and the equation to the perpendicular to the tangent from the focus whose co-ordinates are (ae, o) is y' y = 12X y x x and y in the lines whose equations are (1) and 2, are the same, but x', y' are different, but the latter are the same in both at the point of intersection, therefore if we solve (1) and (2) as simultaneous, supposing x' and y' as unknown, we will have the relation between the co-ordinates of the point of intersection. Hence by the above supposition b2 x (x' — ae) — a2 - (x' — ac) a2 y b2 x a2 y (a-ex)= + a y a'y b2 x (x' -ae) = a2 (a + ex)2 (x2 + 2 aex + a2 e2 + a2 — a2 e2 — x2 + c2 x2) Now the sum of the squares of the co-ordinates (x', y') of the point of intersection being equal to the square of the semi-major axis, the locus is evidently a circle on the major axis as diameter. The equation to the secant PP'T through the points P (x', y') and P' (x", y") in the curve, is B (1) if x' = x" and y' = y", in that case the secant becomes a tangent, and its equation is (1) ... a2 y y' — a2 y'2 = — b2 x x' + b2 x'2 a2 y y' + b2 x x′ = a2 y'2 + b2 x22 = a2 b2 n The supposition of x′ = x′′ and y'=y" makes the points P & P' coincide, and the line PP' instead of cutting the curve in P and P' touches it only in one point, i. e. it becomes a tangent as observed before. If in (1), we make y': o we have b2 x x′ = a2 b2, or x: a::a: x' .. x x' a2, if then we take a third proportional to x and a in the major axis produced, we will have the point at which the tangent cuts the axis. And by joining this point and the point of contact, we will have the tangent drawn. Another way of drawing the tangent is the following: join PS and PH, and draw a line TPt such that the angles HPT and SPt will be equal, that line will be the tangent. 18. y2 + (1 − e2) x2 - 2 mx (1 + e) = o.. (1) 1st e 1. 1-e2 is positive = (2)...... ....... y2 2 mx (1 + e) - (1-e) x2, the co-efficient of x is positive and that of x2 is negative in (2). But the equation (2) is of the same form as the equation to the ellipse the origin being at the vertex or general equation (1) represents an ellipse when e 1. The axes may be found thus, By comparing the equation (2) and (3), we have 2nd. e > 1, e2-1 is positive and the equation becomes (4) ...... y2 =2m x (1 + e) + (e2—1) x2, where the coefficients of both x and x2 are positive equation is of the same form as that of the hyperbola the origin being at the vertex, or 3rd. Let e 1, therefore the general equation y2 + (1-e2) x2—2 mx (1+e)o becomes y2-4 m x = 0, or y2 4 m x which is the equation to the parabola. The principal parameter is evi A C is the centre of the ellipse, S and H the two foci, P any point in the curve, join P and S, and produce PS to meet the curve in P'. The inclination of the radius rector SP is supposed to be measured from the further focus H, and in the direction AP, so that if the inclination of ASP = 0, that of SP' will be the obtuse 180° + A'SP'=180° + 0. SP to the axis or angle ASP' |