Page:The Mathematical Principles of Natural Philosophy - 1729 - Volume 1.djvu/124

the ratio of $\scriptstyle PC^{2}$ to $\scriptstyle CD^{2}$ and of the ratio of $\scriptstyle PC^{2}$ to $\scriptstyle PF^{2}$, that is, vG to $\textstyle {\frac {QT^{2}}{Pv}}$ as $\scriptstyle PC^{2}$ to $\textstyle {\frac {CD^{2}\times PF^{1}}{PC^{2}}}$. Put QR for Pv, and (by lem. 11.) BC x CA for CD x PF PM-C, alſo (the points P and Q coinciding,) 2PC for vG; and multiplying the extremes and means together, we ſhall have $\textstyle {\frac {QT^{2}\times PC^{2}}{QR}}$ equal to $\textstyle {\frac {2BC^{2}\times CA^{2}}{PC}}$. Therefore (by cor. 5. prop. 6.) the centripetal force us reciprocally as $\textstyle {\frac {2BC^{2}\times CA^{2}}{PC}}$; that is (becauſe $\scriptstyle 2BC^{2}\times CA^{2}$ is given) reciprocally as $\textstyle {\frac {1}{PC}}$; that is, directly as the diſŧance PC. Q. E. I.

The ſame otherwiſe.

In the right line PG on the other ſide of the point T; take the point u ſo that Tu may be eqal to Tv; then take uV, ſuch as ſhall be to vG as $\scriptstyle DC^{2}$ to $\scriptstyle PC^{2}$. And becauſe $\scriptstyle Qc^{2}$ is to PvG as $\scriptstyle DC^{2}$ to $\scriptstyle PC^{2}$ (by the conic ſections) we ſhall have $\scriptstyle Qv^{2}=Pv\times uV$. Add the rectangle uPv to both ſides, and the ſquare of the chord of the arc PQ will be equal to the rectangle VPv; and therefore a circle, which touches the conic ſection in P, and paſſes thro' the point Q will pals alſo thro' the point V. Now let the points P and Q meet, and the ratio of uV to vG, which is the ſame with the ratio of $\scriptstyle DC^{2}$ to $\scriptstyle PC^{2}$, will become the ratio of PV to PG or PV to 2PC; and therefore PV will be equal to $\textstyle {\frac {2DC^{2}}{PC}}$. And therefore the force, by which the body P revolves in the ellipſes, will be reciprocally as