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 centripetal force tending to the focus of that figure. Pl. 5. Fig. 1.
Let CA, CB be the ſemi-axes of the hyperbola; PG, KD other conjugate diameters; PF a perpendicula 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 compleat the parallelogram QRPx. It is evident that EP is equal to the ſemi-tranſverſe axe 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 of PS, PI; that is, (becauſe of the parallels IH, PR, and the equal angles IPR. HPZ) if 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 LX QR to L x Iv 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
