m+n 58. Show that the polar reciprocal of the curve x”y” = a with regard to a circle whose centre is at the origin is another curve of the same kind. 59. Show that the first positive pedal of the curve p = pm+1am = p2m+1, = pm+1 is m α and that its polar reciprocal with regard to a circle of radius a whose centre is at the origin is pm+1=amr. 60. Show that the inverse of the curve p=f(r) with regard to a circle whose radius is k and centre at the pole is 61. Show that the pedal of the inverse of p=f(r) with regard to a circle whose radius is k and centre at the origin is 62. Show that the pedal of the inverse of p regard to a circle whose radius is k and centre at the origin is 63. Show that the polar reciprocal of the curve r2 = aTMcos m✪ with regard to the hyperbola r2cos 20 = a2 is 64. In the semicubical parabola ay2= the tangent at any point P cuts the axis of y in M and the curve in Q. O is the origin and N the foot of the ordinate of P. Prove that MN and OQ are equally inclined to the axis of x. 65. At any point of a curve where the ordinate varies as the cube of the abscissa, a tangent is drawn; where it cuts the curve another tangent is drawn; where this cuts the curve a third is drawn, and so on. Prove that the abscissae of the points of contact form a geometrical progression, and also the ordinates. 66. A straight line AOP of given length always passes through a fixed point 0, while A describes a given straight line AT; show that if PT be the tangent at P to the locus of P, the projection of PT on AOP=AO. 67. The point P moves so that OP.O'P=constant, 0, Ο' being fixed points. If OY, O'Y' be the perpendiculars from O and O' on the tangent at P to the locus of P, prove that PY : PY' :: OP2 : 0′P2. 68. O and O' are two fixed points, P any point in a curve where r = OP, r' = O'P, and c is constant. Prove that the dis tance between P and the consecutive curve obtained by changing c to c + dc is ultimately 69. In a system of curves defined by an equation containing a variable parameter investigate at any point the normal distance between two consecutive curves, and determine the form of the equation for a system of parallel curves. [PROFESSOR CAYLEY, Messenger of Mathematics, Vol. V.] CHAPTER VIII. ASYMPTOTES. 208. DEF. If a straight line cut a curve in two points at an infinite distance from the origin and yet is not itself wholly at infinity, it is called an asymptote to the curve. 209. Equations of the Asymptotes. Let the equation of any curve of the nth degree be arranged in homogeneous sets of terms and expressed as xn on + ... = 0. .... (A) To find where this curve is cut by any straight line gives the abscissae of the points of intersection. Applying Taylor's Theorem to expand each of these functional forms, equation (c) may be written This is an equation of the nth degree, proving that a straight line will in general intersect a curve of the nth degree in n points real or imaginary. The straight line y=ux+ẞ is at our choice, and therefore the two constants μ and ẞ may be chosen, so as to satisfy any pair of consistent equations. Suppose we (E) (F) The two highest powers of a now disappear from equation (D), and that equation has therefore two infinite roots. If, then, μ1, Mos ......., μn be the n values of deduced from equation (E) (which is of the nth degree in μ), the corresponding values of ß will in general be given by and then straight lines , Hence, in order to find the asymptotes of any given curve, we may either substitute μx+ß for y in the equation of the curve, and then by equating the coefficients of the two highest powers of x to zero find μ and ß. Or we may assume the result of the preceding article, which may be enunciated in the following practical way :- -In the highest degree terms put x=1 and y=μ [the result of this is to form pn(u)] and equate to zero. Hence find μ. Form pn-1(u) in a similar way from the terms of degree n-1, and differentiate pn(u), then the values of B are found by substituting the several values of μ in the ФП-1(м) formula ('n(u) B= Ex. Find the asymptotes of the cubic Here therefore 2x3 — x2y — 2xy2+y3 + 2x2+xy — y2+x+y+1=0. Þ3(μ) = μ3 — 2μ2 — μ+2=0; (μ-1) (μ + 1)(μ-2)=0; Þ2(μ)=2+μ- μ2, p'z(μ) = 3μ2 — 4μ – 1 ; giving Again, and |