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 parallel 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 intersect in E, will be as (by Cor. 3, Prop. VII.); that is, if the point S is the focus of the ellipsis, and therefore PE be given as reciprocally. 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 the 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 conjugate 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 QRPx. It is evident that EP is equal to the semi-transverse axis AC; for drawing HI, from the other focus H of the hyperbola, parallel to EC, because CS, CH are equal, ES, EI will be also equal; so that EP is the half difference[Pg 118] of PS, PI; that is (because of the parallels IH, PR, and the equal angles IPR, HPZ), of PS, PH, the difference of which is equal to the whole axis 2AC. Draw QT perpendicular to SP; and putting L for the principal latus rectum of the hyperbola (that is, for ), we shall have L × QR to L × Pv as QR to Pv, or Px to Pv, that is (because of the similar triangles Pxv, PEC), as PE to PC, or AC to PC. And L × Pv will be to Gv × Pv as L to Gv; and (by the properties of the conic sections) the rectangle GvP is to Qv2 as PC2 to CD2; and by (Cor. 2, Lem. VII.), Qv2 to Qx2, the points Q and P coinciding, becomes a ratio of equality; and Qx2 or Qv2 is to QT2 as EP2 to PF2, that is, as CA2 to PF2, or (by Lem. XII.) as CD2 to CB2: and, compounding all those ratios together, we shall have L × QR to QT2 as AC × L × PC2 × CD2, or 2CB2 × PC2 × CD2 to PC × Gv × CD2 × CB2, or as 2PC to Gv. But the points P and Q coinciding, 2PC and Gv are equal. And therefore the quantities L × QR and QT2, proportional to them, will be also equal. Let those equals be drawn into , and we shall have L × SP2 equal to . And therefore (by Cor. 1 and 5, Prop. VI.) the centripetal force is reciprocally as L × SP2, that is, reciprocally in the duplicate ratio of the distance SP. Q.E.I. The same otherwise. Find out the force tending from the centre C of the hyperbola. This will be proportional to the distance CP. But from thence (by Cor. 3, Prop. VII.) the force tending to the focus S will be as , that is, because PE is [Pg 119]given reciprocally as SP2. Q.E.I.