which produce to V, v, ſo that TV, tv may be equal to TS, tS. Bifect Vv in O, and erect the indefinite perpendicular OH, and cut the right Line VS infinitely produced in K and k, ſo that VK be to KS, and Vk to kS as the principal axe of the trajectory to be deſcribed is to the diſtance of it's foci. On the diameter Kk deſcribe a circle cutting OH in H; and with the foci S, H, and principal axe equal to VH, deſcribe a trajectory. I ſay the thing is done. For, biſecting Kk in X, and joining HX, HS, HV, Hv, becauſe VK is to KS, as Vk to kS; and by compoſition, as Vk + Vk to KS + kS; and by diviſion as Vk - VK to kS - KS that is, as 2VX to 2KX and 2KX to 2SX, and therefore as VX to HX and HX to SX the triangles VXH, HXS will be ſimilar; Therefore VH will be to SH, as VX to XH; and therefore as VK to KS. Wherefore VH the principal axe of the deſcribed trajectory has the ſame ratio to SH the diſtance of the foci, as the principal axe of the trajectory which was to be deſcribed has to the diſtance of its foci; and is therefore of the ſame ſpecies. And ſeeing VH, vH, are equal to the principal axe, and VS, vS are perpendicularly biſected by the right lines TR, tr; 'tis evident (by lem. 15.) that thoſe right lines touch the deſcribed trajectory. Q. E. F.
Case 3.. About the focus S (Pl. 7. Fig. 6.) it is required to deſcribe a trajectory, which ſhall touch a right line TR in a given point R. On the right line TR let fall the perpendicular ST; which produce to K ſo that TV maybe equal to ST, join VR, and cut the right line VS indefinitely produced in K and k, ſo that VK may be to SK, and Vk to Sk as the principal axe of the ellipſis to be deſcribed, to the diſtance of itſfoci; and on the diameter Kk deſcribing
