore have cos PAB + cos PBA = k, and corresponding to 
 different values of k, we get a series of curves passing through 
 A and B. These represent the lines of force in the plane of the 
 paper due to a magnet whose poles are the points A and B. 
 
 The greatest value of k is 2, since the numerical value of the 
 cosine of an angle is never > 1, and k may have any value between 
 and 2. 
 
 Problem 168. To draw a magnetic curve, the base AB and 
 the constant k being given (Fig. 168). 
 
 On AB as diameter describe a circle ARQB, and on AB take 
 a point M such that 
 
 AM^h.AB, 
 
33G 
 
 MAGNETIC CURVES. 
 
 Draw any line through A cutting the circle in Qj and make 
 AqonA3I = AQ. 
 
 Fig. 168. 
 
 With centre B, and radius BB = Mq describe an arc cutting 
 the circle in R. 
 
 BR will intersect AQ in P a point on the required curve, for 
 
 cos5^P = 4l and cos^J5P-^; 
 A B AB 
 
 . •. cos BAP + cos ABP = , ., — = -^~ — = k. 
 
 AB AB 
 
 Similarly, any additional number of points can be obtained. 
 
MISCELLANEOUS CURVES. 337 
 
 The tangents at A and B may be determined by considering 
 that when F moves down to B the angle BAP becomes zero, and 
 its cosine = unity ; 
 
 .-. cosABT=k-l. 
 In the curve marked 1 in the figure ^ = f , 
 » )> •^ )> ^ = 1> 
 
 >» » " » "' ~ 2' 
 
 it a ^ j> «' — :j) 
 
 j> » 5 9> ^ — -g' 
 
 For curve number 2 therefore M coincides with B^ 
 
 „ 3 „ Jf „ C, the centre 
 
 of the circle on AB. 
 
 For curves Nos. 4 and 5 Jf is at M^ and Jf^ respectively 
 bisecting and quadrisecting AG. 
 
 Corresponding to the value 2 of ^ we get the diameter AB 
 itself for the locus, and corresponding to the value zero we get the 
 productions of the diameter to the right and left of AB. 
 
 Each curve cuts the diameter of the circle perpendicular to 
 
 AB 
 AB at a distance from ^ or ^ = -^— . 
 
 k 
 
 The chain-dotted curves in the figure are equi-potential curves 
 (see next problem) and cut all the lines of force or magnetic 
 curves at right angles. 
 
 Equi-potential Curves. 
 
 If the lines of force due to a magnet, in any plane passing 
 through its poles, are cut normally by a series of curves, these are 
 known as equi-potential curves, and by revolution round the line 
 joining the poles they generate equi-potential surfaces. 
 
 If A and B are the poles of the magnet, and the length 
 AB — c, the distances of any point F on one of the curves, from 
 A and B are known to be connected by the relation 
 
 AF BF'c 
 E. 22 
 
338 
 
 EQUI-POTENTIAL CURVE. 
 
 where h is constant throughout the particular curve considered, 
 i.e. the equation to the series of curves may be written 
 
 r T-j c' 
 
 where r and r^ denote the distances of a point from A and B, 
 
 The value of h of course varies from curve to curve of the 
 series. 
 
 Problem 169. To draw an equi-'potential curve^ the poles A 
 and B and the constant k being given (Fig. 169). 
 
 First determine the points in which the curve cuts the 
 lineil^. 
 
 /^ 
 
 ^ 
 
 -^ 
 
 \ 
 \ 
 
 Fig. 
 
 \ 
 
 169. 
 
 ^^ 
 
 \ \* 
 
 1 
 
 // 
 
 ' 7^ 
 
 
 i 
 
 k / y<^ 
 
 ^ 
 
 V \ 
 
 / 
 
 / / 
 
 |B 
 
 1 
 
 \\\ 
 
 M/k l-^ 
 
 li 1 
 
 \ 
 
 I 
 
 
 ;— — 1 — 
 
 
 
 wt; ^ — 
 
 4\ v^iiu- 
 
 J) 
 
 
 k: 
 
 ^ 
 
 > 
 
 / 
 
 
 
 3SV 
 
 / / 
 
 At the point K we evidently have 
 
 AK + BK = c, i.e. r^r^=c or r^=c — r 
 
MISCELLANEOUS CURVES. 339 
 
 which, combined with the equation 
 
 
 1 1 k 
 
 
 determines the value of r and rj . 
 
 
 We evidently have 
 
 I k 1 1 
 r c r^ c - 
 
 r 
 
 c-kr 1 
 
 or ^^ , 
 
 cr c - r^ 
 
 
 or d'- {2 + k) cr + kr^ = 
 
 a quadratic to determine r or AK, but the smallest of the two 
 roots is the only admissible solution. 
 
 At the point L we have BL - AL = Cy 
 
 i.e. r^-r = c or r, = c + r. 
 
 Ilk 
 
 The equation — = - becomes therefore in this case 
 
 r r^ c 
 
 1_^__J_ c-kr _ 1 
 
 re c + r cr c +r* 
 
 «' 
 or r + cr = Tf 
 
 and one of the roots of this equation is the length AL. 
 
 To find any points on the curve ; on AB determine a length 
 
 AO such that AO = -r , i.e. take 
 
 rC 
 
 AO : AB :: 2 : k. 
 
 Through draw any line Oa and on it make Oa = OA ; set off 
 on aO on each side of a equal lengths aq, aq^ ; and through a 
 draw ap parallel to Aq meeting J.5 in p and also drawapj parallel 
 to Aq^^ meeting AB in p^ . Then Ap and Ap^ are corresponding 
 values of r and r^ for a point on the curve and therefore a circle 
 described with centre A and radius Ap will intersect a circle 
 described with centre B and radius = Api in points F and F^ on 
 the curve. 
 
 22—2 
 
340 EQUI-POTENTIAL CURVE. 
 
 The distances aq, aq^ must be taken within certain limits, 
 since the length Ap which depends on aq cannot be greater than 
 AL or less than AK. These limits can evidently be determined 
 by drawing through A a parallel to Ka meeting aO in y, and 
 similarly drawing Ag parallel to a line through a and point I on 
 AB such that Al =AL. The points q must then be taken 
 between y and g. 
 
 In the figure, the value of h for the curve marked 1 is f , 
 » » » » ^ if ^f 
 
 l» >f tf J> ^ 11 21 
 
 if 17 i> it ^ 11 ii 
 
 and the corresponding values of ^Z" and AL are 
 
 4^ 1 AZ^ ^ AT V33-3 
 
 fori, AK = ^, AL=c. — - — , 
 
 „ 2, AK='^(3~j5), ^Z = |(V5-1), 
 „ 3, AK = ^{5-.Jl7), AL=c, 
 
 „ 4, ^Z=|{9-V65}, AL = ^{^/T7-l), 
 
 These values can of course be determined arithmetically, or 
 graphic methods may be employed. 
 
 Proof. From the similar triangles Opa^ OAq 
 
 Ap OA I .. r^A 1 1 ^ + ^3' 
 
 -^ = 7.— = -r- — \i OA = l or -T- = , 
 
 aq Oq l + aq Ap I .aq 
 
 from the similar triangles Oap^^ Oq^A 
 
 Ap^ _ OA _ I 1 _l-aq^ 
 
 aq~~'Oq^~l-aq^ ^ Ap^~l.aq^ 
 
 1 12. 
 
 .'. -. — = Y Since aq, =aq. 
 
 Ap Ap^ I 
 
MISCELLANEOUS CURVES. 341 
 
 but I by construction = y ; 
 
 J^ 1^ _k 
 
 Ap Ap^ c 
 
 or Ap and Ap^ are corresponding values of r and r^. 
 
 It may be noticed that the line corresponding to Oa of 
 curve 1 , is, for No. 3 the line oa^, the distance between A and 
 the intersection of AB and oa^ being 4 . AB ; that the limits, 
 between which points corresponding to q must be taken are/j and 
 g^, and that the point B on the curve corresponds to s and s^ 
 on oa^ , a^r being parallel to As and a^r^ to -4si ; so that 
 AB^Ar and BB=^Ar,. 
 
 The equi-potential curve corresponding to zero value of k, is 
 the perpendicular to A B through its centre point. 
 
 THE CARTESIAN OVAL. 
 
 This curve owes its name to Descartes who first discussed its 
 properties. M. Chasles, Mr Cayley, Mr Casey and others have 
 since devoted a good deal of attention to it. A short discussion 
 of the curve, treated geometrically, will be found in Chap. xx. of 
 Williamson's Differential Calculus^ 4th Edition, from which the 
 following is mainly taken. 
 
 Def. The locus of a point moving so that the sum or differ- 
 ence of its distances each multiplied by some constant from two 
 fixed points, called the foci, is constant, is called a Cartesian 
 oval. 
 
 If F, F^ are the two fixed points, P the moving point, and 
 FP = r, FP^ = r, and FF^ = c, the equation of the curve may be 
 written in either of the forms 
 
 nr± lr^ = mG (1), 
 
 or r ± Mr^ =K (2), 
 
 where K is some given length and M may be assumed to be less 
 than unity. 
 
342 CARTESIAN OVAL. 
 
 Problem 170. To draio a Cartesian oval, the foci and constants 
 of the curve being given (Fig. 170). 
 
 Let F^ and F^ be the given foci, and the length F,F^ = c^. 
 The line joining F^, F^ is called the axis. 
 
 Fig.170. 
 
 Let the distance of any point F on the curve from F^ he 
 denoted by r^, and from F^ by r^, and suppose the equation of the 
 curve to be written in the second of the above forms, i.e. 
 r^ ± Mr^ = K. 
 
 On the line joining the foci, make F^X= K, and through X 
 draw a line XY making any convenient angle with the axis. 
 On XY determine a length XY such that 
 
 XY : F,X :: I : M. 
 
 With centre F^ and any convenient radius less than F^X 
 describe an arc pp^ cutting the axis in p^ ; draw p^k parallel to 
 FiY meeting XY in k, and with centre F^ and radius = X^ 
 describe an arc cutting the former in ;? ; ^ will be a point on the 
 curve, for 
 
I 
 
 MISCELLANEOUS CURVES. 343 
 
 • « M ' 
 but (p^X^)= K-r^f 
 
 and Xk : p^ \\ XY : F,X :: 1 : if, 
 
 i.e. Z^ or r3= -^- . 
 
 Similarly any additional number of points may be deter- 
 mined. 
 
 Again with centre F^ and any convenient radius greater than 
 F^X describe an arc Qq cutting the axis in q. Draw qm parallel 
 to F^ cutting YX in m, and with centre F^ and radius = Xw 
 describe an arc cutting the former in Q. Q will be a point on the 
 curve, for 
 
 but Xq=r^-K, 
 
 and Xm '. Xq :: XY : F,X :: 1 : Jf, 
 
 I.e. Xirn or r, = ,> . 
 
 The curve consists of two ovals one lying wholly inside the 
 other, the point p belonging to the inner, and Q to the outer. 
 
 The radii F^p, F^Q must be taken within certain limits which 
 may be determined thus : — 
 
 To find the points in which the curve cuts the axis. 
 Let the inner curve cut the axis in v and v^, and the outer in 
 Fand Fj. 
 We have 
 
 r,^Mr^ = K, .: r,^-^ , 
 
 r-K 
 
344 CARTESIAN OVAL. 
 
 the positive sign referring to the inner curve, and the negative 
 sign to the outer. 
 
 At V and V we have 
 
 i.e. K'^ + c. = K^ = -j^> 
 
 or F^v{l + M) = K-M.c^, 
 
 which determines F^v, and 
 
 F V-K 
 F,V+c, = FJ = ^^, 
 
 or FJ(l-M)='K + M.c^, 
 
 which determines F^V. 
 
 Again at v^ and F, we have 
 
 i.e. F^v^ + F^v^=c, = F^v,+^^\ 
 
 or F^v^ (1 -M) = K-M. c^, 
 
 which determines F^v^, and 
 
 FJ.^FJ^^c.^FJ^^-"-^^, 
 
 or i^,F,(l + J/) = /i:+if.C2, 
 
 which determines F^ F, . 
 
 The radii for points on the inner oval must be greater than 
 F{o and less than F^v^ , and for points on the outer greater than 
 FyV^ and less than FJ. 
 
 Geometrical properties of the Curve. 
 
 The curve is evidently symmetrical about the axis. 
 
 Draw any line through F^ cutting the curve in P and Q (on 
 the same side of F^)', describe a circJe round the triangle PQF^ 
 cutting the axis again in F^, then F^P . F^Q =F^F^.F^F^; but 
 F^P .F.Qia constant, for 
 
 F^P\' = F^pY + c/ - 2F^P. c^ . cos F^F^P = ^^^ ^ 
 
I 
 
 I 
 
 MISCELLANEOUS CURVES. 345 
 
 or F^\' (1 -M')-2{K- c^M ' cos F^F^P) F^P - M' . c/ + K' = 0, 
 and F^P and F^Q are the roots of this equation, so that their 
 
 product = — ^j „3 ^ and is constant. Hence F^ is a fixed point 
 
 and it possesses the same properties relative to the curve as F^ and 
 F^-y in other words F^ is a third focus. This may most con- 
 veniently be shewn from the equation of the curve in the form 
 
 where r^ is the distance of any point on it from F^ , r^ its distance 
 from F^ and F^F^ = c^, and n>m,>l. Let F^F^ = c^ and denote 
 the distance of a point from F^ by r^. 
 
 It is easily seen that the triangles FJPF^ and F^FJ^ are 
 equiangular; 
 
 " F,F^ F,P F^F^ F^P' 
 
 . '. the equation nF, Q-l. F^Q = m,F^F^ 
 
 may be written 
 
 n . F^F^ - I . F^P=m. F,P, 
 i.e. m.r^+l. r^r=n.c^ (3), 
 
 which shows that the distances of any point on the inner oval 
 from F^ and F^ are connected by an equation similar in form to 
 (1) and consequently F^ is a third focus of the curve. 
 
 In like manner since the triangles F^QF^ and F^FJP are equi- 
 angular, the equation 
 
 n.F^P + l.F^P = mF^F^ 
 gives n.F,F^ + l.F^Q = m.F^Q, 
 
 or mr^ ~ I . r^ = n. c^ (4), 
 
 or the same holds for the outer oval. 
 
 Combined with the previous result, this shews that the con- 
 jugate ovals of a Cartesian referred to the two internal foci are 
 represented by the equation 
 
 mr, :i=lr =n.c (5), 
 
346 CARTESIAN OVAL, 
 
 and referred to the two extreme foci by 
 
 Similarly it is easily seen that referred to the middle and 
 external foci, they are represented by 
 
 nr^-mr^= ^Ic^ (6), 
 
 where c^ = F^F^ . 
 
 Taking the equation (5) referred to the two internal foci, 
 
 it may be written 
 
 I n 
 ^ ± _ r„ = — c , 
 m - m ^ 
 
 ov r^^ A , r^ = B where A and B are constants. 
 
 With centre F^ and radius = B describe a circle DE. 
 [Evidently comparing equations (1) and (2) we may take 
 
 71 = 1, l-M^ m 
 
 F,F,' 
 
 so that 5 or - c=-^.F,F^, 
 
 m ^ K ^ 
 
 i.e. B : F,F^ :: F,F^ : K.] 
 
 Let any line through F^ meet it in D and the curve in P and 
 Q. Let DF^ meet the circle again in E. 
 
 Now PD = B-PF^=A.PF^, 
 
 QI) = F,Q-B^A.F^Q; 
 
 .-. F^Q : F^P :: QD : DP, 
 
 so that F^D bisects the angle PF^Q. 
 
 Produce PF^ and QF^ to intersect F^E in Q^ and P,. The 
 triangles PF„D and P^F^E are similar and 
 
 P,E _PD ^ . 
 •'• F\P^~ F^P~ '' 
 
 and consequently the point P^ lies on the inner oval. So also the 
 point ^j lies on the outer. 
 
MISCELLANEOUS CURVES. 347 
 
 Again, since F^D bisects the angle PF^Q, 
 
 F,P.F^Q = PD.DQ + F^D\' 
 
 ^A\F^P.F^Q + F'fi\\ 
 1 or (1-A')F^P.F^Q = f;d\% 
 
 F P F D 
 
 and by similar triangles -^ = -^^^ ; 
 
 .-. {l-A')F^Q.F^P^ = F,D.F^E, 
 
 i.e. the rectangle under F^Q and F^P^ is constant ; a theorem due 
 to M. Quetelet. 
 
 If the curve has been constructed from the two internal foci, 
 the external focus can easily be determined, for the angle F^P^F^ 
 = the angle F^PF^ = F^F^Q, i. e. the angle F,P,Q = the angle F.F^Q 
 or a circle through F^P^Q passes also through Fy 
 
 To draw the tangent and normal at any point P. 
 
 Let F^P meet the circle DE (of radius as previously described) 
 in D and let F^D meet the circle through PQFJF^ in P. Then 
 i? is a point on the normal at P and also on the normal at Q. 
 
 They may also easily be drawn without using the circle DE. 
 
 The equation of the curve referred to the extreme foci has 
 been shewn to be 
 
 w, ± Ir^ = mc^. 
 
 On PF^, PF^ measure lengths PL, PM proportional to n and I 
 respectively, i. e. make PL : PM :: n : I. 
 
 Bisect LM in G and G will be a point on the normal at P. 
 
 The normal at Q may be constructed in exactly the same way, 
 one of the two lengths being measured on the corresponding focal 
 radius produced. 
 
 Similarly lengths on PF^, PF^ proportional to m and I deter- 
 mine the normal at P from vectors drawn from the internal foci. 
 
348 BENT SPRING. 
 
 ELASTIC CURVES. 
 
 In the widest sense of the term, an elastic curve is the figure 
 assumed by the longitudinal axis of an originally straight bar under 
 any system of bending forces. It is here restricted to the figure 
 taken by a slender flat spring of uniform section when acted upon 
 by a pair of equal and opposite forces. 
 
 The essential property of the curve under these conditions is 
 that the radius of curvature at any point is inversely proportional 
 to the perpendicular distance of that point from the line of action 
 of the forces. Its equation may therefore be written 
 
 where p is the radius of curvature at any point, y the distance of 
 that point from a fixed line in the plane of the curve and a con- 
 stant. 
 
 A very close approximation to the form of the curve can be 
 easily drawn by considering it as formed of a series of circular 
 arcs — the appropriate radius for each being determined. 
 
 Problem 171. To draw an elastic curve the constant of tlie 
 curve and the distance of the extreme point of the loop from the line 
 of action of the forces being given. 
 
 1st. A bent bow (Fig. 171). 
 
 Let AB be the line of action of the given forces, CD the 
 maximum ordinate of the curve from AB. From any point D in 
 AB draw DC perpendicular to AB and on it make DC = the given 
 maximum ordinate. From C inflect to ^^ a length (7^= the 
 given constant of the curve and draw EO^ perpendicular to CE 
 meeting CD in 0,. Evidently CD .CE :\GE : 0,C, so that 0, 
 is the required centre of curvature at C and may be taken as the 
 centre of a circular arc extending to a reasonably short distance 
 on either side of (7, draw it say to F and since FO^ is the normal 
 at F the centre for the adjacent arc must be taken on FO^. Draw 
 FG parallel to AB meeting CD in G and on DA make DH= CE 
 = the given constant of the curve. HK perpendicular to GH 
 
MISCELLANEOUS CURVES. 349 
 
 meets 00^ in a point K such that 
 
 GD , DH :: DH : DK-, 
 
 Fig.171. 
 
 s 
 
 \l 
 
 1 \ 
 
 1 \ 
 
 1 1 
 
 \ 
 
 / 1 
 
 //I 
 
 A h/ / !D 
 
 1 
 E,''\ B^ 
 
 
 _..i^-^ 
 
 i. e. DK is the required radius of curvature at F, and therefore if 
 FO^ on FO^ be made equal to DK, 0^ may be taken as the centre 
 of a circular arc extending to a reasonably short distance from F 
 as to L. Any number of successive centres may similarly be 
 determined. 
 
 2nd. An undulating figure crossing ^^ at any number of 
 intermediate points. 
 
 a. Let the given constant of the curve be greater than the 
 maximum ordinate (Fig. 172). 
 
 Divide the given length AB into a number of equal parts 
 corresponding to the number of required undulations and at the 
 
350 
 
 BENT SPRING. 
 
 centre of one such segment of the line draw CD perpendicular to 
 AB and equal to the given maximum ordinate, from G inflect to 
 
 AB a length CE equal to the given constant and draw EO^ per- 
 pendicular to CE meeting CD in 0,. 0^ will be the required 
 centre of curvature at C for evidently CD : CE : : CE : CO^ ; 
 and a circular arc may be drawn through C with centre 0^ and 
 extending to a reasonably short distance on either side of (7 as to 
 F. The centre of the adjacent arc must lie on FO^ . Draw F/ 
 
MISCELLANEOUS CURVES. 351 
 
 parallel to AB meeting CD in / and on DC, DB respectively 
 make De — De^ = GE. Through e draw em^ parallel to fe^ , meeting 
 AB in w^ ^^^ ^'"^2 ^^^^ ^® *^® required radius of curvature at F 
 for evidently D/ : De^ :: De i Dm^^ i. e. py = a* where y is the 
 ordinate of F. On FO^ make FO^ = Dm^ and 0^ may be taken 
 as the centre of the arc adjacent to GF. Similarly any number 
 of additional centres may be determined — supposing the second 
 arc extends to G^ draw Gg parallel to AB^ em^ parallel to ge^ and 
 on GO^ make GO^ equal to Dm^, 0^ will be the centre of curvature 
 at G. As the radius of curvature at A is infinite the portion AH 
 may be drawn tangential to the adjacent arc. 
 
 p. Let the given constant be less than the maximum ordinate 
 (Fig. 173). 
 
 Divide w^ AB and draw CD the maximum ordinate as before. 
 On CD describe a semicircle and in it make CE equal to the 
 
 Fig.I73. 
 
 given constant : draw EO^ parallel to ^^ meeting CD in the 
 required centre of curvature at C, The rest of the construction 
 is exactly similar to the above. De = De^ = CE. Ff is parallel to 
 AB and em^ parallel to fe^ determines Dm^ the radius at F. In 
 the figure G is taken on EO^ so that g coincides with and em 
 parallel to O^ei determines Dnig the radius of curvature at G. 
 
852 
 
 BENT SPRING. 
 
 3rd. The points A and B coinciding, which may give, with an 
 endless spring, a figure of 8 (Fig. 174). 
 
 On CD describe a semi-circle; in it make CD equal to the 
 given constant and draw EO^ perpendicular to CD meeting it in 
 
 Fig.l74. 
 
 Oi which will be the required centre of curvature at G. Make 
 De = Dcy = GEy De^ being perpendicular to DC, and successive 
 
 Fig.l75. 
 
MISCELLANEOUS CURVES. 
 
 853 
 
 centres may be determined precisely as before, the curve at D 
 being drawn tangential to the adjacent arc. 
 
 4th. In figs. 171 to 174 inclusive the forces are directed to- 
 wards each other. When they act in directions from each other 
 the spring may form one or more loops, with the ends and inter- 
 mediate portions meeting or crossing AB, as shewn in fig. 175, 
 
 ^ 
 
 E. 
 
 23 
 
oo4 CURVES OF PURSUIT. 
 
 the construction for which is exactly similar to the preceding and 
 which is lettered to correspond. 
 
 5th. If the forces are directed from each other at the points 
 A J B, in two rigid levers AD, BE to which the spring is fixed at 
 D and E, the spring forms one or more looped coils lying alto- 
 gether at one side of the line of action AB (fig. 176). 
 
 The general method of construction is the same as before, but 
 the radius of each arc corresponding to its central portion instead 
 of to one extremity has been determined. 
 
 Let GF be a maximum ordinate ; on it describe a semi-circle 
 and in the semi-circle make GH equal to the given constant : draw 
 HO^ perpendicular to GF meeting it in 0^ the centre of curvature 
 at G. Draw GO^ parallel to GF and at a distance from it equal 
 to one-half the desired length of the loop of the curve, and on it 
 make Gh = GH the given constant : make Gh^ on AB equal to 
 Gh. Take any convenient point K at about the centre point of 
 the intended second arc of the curve and draw Kk parallel to 
 AB meeting GO^ in ^, then Ay/ig drawn through h parallel to kh^ 
 determines Gm^ the required radius of curvature at E. Take any 
 convenient point L on the arc struck through G and join it to 
 the centre G^; make LO^ on LG^ = Gm_^^ and G^ will be the re- 
 quired second centre. Similarly any additional number of centres 
 can be determined. 
 
 CURVES OP PURSUIT. 
 
 When a point A moves so that it is continually directed to- 
 wards a second point B also in motion in some known curve, the 
 locus of A is called a " curve of pursuiV^ 
 
 The problem was first presented in the form — To find the 
 path described by a dog which runs to overtake its master. 
 
 The velocities of the two moving points must of course be 
 known, and the required locus can then be easily traced to any 
 required degree of approximation by supposing the direction of 
 
MISCELLANEOUS CURVES. 
 
 355 
 
 motion to be constant for a short interval and then to be suddenly 
 deflected. 
 
 Problem 172. A moves in a straight line from A to B with 
 constant velocity , and G starts from C with constant velocity double 
 that of A and is constantly directed on A. To find the curve of 
 pursuit (Fig. 177). 
 
 Set off from A along AB any convenient equal distances Al, 
 
 \ 
 
 \ 
 
 V 
 
 V 
 
 V 
 
 < 
 
 I 
 
 f 
 
 I 
 
 i 
 
 23—2 
 
356 EXAMPLES. 
 
 12, 23,... While A advances from -4 to 1 suppose (7's motion to 
 be directed on c the centre point of -41. Then when A arrives 
 at 1, (7 will be at the point I) on Cc such that (7Z> = twice Al — 
 while A advances from 1 to 2 suppose (7's motion to be directed 
 on d, the centre point of 12; then when it is at 2 (7 will be at 
 the point E on Dd such that i>^= twice 12, and similarly any 
 number of successive points can be determined. 
 
 Examples. 
 
 1. Draw a Harmonic Curve given the length A£ of a vibrar 
 tion and a point F on the curve. 
 
 [From F draw FJV^ perpendicular to ^^ meeting it in N. If 
 a is the amplitude of the vibration PiV= a sin 0, 
 
 and e : 27r :: A]^ : AB, 
 
 FN 
 
 .'. a = 
 
 . 27r.AN-' 
 sin 
 
 AB 
 which determines a. 
 
 AN 
 The expression lir . -r-f^ is the circular measure of the angle, 
 
 the sine of which can then be obtained from a trigonometrical 
 table.] 
 
 As a numerical example take ^i5=10'8, AN=\'16, FN'=1'Q7. 
 a then equals 1*96 very approximately. 
 
 2. Draw a Cassini's oval, the foci F, F^, and a point F on 
 the curve being given. 
 
 [Take a mean proportional (k) between the focal distances 
 FF, F^F. h is the constant of the curve. Prob. 154.] 
 
 3. Draw a Cassini's oval, the foci F, F^ and a tangent FT 
 being given. 
 
 [Bisect FF^ in C and draw CT perpendicular to FT meeting 
 it in T. From one of the foci F draw a line meeting CT in Q 
 
MISCELLANEOUS CURVES. 357 
 
 and on CF^ describe a segment of a circle containing an angle 
 equal to the angle CQF (Prob. 30) and cutting FQ in p. The 
 locus of p will intersect the given tangent in its point of contact, 
 and the question reduces to the preceding. The line FQ must be 
 drawn within certain limiting positions in order that the circle 
 may meet it in real points.] 
 
 4. Draw through a focus i^ of a lemniscate a line which shall 
 cut the curve at a given angle a. 
 
 [Let C be the centre and F^ the second focus. On CF^ 
 describe a segment of a circle containing an angle ^ — a, and 
 meeting the curve in P. FP will be the required line.] 
 
 5. Given the centre C, direction of axis CJ, and a point P, 
 on a lemniscate, draw the tangent at the point. 
 
 [Draw CB perpendicular to CA, and GT (between CB and 
 CP) making the angle BCT=sing\e AGP. Bisect CP in D and 
 draw DT perpendicular to GP. T will be a point on the tangent 
 at P.] 
 
 6. Describe a lemniscate with given centre C, given direction 
 of axis GA, and to cut a given right line at a given angle. 
 
 [The direction of a tangent is obviously given. Through G 
 draw a line parallel to this given direction, and the angle between 
 this line and GB, perpendicular to CJ, is three times the angle 
 AGP, where P is the point in which the required tangent 
 meets the given line.] 
 
 7. Describe a lemniscate, with given centre (7, given direction 
 of axis GA, and to pass through a given point P. 
 
 [Draw the tangent and normal at P. Ex. 5. Let the 
 normal meet GA in G. Bisect the angle GPG by PD meeting 
 GA in D. Through P draw lines making equal angles with PD 
 and cutting off equal distances CP, GF^ on GA. (Prob. 19.) F 
 and Pj are the foci of the required curve.] 
 
358 EXAMPLES. 
 
 8. a6, a'ah' are two lines at right angles to eacli other and 
 a! a = ah' = ^ab. ah moves round in the plane of the two lines till 
 h comes to ¥ and a to a', the centre point c of ah moving always 
 along ca and a certain point d of ah describing a circular arc 
 round h'. Determine the position of d and draw the loci of h and 
 a throughout the motion. 
 
 9. A pendulum 5" long vibrates uniformly in an arc of 40**. 
 A fly starting at the bottom crawls at a uniform speed to the 
 top, arriving there in the time taken by a forward and backward 
 swing of the pendulum. Trace the course of the fly. 
 
 10. A train is running in a straight line at 10 miles an 
 hour. The door (30" wide) of one of the carriages is opened 
 with uniform angular velocity till it stands at right angles to the 
 direction of motion in J a second and closed again in the same 
 time. Draw the curve traced out by a point on the edge of the 
 door. Scale, J = 1 foot (Harmonic Curve). 
 
 11. BD is a line If" long. Draw AB, DC perpendicular to 
 BB and each 2" long, the points A and C being on opposite sides 
 of BD. Consider these lines as three bars jointed at B and />, 
 and free to turn in the plane of the paper about the points A and 
 C as centres. Trace the locus of the centre point of BD. 
 
 [The complete locus is a figure of 8, the central portion being 
 very nearly straight lines.] 
 
 12. C is the centre of a circle, A and B are points outside 
 the circle and in its plane. A double string is wrapped round 
 the circle and the free loop is led oflf so that one portion passes 
 round A and the other round B. Shew that any fixed point on 
 the loop describes an hyperbola as the string is unwound by the 
 rotation of the circle. 
 
 As a particular example take CA = 2", CB = 2", diameter of 
 circle f", and one position of the tracing point IJ" from A and 
 2J" from B. 
 
CHAPTER XIII. 
 
 SOLUTION OF EQUATIONS. 
 
 Graphic methods may be applied to the solution of alge- 
 braical and trigonometrical equations, and in certain cases the 
 process is much simpler and more expeditious than the arith- 
 metical or analytical one. This is particularly the case with 
 certain statical questions in which a position of equilibrium is 
 defined by two angles for which two equations are given. " The 
 equation for either variable which results from eliminating the 
 other may be one of high degree, the approximate solution of 
 which by the methods of the Theory of Equations would be very 
 troublesome. In such cases it is often possible to obtain a 
 solution sufficiently accurate for practical purposes by construct- 
 ing curves corresponding to the equations and taking their points 
 of intersection*." 
 
 For example, to find 6 from the equation 
 
 c sin (2^ — a)=a sin $ (1), 
 
 c, a and a being given constants. 
 If we trace the curves r = a sin 6, 
 
 r = G sin (2^ — a), 
 then at their points of intersection the equation (1) is satisfied — 
 the same origin and initial line being of course taken in tracing 
 both loci. 
 
 * Minchin's Statics, 3rd Edition, p. 49. 
 
360 GENERAL METHOD. 
 
 At first a rough tracing only is necessary, the object of this 
 rough preliminary tracing being merely to find the places in the 
 neighbourhood of which the curves really intersect. Then devote 
 very special care to the tracing of the curves in these indicated 
 neighbourhoods and in these alone. We shall thus get a value 
 or values of the unknown variable accurate within certain narrow 
 limits of error due to the draughtsmanship and possibility of 
 measuring given quantities. This is as exact a solution as the 
 graphic method pure and simple enables us to obtain, but by 
 analysis a further step can be taken. We have obtained a near 
 value (say w) of 6, which does not quite satisfy (1), but cd + 8 
 does, where S is a small unknown quantity. If we write w + S 
 for ^ in (1) and then, 8 being very small, put cos 8 = 1, sin 8 = 8, 
 we have 
 
 c sin'(2o) - a) + 2c cos (2o) — a) x h — a sin w + o^ cos w x 8, 
 
 -J a sin o) — c sin (2q> — a) 
 2c cos (2(0 — a) — a cos w ' 
 
 so that 8 and therefore w + 8, or a still nearer value of ^, is known. 
 
 In general, if we have to solve F{6) ~/(0), i. e. any given 
 function of ^ = to some other given function, we may trace the 
 cui-ves r= F(6); T = f (6), 
 
 and get an approximate value o> of ^ from their points of inter- 
 section as above. Then the correction 8 is given by the equation 
 
 the dashes denoting the differential coefficients of the original 
 functions. 
 
 Example: — Solve the equation 2^= 5 sin 0. 
 r= 2^ represents an equiangular spiral, (Prob. 149), 
 
 r = 5 sin 6 represents a circle of diameter 5 units, j)assing 
 through the origin and its centre on line through the origin per- 
 pendicular to the initial line. 
 
SOLUTION OF EQUATIONS. 361 
 
 Let 0) be the circular measure of the angle between the initial 
 line and the radius drawn from the origin to a point of inter- 
 section of these curves, then 
 
 5 cos (o — 2'^log^2 ' 
 CO will be an approximate solution of the original equation ; and 
 CO + S a more exact one. 
 
 Pkoblem 173. To solve the quadratic equation 
 x'-2Ax + £'^0 (Fig. 178). 
 
 Draw two lines Oa, Ob at right angles to each other, and on 
 one of them make Oh = B. 
 
 With h as centre and A as radius describe an arc cutting Oa 
 
 in a, so that Oa — jA^ — B^; and with centre a and radius ab 
 
 p. 
 
 fr- 
 
 Fig.178. 
 
 Or a " d^ 
 
 describe arcs cutting Oa in d and d^. Od and Od^ are lines 
 representing the two values of x in the above equation. If the 
 numerical values of the roots are required they must be measured 
 of course on the same scale which has been used for laying off the 
 lengths A and B. 
 
 If A is numerically less than B the roots become imaginary, 
 and the graphic method is not applicable. 
 
 As a numerical example we may take the equation to deter- 
 mine the length AK in problem 169. 
 
 Here AK is one of the roots of kr^ — 2 + k\cr + c" = 0, 
 
 , 2 + Jc c' . 
 
 or r'--^c.r + ^=0, 
 
362 QUADRATICS. 
 
 which is of the above form if ^ = (?^^ and ^ = -^ 
 
 Suppose h = ^ then A = — and B = '^c, 
 
 where c is a given length. 
 
 Make Oe in fig. 178 = this given length c. 
 
 On 0<x, Oh, take Of=Of^, thenjg^, represents J 2 the length 
 0/ being the unit; make OF on Ob ^ff^. 
 
 With centre f^ and radius = 2.0/* describe an arc cutting Oa 
 in G^ then 06^ represents ^3, the length Of being the unit. 
 Through e draw a parallel to FG meeting Oh in h. 
 
 Since evidently Oh \ Oe w J 2 : ^3, 
 
 05 = the constant B of the equation. 
 
 7c . 
 
 With centre 5 and radius = -^ describe an arc cutting Oa in 
 
 a, and with centre a and radius ah describe arcs cutting Oa 
 
 in d and d^ . Oc?, OtZ^ represent the values of r in the equation, 
 
 Oe 
 and the particular value of AK in Problem 169 is Od = -^. 
 
 Problem 174. To solve the quadratic equation 
 
 x' + 2Ax + B'=0. 
 
 The solution is exactly the same as that of the last problem, 
 but both roots are negative. 
 
 Problem 175. To solve the quadratic equation 
 x'-2Ax-B' = 0{Yig. 179). 
 
 Draw 2 lines Oa, Oh at right angles to each other and on 
 them make Oa = Aj Oh = B : 
 
 then ah = J A'' + B\ 
 
 With centre a and radius ah describe an arc cutting Oa in d 
 
 i 
 
 J 
 
SOLUTION OF EQUATIONS. 
 
 363 
 
 and d^. Od, Od^ represent the roots of the equation, but the 
 smaller one must be taken with negative sign. 
 
 A may be greater or less than B. 
 
 
 b 
 
 \ 
 
 
 FIg.179. 
 
 
 
 P- 
 
 
 
 
 
 
 ^\ 
 
 
 
 \^ 
 
 
 'd 
 
 
 O 
 
 'a 
 
 1 
 
 'd, 
 
 Problem 176. To solve the quadratic equation 
 x'+2Ax-JB' = 0. 
 
 The solution is identical with that of the last problem, but 
 the greater root must be taken with negative sign. 
 
 As a numerical example take the equation to determine the 
 length AL in Problem 169. 
 
 T^ + cr—{ —7- ) = so that A=- and B = — - , 
 
 suppose k = ^. .•. Oh : c -.'. J2 : 1. 
 
 Make Oe = c ; bisect Oe in a and make Oa^^ = Oa so that 
 aa^ :0a:: J 2 : 1. 
 
 Make OF on Ob = aa^ and through e draw eh parallel to aF ; 
 
 then Oh = B. 
 
 With centre a and radius ah describe arcs cutting Oa in d 
 and c?j . Od is the positive root of the equation, and is the length 
 ALm. curve No. 3 of Problem 169. 
 
364 
 
 a cos ^ + 6 sin ^ = c. 
 
 Problem 177. To solve graphically the equation 
 acose + bsiiie = c (Fig. 180). 
 
 Draw 2 lines at right angles to each other as AO, AB. Make 
 AO=^a and AB — h on any convenient scale. Describe a circle 
 
 Fig.180. 
 
 round OAB (its centre will of course be at the middle point of 
 OB) and with centre and radius OD = c describe an arc cutting 
 it in B, the angle AOB is the required angle 0. 
 
 [In the figure a =2*5, 6 = 1-3, c-2'66, the unit being the 
 length Z and ^ = 47 -S".] 
 
 OD c 
 
 Proof. 
 
 cos BOD = 
 
 OB ~ OB ' 
 
 cos (AOD- AOB) = ^^, 
 cos AOD cos AOB + sin AODsm AOB: 
 
 cos AOD ^^^ + sin AOD-^ = -^^, 
 
 OB' 
 
 .: A0D = 6. 
 
 The second point D^ in which the arc described with centre 
 and radius c would cut the circle gives when c is greater than a a 
 second solution, the angle AOD^ being the value of 6 in this case. 
 
 When c is less than a so that D^ falls between and A the 
 second solution corresponds to a cos 0—h sin = c. 
 
SOLUTION OF EQUATIONS. 365 
 
 Problem 178. A and B are two fixed points and P a variable 
 point, the position of which is defined hy the angles PAB (= 6) and 
 PBA [- <fi); draw the locus represented hy the equation 
 
 sin $ + sin 4> = <^, 
 where a is constant. \a may he either positive or negative hut 
 its numerical value cannot he greater than 2.] (Fig. 181.) 
 
 On AB make BG = a . AB, and describe a semi-circle on AB. 
 Draw a line Ap meeting the semi-circle in p and on BA make 
 
 Bh = Bp. With centre A and radius = hC describe an arc cutting 
 the semi-circle in q, and draw Bq cutting Ap in P. P will be a 
 point on the required locus. Similarly any number of points can 
 be determined. 
 
 If a is greater than unity, i.e. if BC is greater than AB, the 
 locus will meet AT, BT^ drawn perpendicular to AB, in points T 
 and T^ determined by inflecting AR, BE in the semi-circle each 
 equal to AG and drawing BR, AR^ meeting AT, BT^ in T and T^ 
 respectively. BT, AT^ are tangents to the required locus at T 
 and T^ . Lines drawn from A to points between R^ and B do not 
 intersect the locus in real points. 
 
 If a is less than unity, i.e. if BG^^ is less than AB, the curve 
 passes through A and B and the tangents at those points can be 
 drawn by inflecting BV, AY^ in the semi-circle each equal to BG . 
 AY, BY^ are tangents to the required curve. In the figure the 
 value of a for the upper curve is \ and for the lower |. There 
 
366 
 
 TWO UNKNOWNS. 
 
 are similar branches on the other side of AB corresponding to 
 negfative values of the angles. 
 
 Proof. 
 
 i.e. 
 
 BhiFAB = ^~^, and BmF£A=4^, 
 AB AB 
 
 . . . , pB + Aq Bb + hC BC 
 sm d + sin<f>= — „ = tt: — = —,. = a- 
 
 AB 
 
 AB AB 
 
 Problem 179. To determine values of r and 6 ivhich simul- 
 taneously satisfy the equations 
 
 r^ cos 29 = a^... (1), and r .sin a-0 = b .sin a. . .(2), 
 
 vjhere is the angle between the radius vector r and a fixed right 
 line and a, b and a are constants. 
 
 Equation (2) may be written j = ; j^ , so that r and b are 
 
 6 sm(a-6^) 
 
 evidently sides of a triangle the opposite angles of which are a 
 
 (or TT—a) and a-d. 
 
 Let OA (fig. 182) be the fixed straight line from which 6 is 
 measured, the origin. On it make OB = b and through B draw 
 
 
 'I 
 
 Flg.l82. 
 
 f 
 
 1 
 I 
 
 1 
 
 ; 
 
 
 i / 
 \ / 
 
 /-*; 
 
 )^ 
 
 \L^ 
 
 ^ 
 
 1 / \ 
 
 o ' 
 
 A 
 
 /B 4 
 
SOLUTION OF EQUATIONS. 367 
 
 BP makiug an angle a with the positive direction of the initial 
 line. BP is the locus represented by (2), for P being any point 
 on it OP = r and BOP = 6, so that 
 
 OP sin OBP sin a 
 
 OB Bin BPO sina^ 
 
 To find points on the second locus. Make OA = a; when 0=0, 
 r = ± a so that the curve passes through A, andA^ on the other side 
 of such that OA^ = a would be a second point on the locus. 
 The curve is symmetrical about OA because negative values of 
 give the same r as the corresponding positive values. Through 
 draw any line Op and make the angle ^0^ = twice the angle 
 AOp. Draw AQ perpendicular to OA meeting OQ in Q, then 
 
 cos 2. A Op = ^-^, and .*. if j9 is a point on the curve 
 
 0P''%-0A^ 
 
 or Op' = OA.OQ. 
 
 Make Oq on OA = OQ and on Oq describe a semi-circle cutting 
 AQ in p^ and make Op= 0])^. Similarly any additional number 
 of points on the curve may be determined, and at the points P 
 and P^ where the line BP intersects the curve the same values of 
 6 and r hold for both. 
 
 As the angle AOp increases the line OQ will not intersect 
 AQ within any reasonable distance; the length OQ may however 
 be determined by bisecting or quadrisecting OA and taking the 
 intersection of the ordinate through the point of division with 
 the line corresponding to OQ — the distance of which from will 
 be the half or quarter of the diameter of the required semi- 
 circle. The length Or, for example, corresponding to the radius 
 vector OE is one-fourth the diameter of the semi-circle which 
 determines r^ on J. ^ and so the length OB, 
 
 The only portions of the second locus which it is necessary 
 to trace, are of course those in the immediate neighbourhood of 
 
368 
 
 a cot {6 — ol) + h cot {(j) — /3) = c. 
 
 the points where it cuts the line, and a trial or two readily 
 shews whereabouts the radii Op should be drawn. 
 
 The second locus is a rectangular hyperbola with centre 
 and transverse axis 2a, and if this were recognised from the 
 equation, the ordinary method of drawing an hyperbola might of 
 course be adopted. 
 
 Problem 180. A and B are two fixed points and P a variable 
 point, whose position is defined by the angles FA B {- 6) and PBA 
 (= <^), what locus is represented by the equation 
 
 a cot (6— a) + b cot (0 — ^) = c, 
 where a, b, c, a, jS are constants ? 
 
 Equations of which the above is the general form frequently 
 occur in statical problems, .and therefore a knowledge of what it 
 represents and how it is liable to modification may be useful 
 (Fig. 183). 
 
 Draw AC, BG making with AB the angles BAC = a and 
 ABG = p. The required locus is a conic circumscribing the tri- 
 
 ^1^^^_ 
 
 ;.I83. 
 
 .^y^ -1^ 
 
 o 1 
 
 \a ,/P 
 
 \ _\iT 
 
 r^- 
 
 angle ABC, the tangents to which at those points are easily 
 drawn. 
 
 The distance of any point T on the tangent at C from BC : its 
 distance from AC produced :: 5 sin a : a sin p. 
 
SOLUTION OF EQUATIONS. 
 
 369 
 
 li Ap on AC =b and pn be drawn perpendicular to AB^ 
 pn^bsiii a, 
 and if Bq on BC = a and qm be drawn perpendicular to AB, 
 
 qm = a sin )8. 
 T can therefore be determined by drawing parallels to BC and 
 AC 8it distances =j97i and qm respectively. 
 
 The tangent at A divides the exterior angle at A so that the 
 distance of any point t from AC : distance from 
 
 AB : : a : (a cot a+b cot /3 + c) sin a. 
 
 The length given by this last term is easily obtained, for if the 
 angle mAB (fig. 184) = a, and Bm perpendicular to Am = a, 
 ^m = acota, draw Bk parallel to Am and make kBC = (3, and Ck 
 perpendicular to Bk = 6, 
 then Bk or mn = b cot jS, 
 
 Fig.184. 
 
 make nl= c, I being taken on the same side of w as ^ if c is 
 negative and on the opposite side if c is positive and the length 
 Al = a cot a+b cot )8 ± c : from I draw Is perpendicular to AB and 
 Is = Al sin a. 
 
 The tangent at A is determined by drawing parallels to AC, 
 AB respectively at distances a and Is intersecting in t. 
 
 Similarly the tangent at B divides the exterior angle at B so 
 that the distance of any point t^ from BC : its distance from AB 
 
 :: b : (a cot a + b cot /3 + c) sin p. 
 The conic is therefore completely determined. 
 
 E. 2.4 
 
370 a cos ^ + 6 cos </> = c and h cot 6 + I cot <f> = m. 
 
 If acota + 6cot)8 + c = the tangents at A and B evidently 
 coincide with the line AB^ and the locus becomes a straight line 
 through C, identical with the tangent at in the general case. 
 
 If a and (3 both equal zero, i. e. if the equation is 
 a cot + h cot <f> = Cj 
 the locus is a right line, which may be constructed as shewn in 
 the next problem; for the point C is evidently in this case some- 
 where on the line AB, and the tangents at A and B again coincide 
 with the line AB. 
 
 •(1). 
 
 (2), 
 
 Problem 181. To solve the equations 
 
 acos6 + bcoB<f> = c 
 
 kcot6 + I cot <fi = m 
 
 where a, 6, c, k, I, m are constants (Fig. 185). 
 
 The second equation represents a right line which may be 
 
 Flg.l85. 
 
 drawn as follows : Draw any straight line AB and produce it to D 
 so that AB :BI)::l-k: k, and dr&w DC so that cotCDB = j^ , 
 i.e. i£l)E = m, EG = l-k. 
 
 [On any line through B make Bd=k, da = l, and draw dD 
 parallel to Aa. This determines D. Make DE = m and EC per- 
 pendicular to DE, = aB, 
 
SOLUTION OF EQUATIONS. 371 
 
 At any point 'p of the line we have, if 'pAq = ^, 'pBq = 4>, 
 
 cot ^ = ~ , cot </) = — , 
 pq pq 
 
 and we want to shew therefore that 
 
 k . Aq + l. Bq = m.pq ••••(«); 
 
 .'. (a) may be written k (Aq + Bq) = I {Dq — Bq), 
 
 i.e. k.AD = l.BD, 
 
 which by construction it does.] 
 
 To find points on the locus represented by (1). With the 
 points A and B as centres describe two circles S and T of radii 
 
 - . AB and - . AB respectively. Draw any common ordinate NLQ, 
 
 meeting aS' in Z and T in N ] then the lines AL and BN intersect 
 in a point, P, on the required locus ; for 
 
 AQ + QB=AB, 
 
 or AL cos 6 + ^iTcos <j> = AB, if BAL is and ABN is <f>, 
 
 r 
 
 or - . AB . cob6 + - AB cos d> = AB. 
 
 c c ^ 
 
 which is the given equation. 
 
 If the line DC meet the curve in E and B^ the angles BAB, 
 R^AB are the required values of By and the angles RBA, B^BA 
 those of <^. 
 
 There is a precisely similar loop on the other side of AB. 
 
 In the particular case in which a = h the locus is the Magnetic 
 Curve. (Prob. 168.) 
 
 Problem 182. To find 6 and cfi/rom the equations 
 
 - — 7, + -7 — - = c...(l), and cos ^ = A;cos<f)., (2), 
 smc' sm</) ^ ' T \ ' 
 
 where a, b, c, k are constants (Fig. 186). 
 
372 
 
 sin 6 sin </> 
 
 = c and cos ^ = A; cos <^. 
 
 Take two points A and B sucli that ^^ ^ a + 6; make AO^a, 
 OB = h and draw OD perpendicular to J.^; with A as centre and 
 
 Fig.l86. 
 
 c as radius describe a circle, and draw any radius AC meeting 
 OD in L; inflect BJ^LC (J" being on OD); then P, the point of 
 intersection oi AG and BJi^ a point on the locus represented by 
 (1), the angles 6 and <^ being ALO and BJO respectively. 
 
 There is a precisely similar loop on the other side of AB. 
 
 Again the equation cos ^ = ^ cos <f> gives sin PAB = ^ . sin PBA 
 or PB - h . PA, i.e. P is the vertex of a triangle on a given base 
 AB and with sides in a given ratio (Problem 17), i.e. the locus 
 represented by the second equation is a circle whose diameter 
 QQ^ is the line joining the points which divide AB internally 
 and externally in the ratio 1 : k; i.e. 
 
 AQ : QB :. \ : k ;: AQ, : Q^B. 
 
 The values of 6 and <^ which satisfy both equations are those 
 belonging to the points of intersection of this circle and the 
 previous curve. 
 
SOLUTION OF EQUATIONS. 37S 
 
 Examples. 
 
 1. Solve the equation = ^ . 
 
 X 2 
 
 oc 
 [Trace the loci y = sin x (harmonic curve) and y = - (a straight 
 
 line through the origin) : the values of x corresponding to their 
 points of intersection are solutions.] 
 
 2. Solve the equation sin x = ax + b. 
 
 [The intersections of the harmonic curve y = sin x and of the 
 straight line y = ax + h where a and h are constants.] 
 
 3. Solve the equation 2=5 sin 6. 
 
 [The intersections of the equiangular spiral r = 2 and of the 
 circle r = 5 sin 0.'\ 
 
 4. Find 6 and <jS> from the equations 
 
 tan ^ = 9*tan^ (1), 
 
 a cos ^ = 6 cos <^ + c (2), 
 
 where a, b, c and n are given constants. 
 
 [The 2nd equation represents a locus identical with (1) of 
 Problem 181, attention being paid to the usual conventions as to 
 sign. 
 
 The 1st equation represents a right line perpendicular to AB 
 (fig. 185), the base of this locus, and meeting it in D so that 
 AD^n.BI).] 
 
 5. Find and <f) from the equations 
 
 I cos 6 +71 cos (fi = a — m cos a, 
 I sin^ — Qi sin <^ = m sin a 
 where I, m, n, a and a are constants. 
 
 [Draw two lines AB, AC including an angle a, and make 
 AB = a and AC = m. With centres B and C and radii == n and I 
 
374 EXAMPLES. 
 
 respectively describe arcs intersecting in D on the same side of 
 BC as AB. Let CD meet AB in E. BED is the required value 
 of 6 and DBE that of </>.] 
 
 6. Determine 6 from the equation 
 
 a cos A . cos (X + 2^) = c . cos (a + 6) 
 where a, c, X and a are given constants. 
 
 [The locus represented by the right-hand side of the above 
 equation is a circle of radius c, the origin (0) being the extremity 
 of a diameter, and the initial line making an angle a therewith. 
 
 To draw the locus represented by the left-hand side : — draw a 
 line through the origin making an angle X with the initial line, 
 and on it measure a length OL = a. Draw ZiV perpendicular to 
 the initial line meeting it in iV so that OiV = a cos X. With 
 centre and radius Oli describe a circle. From any point Q 
 on this circle draw QM perpendicular to OL meeting it in M. 
 Draw OF bisecting the angle JVOQ and make OP = OM. P will 
 be a point on the second locus, and any additional number of 
 points may be similarly determined. Let the two loci intersect 
 in Xj and the angle between OX and the initial line is the 
 required angle 6.'] 
 
 This equation defines the position of equilibrium of a uniform 
 rectangular board resting in a vertical plane against two equally 
 rough pegs in a horizontal line. 
 
 THE END. 
 
 CAHBBIDOE : PRINTED BY C. J. CLAY, M.A. & SON, AT THE UNIVERSITY PRESS. 
 
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