Case 2. But if PR and PT are supposed to be in a given ratio one to the other, then by going back again, by a like reasoning, it will follow that the rectangle DE DF is to the rectangle DG DH in a given ratio; and so the point D (by Lem. XVIII) will lie in a conic section passing through the points A, B, C, P, as its locus. Q.E.D.
Cor. 1. Hence if we draw BC cutting PQ in r and in PT take Pt to Pr in the same ratio which PT has to PR; then Bt will touch the conic section in the point B. For suppose the point D to coalesce with the point B, so that the chord BD vanishing, BT shall become a tangent, and CD and BT will coincide with CB and Bt.
Cor. 2. And, vice versa, if Bt is a tangent, and the lines BD, CD meet in any point D of a conic section, PR will be to PT as Pr to Pt. And, on the contrary, if PR is to PT as Pr to Pt, then BD and CD will meet in some point D of a conic section.
Cor. 3. One conic section cannot cut another conic section in more than four points. For, if it is possible, let two conic sections pass through the five points A, B, C, P, O; and let the right line BD cut them in the points D, d, and the right line Cd cut the right line PQ in q. Therefore PR is to PT as Pq to PT: whence PR and Pq are equal one to the other, against the supposition.
LEMMA XXI.
- If two moveable and indefinite right lines BM, CM drawn through given points B, C, as poles, do by their point of concourse M describe a third right line MN given by position; and other two indefinite right lines BD, CD are drawn, making with the former two at those given points B, C, given angles, MBD, MCD: I say, that those two right lines BD, CD will by their point of concourse D describe a conic section passing through the points B, C. And, vice versa, if the right lines BD, CD do by their point of concourse D describe a conic section passing through the given points B, C, A, and the angle DBM is always equal to the given angle ABC, as well as the angle DCM always equal to the given angle ACB, the point M will lie in a right line given by position, as its locus.
For in the right line MN let a point N be given, and when the moveable point M falls on the immoveable point N, let the moveable point D fall on an immovable point P. Join CN, BN, CP, BP, and from the point P draw the right lines PT, PR meeting BD, CD in T and R, and making the angle BPT equal to the given angle BNM, and the angle CPR
