Hence we obtain the following construction for drawing the diameter conjugate to any given one. Let the ordinate at the given point P, when produced, meet the semicircle on the axis major at Q, join CQ, and erect CQ'perpendicular to it; then the perpendicular let fall on the axis from Q' will pass through P', a point on the conjugate diameter. P' M' M Hence, too, can easily be found the co-ordinates of P' given in Art. 176, for, since From these values it appears that the areas of the triangles PCM, P'CM', are equal. Ex. 1. To express the lengths of two conjugate semidiameters in terms of the angle 4. Ans. a'2 = a2 cos2 + b2 sin2; b'2 = a2 sin2 + b2 cos2. Ex. 2. To express the equation of any chord of the ellipse in terms of ø and p' (see p. 93), Ans. a y cos(+)+sin(+9)= cos § (p − 4'). Ex. 3. To express similarly the equation of the tangent. x sin = 1. α Ex. 4. To express the length of the chord joining two points a, ß, D = 2 sin § (a — ẞ) { a2 sin2 } (a + ß) + b2cos2¦ (a + ẞ) } }. But (Ex. 1) the quantity between the parentheses is the semidiameter conjugate to that to the point (a + ẞ); and (Ex. 2, 3) the tangent at the point (a + ß) is parallel to the chord joining the points a, ß; hence, if b' denote the length of the semidiameter parallel to the given chords, D = 26' sin } (a — ß). Ex. 5. To find the area of the triangle formed by three given points a, ß, y. 2 = ab{sin (a− ß) + = ab{2sin(a-3) cos(a-3)-2 sin =4ab sin (a-3) sin (3-y) sin (y-a) Σ=2ab sin (a-B) sin(37) sin (ya). Ex. 6. To find the radius of the circle circumscribing the triangle formed by three given points a, ß, y. If d, e, ƒ be the sides of the triangle formed by the three points, R= def 42 = b'b"b"" ab where b', b", b′′ are the semidiameters parallel to the sides of the triangle. If c', c", c" ' be the parallel focal chords, then (see p. 193) R2= c' c"c"" (These expressions are due to Mr. MacCullagh, Dublin Exam. Papers, 1836, p. 22; see also Crelle, vol. XL. p.31.) Ex. 7. To find the equation of the circle circumscribing this triangle. cos (a+B) cos (B+ y) cos (y + a)— 2(b2-a2)y b = {cos (a + B) 2 2 sin (a+ẞ) sin (B+ y) sin (y +a): + cos (B+ y) + cos(y + a)}. From this equation the co-ordinates of the centre of this circle are at once obtained. Ex. 8. To find the locus of the intersection of the focal radius vector FP with the radius of the circle CQ. Let the central co-ordinates of P be x'y', of O xy, then we have, from the similar triangles, FON, FPM, = b sin o x + c x2 + c a (e + cos $) Now, since is the angle made with the axis by the radius vector to the point O, we at once obtain the polar equation of the locus by writing p cosp for x, p sind for y, and we find Hence (Art. 199) the locus is an ellipse, of which C is one focus, and it can easily be proved that F is the other. Ex. 9. The normal at P is produced to meet CQ; find the locus of their intersection. The equation of the normal is (Art. 184) but we may, as in the last example, write p cos and p sin for x and y, and the equation becomes Ex. 10. It is useful in astronomy to express the angle PFC in terms of the angle. It will be found that tan PFC = tan 10. + Ex. 11. If from the vertex of an ellipse a radius vector be drawn to any point on the curve, find the locus of the point where a parallel radius through the centre meets the tangent at the point. The tangent of the angle made with the axis by the radius vector to the vertex ; therefore, the equation of the parallel radius through the centre is and the locus of the intersection of this line with the tangent is, obviously, α = 1, the tangent at the other extremity of the axis. The same investigation will apply, if the first radius vector be drawn through any point of the curve, by substituting a' and b' for a and b ; the locus will then be the tangent at the diametrically opposite point. 237. The methods of the preceding Articles do not apply to the hyperbola. For the hyperbola, however, we may substitute y' = b tan 6, since x' = a sec p, This angle may be represented geometrically by drawing a tangent MQ from the foot of the ordi nate M to the circle described on 1. the transverse axis, then the angle QCM = 4, since CM = CQ secQCM. = P M We have also QM= a tan o, but PM b tan p. Hence, if from the foot of any ordinate of a hyperbola we draw a tangent to the circle described on the transverse axis, this tangent is in a constant ratio to the ordinate. 238.* Since the equation of the conjugate hyperbola is * This Article is taken from a paper by Mr. Turner in the Cambridge and Dublin Math. Jour., vol. i. p. 122. any point on the conjugate hyperbola may be expressed by y" = b sec p', and x" = a tan q'. Now if 0 be the angle made by any diameter with the axis of Р 239. Any two figures are said to be similar and similarly placed, if radii vectores drawn to the first from a certain point O are in a constant ratio to parallel radii drawn to the second from another point o. If it be possible to find any two such points O and o, we can find an infinity of others; for, take any point C, draw oc parallel C с to OC, and in the constant ratiop, then from the similar triangles OCP, ocp, cp is parallel to CP and in the given ratio. In like manner, any other radius vector through c can be proved to be proportional to the parallel radius through C. Iftwo central conic sections be similar, all diameters of the one are constantly proportional to the parallel diameters of the other, since the rectangles OP OQ, op oq, are proportional to the squares of the parallel diameters (Art. 152). 240. We now propose to investigate the condition that two conic sections, whose equations are given, Ax2 + Bxy + Cy2 + Dx + Ey + F = 0, ax2 + bxy + cy2 + dx + ey +f should be similar, and similarly placed. = 0, The equation of the first, referred to its centre as origin, must (Art. 155) be of the form Ax2 + Bxy + Cy2 = F', and the square of R2 = F A cos20+ B cos 0 sin 0+ C sin20 the square of a parallel semidiameter of the second is Hence, two conic sections will be similar, and similarly placed, if the coefficients of the highest powers of the variables are the same in both, or only differ by a constant multiplier. 241. It is evident that the directions of the axes of similar conics must be the same, since the greatest and least diameters of one must be parallel to the greatest and least diameters of the other. If the diameter of one become infinite, so must also the parallel diameter of the other, that is to say, the asymptotes of similar and similarly placed hyperbola are parallel. The same thing follows from the result of the last Article, since (Art. 157) the directions of the asymptotes are wholly determined by the highest terms of the equation. be Similar conics have the same eccentricity; for m2a2 m2b2 m2a2 Similar and similarly placed conic sections have hence sometimes been defined as those whose axes are parallel, and which have the same eccentricity. If two hyperbola have parallel asymptotes they are similar, |