fuch manner that all the right lines drawn from the focus S to the several points of the fection, and the perpendiculars let fall from the fame points on the right line GF, are in that given ratio.

That excellent geometer M. De la Hire has folved this problem much after the fame way, in his Conics, prop. 25, lib. 8.

SECTION V.

How the orbits are to be found when neither focus is given. LEMMA XVII.

If from any point P of a given conic fection, to the four produced fides AB, CD, AC, DB of any trapezium ABDCinfcribed in that fection, as many right lines PQ, PR, PS, PT are drawn in given angles, each line to each fide; the rectangle PQ × PR of thofe on the oppofite fides AB, CD, will be to the rectangle PS × PT of thofe on the other two oppofite fides AC, BD, in a given ratio.

CASE 1. Let us fuppofe, firft, that the lines drawn to one pair of oppofite fides are parallel to either of the other fides; as PQ and PR (Pl. 8, Fig. 4) to the fide AC, and PS and PT to the fide AB. And farther, that one pair of the oppofite fides, as AC and BD, are parallel betwixt themselves; then the right line which bifects thofe parallel fides will be one of the diameters of the conic fection, and will likewife bifect RQ. Let O be the point in which RQ is bifected, and PO will be an ordinate to that diameter. Produce PO to K, fo that OK may be equal to PO, and OK will be an ordinate on the other fide of that diameter. Since, therefore, the points A, B, P, and K are placed in the conic fection, and PK cuts AB in a given angle, the rectangle PQK (by prop. 17, 19, 21, & 23, book 3, 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.

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CASE 2. Let us next fuppofe that the oppofite fides AC and BD (Pl. 8, Fig. 5) of the trapezium are not parallel

Draw Bd parallel to AC, and meeting as well the right line ST in t, as the conic fection 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 fimilar triangles BTt, DBN), Bt or PQ is to Tt as DN to NB. And fo Rr is to AQ or PS as DM to AN. Wherefore, by multiplying the antecedents by the antecedents, and the confequents by the confequents, as the rectangle PQ > Rr is to the rectangle PS × Tt, fo will the rectangle NDM be to the rectangle ANB; and (by cafe 1) fo is the rectangle PQ x Pr to the rectangle PSx Pt; and by divifion, so is the rectangle PQ × PR to the rectangle P S x PT. Q.E.D.

CASE 3. Let us fuppofe, laftly, the four lines PQ, PR, PS, PT (Pl. 8, Fig. 6), not to be parallel to the fides 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 alfo given; and therefore the compounded ratios PQ × ̧PR to Pq x Pr, and PS x PT to Ps x Pt are given. But from what we have demonftrated before, the ratio of Pq x Pr to Psx Pt is given; and therefore also the ratio of PQ × PR to PS x PT. Q.E.D

LEMMA XVIII.

The fame things fuppofed, if the rectangle PQ × PR of the lines drawn to the two oppofite fides of the trapezium is to the rectangle PS × PT of thofe drawn to the other two fides in a given ratio, the point P, from whence thofe lines are drawn, will be placed in a conic fection defcribed about the trapezium. (Pl. 8, Fig. 7.)

Conceive a conic fection to be defcribed paffing through the points A, B, C, D, and any one of the infinite number of points P, as for example p; I fay, the point P will be always placed in this fection. If you deny the thing, join AP cutting this conic fection fomewhere elfe, if poffible, than in P,as in b. Therefore if from thofe points p and b, in the given angles to the fides of the trapezium, we draw the right lines pq, pr, ps, pt, and bk, bn, bf, bd, we shall have, as bk x bn to bf x bd, fo (by lem. 17) pq × pr to ps × pt; and fo (by supposition) PQ x PR to PS x PT. And because of the fimilar trapezia

bkAf, PQAS, as bk to bf, fo PQ to PS. Wherefore by dividing the terms of the preceding proportion by the correfpondent terms of this, we fhall have bn to bd as PR to PT. And therefore the equiangular trapezia Dn bd, DRPT are fimilar, and confequently their diagonals Db, DP do coincide. Wherefore b falls in the interfection of the right lines AP, DP, and confequently coincides with the point P. And therefore the point P, wherever it is taken, falls to be in the affigned conic fection. Q.E.D.

COR. Hence if three right lines PQ, PR, PS, are drawn from a common point P, to as many other right lines given in pofition, AB, CD, AC, each to each, in as many angles refpectively given, and the rectangle PQ × PR under any two of the lines drawn be to the fquare PS2 of the third in a given ratio; the point P, from which the right lines are drawn, will be placed in a conic fection that touches the lines AB, CD in A and C; and the contrary. For the pofition of the three right lines AB, CD, AC remaining the fame, let the line BD approach to and coincide with the line AC; then let the line PT come likewife to coincide with the line PS; and the rectangle PS × PT will become PS2, and the right lines AB, CD, which before did cut the curve in the points A and B, C and D, can no longer cut, but only touch, the curve in those coinciding points.

SCHOLIUM.

In this lemma, the name of conic fection is to be understood in a large fenfe, comprehending as well the rectilinear fection through the vertex of the cone, as the circular one parallel to the bafe. For if the point p happens to be in a right line, by which the points A and D, or C and B are joined, the conic fection will be changed into two right lines, one of which is that right line upon which the point p falls, and the other is a right line that joins the other two of the four points. If the two oppofite angles of the trapezium taken together are equal to two right angles, and if the four lines PQ, PR, PS, PT are drawn to the fides thereof at right angles, or any other equal angles, and the rectangle PQ × PR under two of the lines. drawn PQ and PR, is equal to the rectangle PS × PT under

the other two PS and PT, the conic fection will become a circle. And the fame thing will happen if the four lines are drawn in any angles, and the rectangle PQ x PR, under one pair of the lines drawn, is to the rectangle PS × PT under the other pair as the rectangle under the fines of the angles S, T, in which the two laft, lines PS, PT are drawn to the rectangle under the fines of the angles Q, R, in which the two firft PQ, PR are drawn. In all other cafes the locus of the point P will be one of the three figures which pafs commonly by the name of the conic fections. But in room of the trapezium ABCD, we may fubftitute a quadrilateral figure whose two oppofite fides cross one another like diagonals. And one or two of the four points A, B, C, D may be fuppofed to be removed to an infinite diftance, by which means the fides of the figure which converge to thofe points, will become parallel: and in this cafe the conic fection will pafs through the other points, and will go the fame way as the parallels in infinitum. LEMMA XIX.

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To find a point P (Pl. 8, Fig. 8) from which if four right lines PQ, PR, PS, PT are drawn to as many other right lines AB, CD, AC, BD given by pofition, each to each, at given angles, the rectangle PQ × PR, under any two of the lines drawn, Shall be to the rectangle PS × PT, under the other two, in a given ratio.

Suppofe the lines AB, CD, to which the two right lines PQ, PB, containing one of the rectangles, are drawn to meet two other lines, given by pofition, in the points A, B, C, D. From one of thofe, as A, draw any right line AH, in which you would find the point P. Let this cut the oppofite lines BD, CD, in H and I; and, because all the angles of the figure are given, the ratio of PQ to PA, and PA to PS, and therefore of PQ to PS, will be alfo given. Subducting this ratio from the given ratio of PQ × PR to PS x PT, the ratio of PR to PT will be given; and adding the given ratios of PI to PR, and PT to PH, the ratio of PI to PH, and therefore the point P will be given. Q.E.I.

COR. 1. Hence also a tangent may be drawn to any point D of the locus of all the points P. For the chord PD, where