330 PHILOSOPHICAL TRANSACTIONS. [anNO 1670, 



KL, and the angles SKL, SLK. Again, in the triangle KLP, are given the 

 side KL, the angle K PL, the difference of the observed longitudes of the 

 planet, and PKL, the difference of the angles SKLlast found, and SKP the 

 elongation of the planet from the sun in the first observation, to find LP: then, 

 in the triangle LSP, the sides LS, LP, and the angle P L S, the elongation of the 

 planet from the sun in the second observation, are given, to find the side SP, and 

 the angle LSP; which being found, then as SP is to LP, so is the tangent of the 

 latitude observed from L, to the tangent of the inclination or latitude at the sun ; 

 and as the cosine of the inclination is to the radius, so is SPthe curtate distance, 

 to the planet's true distance from the sun. And thus we find the position and 

 length sought. Now it remains to show, how from three given distances from 

 the sun, together with the intercepted angles, the mean distance may be found, 

 with the eccentricity of the ellipsis. 



Let S, tig. 8, be the sun, and SA, SB, SC three distances in due position; 

 and drawing AB, BC, let AB be the distance of the foci of an hyperbola, and 

 SA — SB = EH the transverse diameter: which being supposed, let the hy- 

 perbolic line be described, whose internal focus is the point A, the extremity of 

 the longer line S A: after the same manner, let B and C be the foci of the other 

 hyperbola, whose diameter SB — SC = KL; from which let the hyperbolic 

 line be described, having its internal focus in the point B. I say these two hy- 

 perbolas, thus described, intersect each other in the point F, which is the other 

 focus of the ellipsis; and drawing the line FA, FB, or FC ; SA -f- FA, SB + 

 FB, or SC + FC will be equal to the transverse diameter, and SF is the distance 

 of the foci ; which being supposed, the description of the ellipsis is very easy. 

 But since the reason of this construction is not clear to every one, it will not be 

 improper to illustrate it a little; therefore I say, that from the known property 

 of the ellipsis, SB + FB = SA + FA, and transposing the members of the 

 equation, FB — FA=:SA — SB; so that though FB and FA be miknown, 

 yet their difference is equal to SA — SB, that is, EH; and since from the 

 nature of the hyperbola, any two lines drawn from the foci to any point of the 

 curve constantly differ by the quantity of the transverse diameter ; it is plain 

 that the point F is somewhere in the curve of the hyperbola, whose transverse 

 diameter is equal to S A — SB, and the foci A and B. After the same manner 

 it may be demonstrated that the point F is in the hyperbola, whose diameter is 

 SB — SC, and foci B and C. Therefore it must necessarily be in the intersection 

 of these two hyperbolas, which since they intersect each other in one point only, 

 clearly show where the other focus of the required ellipsis is. 



Now to perform the same analytically, suppose it done, and let FB = a, S A 

 -SB = FB- FA = /', AB='c, SB-SC= FC - FB = f/, BC =/, and 



