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

Cor. 2. And becauſe SA is given, $\scriptstyle SN^{2}$ will be as PS.

Cor. 3. And the concourſe 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.

Proposition XIII. Problem VIII.

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.

Plate 5, Figure 3

Retaining the conſtruction 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 ſucceed draw QR parallel and QT perpendicular to SP, as alſo Qv parallel to the tangent, and meeting the diameter PG in v, and the diſtance SP in x. Now, becauſe of the ſimilar triangles Pxv, SPM, and of the equal ſides SP, SM of the one, the ſides Px or QR and Pv of the other will be alſo equal. But (by the conic ſections) the ſquare of the ordinate Qv is equal to the rectangle under the latus rectum and the ſegment 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 caſe. becomes equal to the rectangle 4PS x QR. But (becauſe of the ſimilar 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