Page:The Mathematical Principles of Natural Philosophy - 1729 - Volume 1.djvu/119

 PR in Z; and latly, thro' the point Q draw LR parallel to SP, meeting the circle in L, and the tangent PZ in R. And, becaue of the {{ls]]imilar triangles ZQR, ZTP, VPA, we hall have $$\scriptstyle RP^2$$, that is, QRL, to $$\scriptstyle QT^2$$, as $$\scriptstyle AV^2$$ to $$\scriptstyle PV^2$$. And therefore $$\textstyle \frac {QRL \times PV^2}{AV^2}$$ is equal to $$\scriptstyle QT^2$$. Multiply thoe equals by $$\textstyle \frac {SP^2}{QR}$$ and the points P and Q coinciding, for RL write PV; then we {{ls}}hall have $$\textstyle \frac {SP^2 \times PV^2}{AV^2} = \frac {SP^2 \times QT^2}{QR}$$. And therefore (by cor. 1. and 5. prop. 6.) the centripetal force is reciprocally as $$\textstyle \frac {SP^2 \times PV^2}{AV^2}$$, that is (becau{{ls}}e $$\scriptstyle AV^2$$ is given) reciprocally as the {{ls}}quare of the di{{ls}}tance of altitude SP, and the cube of the chord PV conjunctly. Q. E. I.

{{center|The {{ls}}ame otherwi{{ls}}e}}

On the tangent PR produced, let fall the perpendicular ST: and (becau{{ls}}e of the {{ls}}imilar triangles STP, VPA) we {{ls}}hall have AV to PV as SP to ST, and therefore $$\textstyle \frac {SP \times PV}{AV} = ST$$, and $$\textstyle \frac {SP^2 \times PV^2}{AV^2}$$, that is, (becau{{ls}}e AV is given) reciprocally as $$\scriptstyle SP^2 \times PV^2$$. Q. E. I.