Page:The Mathematical Principles of Natural Philosophy - 1729 - Volume 1.djvu/134

 L x Pv will be to Gv x Pv as L to Gv; and (by the properties of the conic ections) the rectangle GvP is to $$\scriptstyle Qv^2$$ as $$\scriptstyle PC^2$$ to $$\scriptstyle Cd^2$$; and (by cor. 2. lem. 7.) $$\scriptstyle Qv^2$$ to $$\scriptstyle Qx^2$$, the points Q and P coinciding, becomes a ratio of equality; and $$\scriptstyle Qx^2$$Qx or $$\scriptstyle Qv^2$$ is to $$\scriptstyle QT^2$$ as $$\scriptstyle EP^2$$ to $$\scriptstyle PF^2$$, that is, as $$\scriptstyle CA^2$$ to $$\scriptstyle PF^2$$, or (by lem. 12.) a $$\scriptstyle CD^2$$ to $$\scriptstyle CB^2$$: and, compounding all thoe ratio's together, we hall have L x QR to $$\scriptstyle QT^2$$ as $$\scriptstyle AC \times L \times PC^2 \times CD^2$$ or $$\scriptstyle 2CB^2 \times PC^2 \times CD^2$$ to $$\scriptstyle PC \ times Gv \times CD^2 \times CB^2$$, or as 2PC to Gv. But the points P and Q coinciding, 2PC and Gv are equal. And therefore the quantities L x QR and $$\scriptstyle QT^2$$, proportional to them, will be alo equal. Q. E. I.

Let thoe equalbe drawn into$$\textstyle \frac {SP^2}{QR^2}$$, and we hall have $$\scriptstyle L \times SP^2$$ to $$\textstyle \frac {SP^2 \times QT^2}{QR}$$. And therefore (by cor. 1 & 5. prop. 6.) the centripetal force is reciprocally as $$\scriptstyle L \times SP^2$$, that is, reciprocally in the duplicate ratio of the ditance SP. Q. E. I.

The ame otherwie.

Find out the force tending from the centre C of the hyperbola. This will be proportional to the ditance CP. But from thence (by cor. 3. prop. 7.) the force tending to the focus S will be as $$\textstyle \frac {PE^2}{SP^2}$$, that is, becaue PE is given, reciprocally as$$\scriptstyle SP^2$$. Q. E. I.

And the ame way it may be demonŧrated, that the body having its centripetal changed into a centrifugal force, will move in the conjugate hyperbola.