VOL. VIII.] PHILOSOPHICAL TRANSACTIONS. lOl 



their equations may be discerned by only the variations of the signs + ^nd — . 

 One case only is excepted, viz. when d AB is a right angle; the equation for 

 which is obtained, after expunging out of the former equation all the terms in 

 which n is found, since it vanishes, viz. the equation 2zbaa — qqba = bzqq 

 — zqqe; or, substituting {or2zbaa its value zbqq — qqba = 2zbee — 

 zqqe. 



But it is to be observed, that though by referring the analysis to the line DA, 

 two hyperbolas offer themselves in the equation ; and as many more different 

 from the former, when the calculation is referred to dA; yet in both cases the 

 very same parabola arises : the reason of which you will easily perceive on a little 

 consideration. 



Let us now. Sir, apply the foregoing analysis to all the problems which are 

 usually proposed about the reflexion of spherical specula. Let there be then, as 

 before, a circle, whose centre is A, (fig. 8), Da given point, from which DE is 

 an incident ray, and EQ its reflected one. Joining DA, to which draw the tan- 

 gent EC, and the perpendicular EI, also the line QEB produced; then denot- 

 ing the parts as before, viz. DA = z, C A = x, AE = q, BA = y, AI = a, 

 IE = e; because the three DA, CA, BA are harmonical proportionals, and 

 the three C A, AE, A I are geometricals, we have always the equation z/ =. 



Zqq 



• , on whatever point of the circle the ray DE falls. So that, if the 



Zza — qq' ... 



point E be required, on which if the ray DE fall, it may be reflected parallel to 

 the diameter LAV, a perpendicular to DA; the reflected ray QE produced 

 will pass through I, as is plain ; and I and B will coincide. Hence a ■=. y z=z 



'^n ^^ „„ 9^« _ 



-, or a a — 



4 qq, and the problem is solved by plain loci. 



220 — qq 2z 



If the point be required from which a ray may be reflected parallel to any 

 other line, as AK, drawn from the centre A; from the point L draw the tangent 

 KL = c?; then it is plain that the triangles AKL, EIB will be similar, since all 

 the sides of the one are parallel to all those of the other. Therefore AL is to 

 LK as EI to IB, or q -.d '.-. e:a — y, and ^"~ = ?/ = r—^ — ; hence zq^ 



■' ^^ q '^ 2 za — qq ^ 



= 2 qzaa — 2 zdae — q^a -{- qqde ; or, putting qq — ee for aa, zq^ = 2 zq^ 



— 2 zqee — 2 zdae — q^a -\- qqde. And each equation belongs to an hyper- 

 bola about the asymptotes, which with the given circle solves the problem. 



Let it be now proposed to make the reflected ray pass through the given 

 point N, as in Alhazen's problem, or that being produced towards the point of 

 reflexion E, it may pass through the given point N. From N draw NO = n 

 perpendicular to AL, and put AO = Z?. Then it is plain that A O will be to the 

 difference between ON and AB_, as EI to IB; that is, b :n — y :: e : a — y. 



