III. OF NATURAL PHILOSOPHY. 117 L X ^R t0 L X Py aS ^R t0 PV that 1S> US PE or AC to PC ; and L X Pv to GvP as L to Gy ; and GvP to Qi>2 as to CD- ; and by (Corol. 2, Lem. VII) the points Q, and P coinciding, Qv* is to Q,r- in the ratio of equality ; and Q,.r2 or Qv2 is to Q,T2 as EP2 to PF2, that is, as CA2 to PF2, or (by Lem. XII) as CD'2 to CB2. And com pounding all those ratios together, we shall have L X QR to Q,T2 as AC X L X PC2 X CD2, or 2CB2 X PC2 X CD2 to PC X Gv X CD2 X CB2, or as 2PC to Gv. But the points Q and P coinciding, 2PC and Gr are equal. And therefore the quantities L X QR and Q,T2, proportional SP2 to these, will be also equal. Let those equals be drawn into-p^B"? and L SP2 X QT2 X SP2 will become equal to -- ^p — — . And therefore (by Corol. 1 and 5, Prop. VI) the centripetal force is reciprocally as L X SP2, that is, re ciprocally in the duplicate ratio of the distance SP. Q.E.I. The same otherwise. Since the force tending to the centre of the ellipsis, by which the body P may revolve in that ellipsis, is (by Corol. 1, Prop. X.) as the distance CP of the body from the centre C of the ellipsis ; let CE be drawn paral lel to the tangent PR of the ellipsis : and the force by which the same body P may revolve about any other point S of the ellipsis, if CE and PS inPE3 tersect in E, will be as ^T3, (by Cor. 3, Prop. VII.) ; that is, if the point S is the focus of the ellipsis, and therefore PE be given as SP2 recipro cally. Q.E.I. With the same brevity with which we reduced the fifth Problem to the parabola, and hyperbola, we might do the like here : but because of the dignity of the Problem and its use in what follows, I shall confirm the other cases by particular demonstrations. PROPOSITION XII. PROBLEM VII. Suppose a body to move in an hyperbola ; it is required to find lite law of the centripetal force tending to the focus of that figure. Let CA, CB be the semi-axes of the hyperbola ; PG, KD other con jugate diameters ; PF a perpendicular to the diameter KD ; and Qv an ordinate to the diameter GP. Draw SP cutting the diameter DK in E, and the ordinate Qv in x, and complete the parallelogram QRP.r. It is evident that EP is equal to the semi-transverse axis AC ; for drawing HE, from the other focus H of the hyperbola, parallel to EC, because CS, TH are equal, ES El will be also equal ; so that EP is the half difference