144 THE MATHEMATICAL PRINCIPLES BOOK I.J Cither between the points, h, i, k, /, or without them. If one of the points a, b, falls between the points h, i, and the other xvithout the points h, I, the Problem is impossible. PROPOSITION XXVI. PROBLEM XVIII. To describe a trajectory that shall pass through a given point, and touch four right lines given by position. From the common intersections, of any two of the tangents to the common intersection of the other two, draw an indefinite right line ; and taking this line for the first ordinate radius; / 'xs o transform the figure (by Lem. XXII) into a new figure, and the two pairs of tangents, each of which before concurred in the first ordinate radius, will now become parallel. Let hi and kl, Al l\ ik and hi, be those pairs of parallels completing the parallelogram hikl. And let p be the point in this new figure corresponding to the given point in the first figure. Through O the centre of the figure draw pq.: and O? being equal to Op, q will be the other point through which the conic sec tion must pass in this new figure. Let this point be transferred, by the inverse operation of Lem. XXII into the first figure, and there we shall have the two points through which the trajectory is to be described. But through those points that trajectory may be described by Prop. XVII. LEMMA XXIII. If two right lines, as AC, BD given by position, and terminating in given points A, B, are in a given ratio one to the other, and the right line CD, by which the, indetermined points C, D are joined is cut in K in a given ratio ; I say, that the point K will be placed in a right line given by position. For let the right lines AC, BD meet in E, and in BE take BG to AE as BD is to AC, and let FD be always equal to the given line EG ; and, by construction, EC will be to GD, that is, to EF, as AC to BD, and therefore in a given ratio ; and therefore the %•-'''' ,.---'' I \ triangle EFC will be given in kind. Let E K cT^"^ CF be cut in L so as CL may be to CF in the ratio of CK to CD ; and because that is a given ratio, the triangle EFL will be given in kind, and therefore the point L will be placed in the right line EL given by position. Join LK, and the triangles CLK, CFD will be similar ; and because FD is a given line, and LK is to FD in a given ratio, LK will be also given