Fig. 31
Since AD and BC are parallel tangents, PQ is a diameter and the conjugate diameter is parallel to AD. The middle point of PQ is the center of the conic. We take now for the axis of abscissas the diameter PQ, and the conjugate diameter for the axis of ordinates. Join A to Q and B to P and draw a line through S parallel to the axis of ordinates. These three lines all meet in a point N, because AP, BQ, and AB form a triangle circumscribed to the conic. Let NS meet PQ in M. Then, from the properties of the circumscribed triangle (§ 89), M, N, S, and the point at infinity on NS are four harmonic points, and therefore N is the middle point of MS. If the coördinates of S are (x, y), so that OM is x and MS is y, then MN = y/2. Now from the similar triangles PMN and PQB we have
BQ : PQ = NM : PM,
and from the similar triangles PQA and MQN,
AP : PQ = MN : MQ,
whence, multiplying, we have
±b2/4 a2 = y2/4 (a + x)(a - x),
where
