Let x AB = a. Then u 0, and the curve passes through B, which is double. Let x=. infinite. Then u∞. and the branches BZ, Bz, are The curve is limited at A, because all negative values of x give imaginary values of u. To find the position of the tangent at singular points, we have tan. = =± are 1 dy x = 0; then, tan. ◊ = ± ∞, and the tangents at A (AT, At) .. the curve at A is concave towards AC. Let x = a. Then tan. 0 = 0, and the tangents at P Then tan. 1, or the tangents at B are inclined to AC at an angle of 45°. 41. To find the nature of the curve AP (Fig. 31) such that AB being drawn from the vertex making 4 BAC 45°, and cutting the ordinate PM in Q, Let x = 0, then y = 0.. the curve passes through A, and Let xoo, then y = ∞ S is infinite towards C. y' dy.y + 2e 3y + ƒ 2e dy + 2e ̄žy 4e + C 12 + 2e 1⁄2 × ( y − 2) + 4, which expresses the 2 area comprehended between the vertex, and any ordinate y generally. 42. BM (Fig. 32) being any chord of the O whose centre is C, and CF any radius cutting the chord in E, and circumference in F; then if EP be drawn BM and the locus of the point P. EF, required Draw CAD BM, and put CF=r, CA = a, AE = x, EP=y. y Then = EF = ±(CF – CE) = ± (r − √√ a2 + x2), the equation to the curve. The signs show whether E falls within or without the parallel to BM, and at M or B, tan. 0 = F are no points of contrary flexure. Ꮖ ; at D the tangent is If we transfer the origin of co-ordinates to D, and the line of abscissæ to DC, by putting and y' =PN= x, which will give, by substitution in the above equation, and proper reductions, y'2 = 2ax' + x22, we easily recognise the locus to be the equilateral hyperbola, whose axes are = a. Here we may remark, that as most known curves are defined by co-ordinates originating in the vertex, or other points symmetrically placed with respect to the branches, in investigating the nature of a curve, it may frequently aid us to refer the co-ordinates to that point. 43. The problem may be generalized by stating it; required the curve to which a straight line, cutting from two straight lines which meet in any angle two segments whose suma, is always a tangent. Let BC, B'C' (Fig. 33,) be any two positions immediately consecutive of the line, cutting off AB + AC = AB' + AC′ = a, and intersecting in the point P. Let the equation of the line AB, (x and y being measured from A along AX, and AY) be Now at the point P, the co-ordinates of the two lines CB, C'B', being precisely the same, the variation, due to their change of position, that will take place in equation (b), must arise from N only; and differentiating on that supposition, we get And substituting in equation (a) we get, after proper reductions, y = x ± 2 √ ax + a = (√x ± √ a)2 (c) which, since P is evidently a point in the curve, is the equation of the curve (viz. a Parabola) referred to the co-ordinates AX, AY. This Theory, which is closely allied to that of particular solutions in the Integral Calculus, is worthy the attention of the Student. The next problem affords another illustration of it. 44. Given the angle A (Fig. 33) and area cut off by CB, to find the curve to which CB is always a tangent. Let the line C'B' be the immediately consecutive position of CB, and let CB be denoted, as in the preceding problem, by y = Mx + N ..... (d) Let, also, the given area ABC = a2. Now, supposing CB to undergo a minute variation in position, at its intersection with the curve, will remain the same, r and y, and N (being a function of the angle PCX) alone will vary. Taking .. the differential of (ƒ) on that supposition, we get NdN x sin. A x + dN0 2x2 sin. A ; Hence equation (d) becomes, by substitution, the equation to the curve, which is therefore an hyperbola whose asymptotes are AX, AY. = If A a right-angle, the curve is a rectangular hyperbola. 45. The section of a prolate spheroid made by a plane passing through the focus of the generating ellipse, is an ellipse having the same focus. The truth of this proposition may be shewn by either of the methods explained in pages 8, &c., vol. ii. As the student will find little or no difficulty in using either of them, beyond that of actual computation in obtaining an equation of the second degree, and afterwards discussing it according to Lacroix Traité de Trig., pp. 156, &c., and other elementary writers, we shall prefer giving a solution founded on principles different from those referred to, and tending to abridge the usual labour in such inquiries. Let DPE (Fig. 34) be the spheroid generated by the revolution of the ellipse DE' E about its axis-major DE, and let DPE'QD' 2 be the intersection of a plane, passing through the focus S, with the surface of the spheroid; D'PE'QD' is an ellipse whose focus is S. |