Page:The Mathematical Principles of Natural Philosophy - 1729 - Volume 1.djvu/214

 perpendicular on the axis; and drawing PH, there will be $$\scriptstyle AG^2 + GH^2$$$$\scriptstyle(= HP^2 = \overline {AO - AG} \vert ^2 + \overline {PO - GH} \vert ^2)$$$$\scriptstyle = AO^2 + PO^2 - 2 GAO$$$$\scriptstyle + 2GH + PO + AG^2 + GH^2$$ Whence $$\scriptstyle 2GH \times PO $$$$\scriptstyle (= AO^2 + PO^2 - 2GAO) $$$$\scriptstyle= AO^2 + \frac 34 PO^2$$. For $$\scriptstyle AO^2$$ write $$\scriptstyle AO \times \frac {PO^2}{4AS}$$; Then dividing all the term by 3PO and multiplying them by 2AS, we hall have $$\scriptstyle \frac 43 GH \times AS $$$$\scriptstyle(= \frac 12AO \times PO + \frac 12AS \times PO =$$$$\scriptstyle \frac {AO + 3AS}{6} \times PO =$$$$\scriptstyle \frac {4AO - 3 SO}{6} \times PO = $$ to the area of $$\scriptstyle \overline {APO - SPO} \vert = $$ to the area APS but GH was 3M and therefore $$\scriptstyle \frac 43 GH \times AS$$ is $$\scriptstyle 4AS \times M$$ Wherefore the area cut of APS is area that was, to he cut of 4ASxM. Q. E. D.

Hence GH is to AS, as the time in which the body decribed the arc AP to the time in which the body decribed the arc between the vertex A and the perpendicular erected from the focus S upon the axis.

And uppoe a circle ASP perpetually to pas through the moving body P, the velocity of the point H, is to the velocity which the body had in the vertex A. as 3 to 8; and therefore in the ame ratio is the line GH to the right line which the body, in the time of its moving from A to P, would decribe with that velocity which it had in the vertex A.

Hence alo, on the other hand, the time may be found, in which the body has decribed any aigned arc AP. Join AP on its middle point erect a perpendicular meeting the right line GH in H.