Page:The Mathematical Principles of Natural Philosophy - 1729 - Volume 1.djvu/187

 through which the conic ection ought to pas in this new figure; and compleating the parallelogram hikl, let the right lines hi, ik, kl be o cut in c, d, e, that be may be to the quare root of the rectangle ahb, ic to id, and ke to kd, as the um of the right lines hi and kl is to the um of the three lines. the firt whereof is the right line ik, and the other two are the quare roots of the rectangles ahb and alb; and c, d, e, will be the points of contact. For by the properties of the conic ections $$\scriptstyle hc^2$$ to the rectangle ahb, and $$\scriptstyle ic^2$$ to $$\scriptstyle id^2$$, and $$\scriptstyle ke^2$$ to $$\scriptstyle kd^2$$ and $$\scriptstyle cl^2$$ to the rectangle alb, are all in the ame ratio; and therefore hc to the quare root of ahb, ic to id, kc to kd, and el to the quare root of alb, are in the ubduplicate of that ratio; and by compoition in the given ratio of the um of all the antecedents hu + kl, to the um of all the conequents $$\scriptstyle \sqrt {ahb} + ik + \sqrt {alb}$$. Wherefore from that given ratio we have the points of contact c, d, c, in the new figure. By the inverted operations of the lat, lemma, let thoe points be tranferred into the firt figure, and the trajectory will be there decribed by prob. 14. Q. E. F. But according as the points a, b, fall between the points h, l, or without them, the points c, d, e, mut be taken either between the points b, i, k, l, or without them. If one of the points h, l, falls between the points h, l, and the other without the points h, l, the problem is impoible.