The earliest questions the solutions of which involved the Theory of Envelops or Ultimate Intersections were those which related to evolutes of curves, investigated by Huyghens *, and those relating to Caustics, a subject introduced by Tschirnbausent; but these authors did not follow any general analytical method for the solution of such problems. Leibnitz was the first who considered the general theory of questions of this kind, so well adapted for exemplifying the utility of his Calculus; and in two memoirs in the Acta Eruditorum , he gave a general process for the solution of all problems which depended on the successive intersections of lines whether straight or curved, the position or magnitude of which were changed according to some law. This method is the same as that usually employed, no important modification having been subsequently introduced, and may be stated in the following manner. If u = f (, y, z, a, b, c ...) = 0 be the equation to a surface, a, b, c ... being parameters determining its position and magnitude, the envelop of all the surfaces formed by the variation of a, b, c... is found by eliminating these quantities between the equations du du du U = 0, 0... da db dc When, as is often the case, there are one or more equations of condition between the parameters, the method of indeterminate multipliers may frequently be conveniently employed. The same method of course applies to lines in two dimensions. Opera, Vol. I. p. 89. 1692, p. 168, and 1694, p. 311. Ex. (1) Find the equation to the curve which touches all the straight lines determined by the equation m S whence a and substituting this value we have y= 2 (mx), or yo = 4mx, the equation to a parabola. (2) Find the equation to the curve which touches all the lines determined by the equation y = ax +r (1 + a')!, when a is supposed to vary. Here U = y - ax – 1 (1 + a')}, du + (1 + a2) | a? (1 + a*)! (1 + a')}} go2 pa? therefore ya also x2 1 + aa Then y (3) Find the envelop of the series of parabolas whose equation is y = a (x – a), a being the variable parameter. du da V or a = ; + = 1. 22 y the equations to two straight lines. (4) To find the envelop of the series of ellipses defined by the equation y? du = 0; a” (k – a)3 kri ky whence k – a = +y and on substituting these values in the original equation we find as the equation to the envelop a'i + y= kl. vi + yi : (5) The straight line PQ (fig. 41) slides between the rectangular axes Ax, Ay; find the locus of its ultimate intersections. Let AP = a, AQ = b, PQ = C; then the equation to PQ is ydo + a, b being subject to the condition a + 12 = c. Differentiating with respect to a and l, xda = 0) (1), ada + bdb = 0 (2), 62 ^ (1) – (2) = 0 gives on equating to zero the coefficients of each differential. Y 12 Multiply by a, b, respectively, and add ; then by the first two equations, Y λ X = (x + 12 = c; =ll, therefore a’ = cʻx, 13 = cʻy, and, substituting these values of a and b in the equation of condition, we obtain gỗ + 3 = cả as the locus of ultimate intersections of PQ. (6) If the equation to a straight line be y 1, b a and b being subject to the condition 6 1, + a a + m n the locus of its ultimate intersections is . y + 1, n 4c which is the equation to a parabola referred to two tangents as axes. (7) Find the envelop to the series of parabolas determined by the equation . y = A x – (1 + a') where a is the variable parameter. The result is a parabola, the equation to which is a ? = 4c (c - y). This is the equation to the curve touched by the parabolas described by projectiles discharged from a given point with a constant velocity, but at different inclinations to the horizon. The problem was proposed by Fatio to John Bernoulli, who solved it, but not by any general method : it was the first case which was brought forward of the locus of the ultimate intersections of curved lines. - Commercium Epistolicum of Leibnitz and Bernoulli, Vol. 1. p. 17. (8) Find the curve which is constantly touched by the circles determined by the equation (x – a)? + yé = b?, a and b being the co-ordinates of a parabola, so that 12 = 4ma. The resulting equation is y = 4 m (x + m), which is the equation to an equal parabola, the vertex of which is shifted through a distance – m. (9) Find the envelop of the series of ellipses defined by the equations yo R? 62 ma na ข + = 1. + = 1. n M The equations to four straight lines in the space contained by which all the ellipses lie. (10) Find the locus of the ultimate intersections of chords joining the extremities of conjugate diameters of an ellipse the axes of which are a and b. If a', y be the co-ordinates of the extremity of a dia b meter, y', -- r' are the co-ordinates of the extremity of its conjugate. Hence the equation to the line joining their extremities is a a r' and y' being connected by the equation to the ellipse The resulting equation of the locus of the ultimate intersections is 2.x2 2y = 1; a 1,2 the equation to an ellipse, the axes of which are + jä |