Page:The Mathematical Principles of Natural Philosophy - 1729 - Volume 1.djvu/136



And becaue SA is given, $$\scriptstyle SN^2$$ will be as PS.

And the concoure of any tangent PM with the right line SN, drawn from the focus perpendicular on the tangent, falls in the right line AN, that touches the parabola in the principal vertex.

If a body moves in the perimeter of a parabola: it it required to find the law of the centripetal force tending to the focus of that figure. Pl. 5. Fig. 3.

Retaining the contruction of the preceding lemma, let P be the body in the perimeter of the parabola; and from the place Q into which it is next to ucceed draw QR parallel and QT perpendicular to SP, as alo Qv parallel to the tangent, and meeting the diameter PG in v, and the ditance SP in x. Now, becaue of the imilar triangles Pxv, SPM, and of the equal ides SP, SM of the one, the ides Px or QR and Pv of the other will be alo equal. But (by the conic ections) the quare of the ordinate Qv is equal to the rectangle under the latus rectum and the egment Pv of the diameter, that is, (by lem. 13.) to the rectangle 4PS x Pv, or 4PS x QR; and the points P and Q coinciding, the ratio of Qv to Qx (by cor. 2 lem. 7.) becomes a ratio of equality. And therefore $$\scriptstyle Qx^2$$, in this cae. becomes equal to the rectangle 4PS x QR. But (becaue of the imilar triangles QxT, SPN) $$\scriptstyle Qx^2$$ is to $$\scriptstyle QT^2$$ as $$\scriptstyle PS^2$$ to $$\scriptstyle SN^2$$, that is (by cor. 1. lem. 14.) as PS to SA; that is, as 4PS x QT to 4SA x QR, and therefore