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in H, and AH will be its diameter or latus transversum, to which the latus rectum will be as BG² to AG $\scriptstyle \times$ GH. If AG nowhere meets the locus, the line AH being infinite, the locus will be a parabola; and its latus rectum corresponding to the diameter AG will be $\scriptstyle {\frac {BG^{2}}{AG}}$. But if it does meet it anywhere, the locus will be an hyperbola, when the points A and H are placed on the same side the point G; and an ellipsis, if the point G falls between the points A and H; unless, perhaps, the angle AGB is a right angle, and at the same time BG² equal to the rectangle AGH, in which case the locus will be a circle.

And so we have given in this Corollary a solution of that famous Problem of the ancients concerning four lines, begun by Euclid, and carried on by Apollonius; and this not an analytical calculus, but a geometrical composition, such as the ancients required.

LEMMA XX.

If the two opposite angular points A and P of any parallelogram ASPQ touch any conic section in the points A and P; and the sides AQ, AS of one of those angles, indefinitely produced, meet the same conic section in B and C; and from the points of concourse B and C to any fifth point D of the conic section, two right lines BD, CD are drawn meeting the two other sides PS, PQ of the parallelogram, indefinitely produced in T and R; the parts PR and PT, cut off from the sides, will always be one to the other in a given ratio. And vice versa, if those parts cut off are one to the other in a given ratio, the locus of the point D will be a conic section passing through the four points A, B, C, P.

Case 1. Join BP, CP, and from the point D draw the two right lines DG, DE, of which the first DG shall be parallel to AB, and meet PB, PQ, CA in H, I, G; and the other DE shall be parallel to AC, and meet PC, PS, AB, in F, K, E; and (by Lem. XVII) the rectangle DE $\scriptstyle \times$ DF will be to the rectangle DG $\scriptstyle \times$ DH in a given ratio. But PQ is to DE (or IQ) as PB to HB, and consequently as PT to DH; and by permutation PQ is to PT as DE to DH. Likewise PR is to DF as RC to DC, and therefore as (IG or) PS to DG; and by permutation PR is to PS as DF to DG; and, by compounding those ratios, the rectangle PQ $\scriptstyle \times$ PR will be to the rectangle PS $\scriptstyle \times$ PT as the rectangle DE $\scriptstyle \times$ DF is to the rectangle DG $\scriptstyle \times$ DH, and consequently in a given ratio. But PQ and PS are given, and therefore the ratio of PR to PT is given.   Q.E.D.