Page:The Mathematical Principles of Natural Philosophy - 1729 - Volume 1.djvu/133



With the ame brevity with which we reduced the fifth problem to the parabola and hyperbola, we might do the like here: But becaue of the dignity of the problem and its ue: in what follows. I hall confirm the other caes by particular demontrations.

Suppoe 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 emi-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 emi-tranvere axe AC; for, drawing HI, from the other focus H of the hyperbola, parallel to EC, becaue CS, CH are equal, ES, EI will be alo equal; o that EP is the half difference of PS, PI; that is, (becaue 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 $$\textstyle \frac {2BC^2}{AC^2}$$) we hall have LX QR to L x  Iv as QR to Pv, or Px to Pv, that is, (becaue of the imilar triangles Pxv, PEC) as PE to PC, or AC to PC. And