TOL. VIII.] PHILOSOPHICAL TRANSACTIONS. 99 



served that in this case a parabola is to be described; which, however, is not to 

 be so understood, as if it could not be solved by an hyperbola, since it may ad- 

 mit both of an hyperbola and an ellipsis, even an infinite number, proceeding 

 by our method ; only this case admits of the parabola, which the other cases do 

 not. And the same limitation is to be made when he says, that his construction 

 takes place in every case when the problem is a solid one; for his meaning is, 

 that by a small alteration an hyperbola may always be found that will answer the 

 question : which will be evident on comparing the cases as above constructed 

 with his construction. But to return to the case of equality : and that I may 

 not seem to have made a rash assertion, you have not one, but two parabolas, 

 and opposite hyperbolas besides, that answer the problem. As before, let EB 

 be the given points, (fig. 5) and a circle from the centre A, also another 

 through the three points A, E, B, whose tangent is AO, and centre D. Draw- 

 ing the diameter NADX, constitute the three proportionals XA, NA, ZA, 

 the half of which is AL. Make these other three proportionals 20A, NA, 

 I A, the half of which is.KA, and complete the rectangle LAOV; also produce 

 LV to S, till VS be a third proportional to AI, OV; then with the axis SL, 

 the latus rectum A I, and vertex S, describe a parabola ; so will this cut the 

 circle in the points P, P, as required. The same thing may be effected by 

 another parabola, if, having completed the rectangle DAHC, and produced 

 KC to T, so that CT be a third proportional to AZ, DC, it be described 

 about the axis TK, with the vertex T, and latus rectum ZA; for this will meet 

 the circle in the same points P, P (fig. 6). But the construction by the opposite 

 sections is still easier ; for making, as before, the three proportionals XA, NA, 

 ZA, demit the perpendicular ZI a third proportional to 2AO and AN. Hence 

 ZI will be greater than ZA, since 2AO is less than XA ; then at the point I 

 let the right lines IQ, IM be inclined to IZ in half a right angle, and inde- 

 finitely produced both ways ; lastly, about these as asymptotes through A de- 

 scribe an hyperbola, and the other opposite one ; then the latter will answer the 

 problem for a convex speculum, and the former for a concave one. For since, 

 as was shown, ZI is always greater than ZA, the line IM will never pass 

 through A. Therefore a case will not here occur, in which from this con- 

 struction, as in the foregoing, the problem may be solved by the asymptotes 

 themselves : and yet this also sometimes admits of a plane locus, viz. when it 

 happens that the right line XO drawn to the centre D touches the circle NPP;. 

 for then the point of contact itself solves the question. And so much con- 

 cerning a problem, which has exercised the genius of many persons, the solu- 

 tion of which I completed some years since. 



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