For, supposing P to be the intersection of the curve D'PQ with the generating ellipse in any position DPE, SP will be a radiusvector, issuing from the pole S, to both curves; and D'E' being the intersection of the plane of D'PQ with that of the ellipse DED', given in position, PSE and PSE will be the corresponding angles. Take SB SA = SC = 1, and in the planes D'PE', DPE, DE'E describe the respective arcs BA = 0, ACB. BC= a, Now, by trig. we have cos.a And eccentricity a sections.) Р PSP = = .. substituting for cos. a, we have, a.(1-2) = d of the ellipse DPE (see any good work on conic 1-e sin.ẞcos.A. sin.0-ecos.B.cos.0 between p and 0, expressing the nature of the section D'PE'Q. Now to recognise the precise form of the curve D'PE'Q, represented in plano by Fig. 34. a, we will transfer the origin of from Se' to the maximum value of SP viz., Se, by putting '+s, s being the interval between Se' and Se. 04 sin. B cos. A. sin. + cos. B. cos. 0, e2), where a = 2 1-ecos.a Butt an. s = gives tan. cos. A. tan. B = tan. (—s) = 1 1- cos. s cos, s (a), an equation - cos. 8 sin. 6) 0, e cos. ß cos. 0) and cos. cos. (-s) = cos. s (cos. ' + sin. ' tan. s) =cos. s (cos. ' cos. A tan. B. sin. '). Hence, substituting in equation (a), we get, a.(1 − e) 1-ecos. ẞ cos.(s)(1 + cos.o A.tan.23)cos.' cos. A. tan.o ß, tan. s. cos. ' tan. s) a'(1-e'1 1- e' cos. 0' whose focus is S, whose 1 1+cos. A. tan. B 1-e cos. B√1+ cos. A. tan. B x cos. ' axis major 2 = 46. Given two ordinates ß, B′, and the difference of the corresponding abscissæd, of the logarithmic curve, to construct it. The equation to the logarithmic curve being y = a*, if we can determine a in terms of known quantities B, B', d, the construction will be effected by assuming any abscissa at pleasure, and finding the corresponding ordinate from the equation. Let u be the abscissa corresponding to B; then u + ♪ is the abscissa due to B', and by the equation we have B = a", B' = a"+"; shews that the curve D'PE'Q is an ellipse e cos.ß/1+cos.'A, tan'ß a.(1-e) e cos. B (1 + cos.' A. tan.2 ß). 1 eccentricity = which being Required the curve whose perpendicular, drawn from 47. a given point upon the tangent, is constant. Let the line of abscissæ AM pass through the given point S. Let also AM = x, PM = y, AS a, and Sy=b= upon the tangent PT. b: ST: y: PT ::: or 1 + y = dx2 dy2 But p.dx = y. bp ( √1+ p2 bpdp √1+ p2 Hence y = b. √√1+ or y b. √1+p2 + (x a) p....(a) which being reducible to Clairaut's Formula, may be integrated as follows: dy = + y2 dx2 dy? ydx dy x + a. SM. + p.dx + (x − a) dp. xa) dp = 0, :.dp=0 bp x-a a=0} + √1+p2 √ b3 (x — a) The first value gives, by substituting in (eq.a), the general solution p = or p = y = b √ 1 + c2 + c (x − a),....(b) The other gives, after proper reductions, b2 - (x-a) √ b2 — (x − a)2 or y2 = b2-a2+2ax-x...(c) an equation involving no arbitrary constant and not deducible from (6), and therefore, a particular solution. This latter result evidently shews the curve to be a circle. It is, in fact, the locus of the intersections of all the = √ b3 — (x − a)3 straight lines made by giving every possible value to the constant c in the general solution (b). See No. 44, Vol. II. 48. Given two distances r, r' from the focus of a parabola and the angle between them a, to construct it. Let the angular distance of r from the axis be ; then the between r' and the axis is 0 + a. Let also the parameter of the parabola, to be determined. P Now the polar equation to the parabola, deduced very readily from y2= px, being 1 = p 2 1 + cos. Q r = and r' = Р 4 cos.2 P 4 cos. 2 2 0+ a 2 COS. COS. COS. = = COS. 1 1+ m2 0+ a 2 0 2 2 Р 4 a .cos. "" we have sin. tan. 1-2 cos. Ө 2 a 2 0 2 1 sin. .sin.. a 2 a sin.2 2 =m sin. 2 P Hence the equation between the radius-vector and its of inclination to the axis p, which affords the required construction, is a 7 1 P = — ) 2 . · cos." a sin.2 y' = 49. A and B being two given points (Fig. 36), and C another such that AC: CB :: n 1; then the locus of C is a circle. y2 = For, take AD: DB :: n : 1 and DN = x, CN= y, AB = a. Then AC+ CB2 = (n2 + 1) CB2 = (n2 + 1) (y2 + BN2) = AN' + BN2 + 2ya :. na — 1) y2 + no. (BD x)2 = (AD + x)o. na But BD = n+1 (1-2 cos. a n+1' Hence, by substitution and reduction, we get 2na 1 (1-2 cos. x (a2 vhich is the equation to a circle whose radius is n2 and AD = x2 50. To find the polar equation to an ellipse, the pole being in the centre. The equation between the rectangular co-ordinates x, y, measured from the centre, is a and b being the principal semi-axes. Let be the radius-vector measured from the centre, and its P |