Page:Newton's Principia (1846).djvu/123

 $$\scriptstyle \frac{2BC^{2}}{AC}$$), we shall have L $$\scriptstyle \times$$ QR to L $$\scriptstyle \times$$ Pv as QR to Pv, that is, as PE or AC to PC; and L $$\scriptstyle \times$$ Pv to GvP as L to Gv; and GvP to Qv² as PC² to CD²; and by (Corol. 2, Lem. VII) the points Q and P coinciding, Qv² is to Qx² in the ratio of equality; and Qx² or Qv² is to QT² as EP² to PF², that is, as CA² to PF², or (by Lem. XII) as CD² to CB². And compounding all those ratios together, we shall have L $$\scriptstyle \times$$ QR to QT² as AC $$\scriptstyle \times$$ L $$\scriptstyle \times$$ PC² $$\scriptstyle \times$$ CD², or 2CB² $$\scriptstyle \times$$ PC² $$\scriptstyle \times$$ CD² to PC $$\scriptstyle \times$$ Gv $$\scriptstyle \times$$ CD² $$\scriptstyle \times$$ CB², or as 2PC to Gv. But the points Q and P coinciding, 2PC and Gv are equal. And therefore the quantities L $$\scriptstyle \times$$ QR and QT², proportional to these, will be also equal. Let those equals be drawn into $$\scriptstyle \frac{SP^{2}}{QR}$$, and L $$\scriptstyle \times$$ SP² will become equal to $$\scriptstyle \frac{SP^{2}\times QT^{2}}{QR}$$. And therefore (by Corol. 1 and 5, Prop. VI) the centripetal force is reciprocally as L $$\scriptstyle \times$$ SP², that is, reciprocally in the duplicate ratio of the distance SP.  Q.E.I.

Since the force tending to the centre of the ellipsis, by which the body P may revolve in that ellipsis, is (by Corol. 1, Prop. X.) as the distance CP of the body from the centre C of the ellipsis; let CE be drawn parallel to the tangent PR of the ellipsis; and the force by which the same body P may revolve about any other point's of the ellipsis, if CE and PS intersect in E, will be as $$\scriptstyle \frac{PE^{3}}{SP^{2}}$$ (by Cor. 3, Prop. VII.); that is, if the point S is the focus of the ellipsis, and therefore PE be given as SP² reciprocally. Q.E.I.

With the same brevity with which we reduced the fifth Problem to the parabola, and hyperbola, we might do the like here: but because of the dignity of the Problem and its use in what follows. I shall confirm the other cases by particular demonstrations.

Let CA, CB be the semi-axes of the hyperbola; PG, KD other conjugate diameters; PF a perpendicular to the diameter KD; and Qv an ordinate to the diameter GP. Draw SP cutting the diameter DK in E, and the ordinate Qv in x, and complete the parallelogram QRPx. It is evident that EP is equal to the semi-transverse axis AC; for drawing HI, from the other focus H of the hyperbola, parallel to EC, because CS, CH are equal, ES, EI will be also equal; so that EP is the half difference