Find the Trajectory of Points

Lemma 17

If from any point of a given conic section, to the 4 produced sides AB, CD, AC, DB, of any trapezium ABDC inscribed in that section, as many right lines PQ, PR, PS, PT are drawn in given ang ei, each line to each side.

The rectangle PQ, X PR of those on the opposite sides AB, CD, will be to the rectangle PS X PT of those on tie other two opposite sides AC, BD, in a given ratio.

CASE 1

Let us suppose, first, that the lines drawn to one pair of opposite sides are parallel to either of the other sides; as PQ and PR to the side AC, and PS and PT to the side AB. And farther, that one pair of the opposite sides, as AC and BD, are parallel betwixt themselves;

Then the right line which bisects those parallel sides will be one of the diameters of the conic section, and will likewise bisect RQ. Let O be the point in which RQ is bisected, and PO will be an ordinate to that diameter. Produce PO to K, so that OK may be equal to PO, and OK will be an ordinate on the other side of that diameter.

Since, therefore, the points A, B, P and K are placed in the conic section, and PK cuts AB in a given angle, the rectangle PQK (by Prop. XVII., XIX., XXI. and XXIII., Book III., of Apollonius’s Conics) will be to the rectangle AQB in a given ratio. But QK and PR are equal, as being the differences of the equal lines OK, OP, and OQ, OR; whence the rectangles PQK and PQ × PR are equal; and therefore the rectangle PQ × PR is to the rectangle AQB, that is, to the rectangle PS x PT in a given ratio. Q.E.D

Case 2

Let us next suppose that the opposite sides AC and BD of the trapezium are not parallel. Draw Bd parallel to AC, and meeting as well the right line ST in t, as the conic section in d.

Join Cd cutting PQ in r, and draw DM parallel to PQ, cutting Cd in M, and AB in N. Then (because of the similar triangles BTt, DBN), Bt or PQ is to Tt as DN to NB. And so Rr is to AQ or PS as DM to AN. Wherefore, by multiplying the antecedents by the antecedents, and the consequents by the consequents, as the rectangle PQ × Rr is to the rectangle PS × Tt, so will the rectangle NDM be to the rectangle ANB; and (by Case 1) so is the rectangle PQ × Pr to the rectangle PS × Pt.

By division, so is the rectangle PQ × PR to the rectangle PS × PT. Q.E.D.

Case 3

Let us suppose, lastly, the four lines PQ, PR, PS, PT, not to be parallel to the sides AC, AB, but any way inclined to them. In their place draw Pq, Pr, parallel to AC; and Ps, Pt parallel to AB; and because the angles of the triangles PQq, PRr, PSs, PTt are given, the ratios of PQ to Pq, PR to Pr, PS to Ps, PT to Pt will be also given; and therefore the compounded ratios PQ × PR to Pq × Pr, and PS × PT to Ps × Pt are given.

But from what we have demonstrated before, the ratio of Pq × Pr to Ps × Pt is given; and therefore also the ratio of PQ × PR to PS × PT. Q.E.D.