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Or rather, having taken AS equal to 2/3 of AT, I observe that 3/2 of GD ought to be equal to 3/2 of SB, plus BQ; and, deducting both of them from FD or FQ, that FD less 3/2 of GD ought to be equal to FB less 3/2 of SB. And this last difference is a given length: and all that is required is to draw the straight line FD from the given point F to meet VG so that it may be thus.
Or rather, by deducting from each the length of LG, which is also given, it will merely be needful to adjust FD up to the straight line VG in such a way that FD together with 3/2 of DG is equal to a given straight line, which is a quite easy plane problem: and the point D will be one of those through which the curve BDK ought to pass.
For it can similarly be demonstrated, by taking any other point in the curve, such as G, that the excess of AG over AD, namely VG, is to the excess of BD over DG, namely DP, in this same ratio of 3 to 2. And following this principle Mr. Des Cartes constructed these curves in his Geometric; and he easily recognized that in the case of parallel rays, these curves became Hyperbolas and Ellipses.
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