Mr. Lubbock on a Property of the Parabola. 10 3 



{^1 + ^2 + ^3+ Y -2^+ (j/1 + 3/2+^3-2 Y)cosxi/J 



[•^l3/2-^2i/l+^3i'l--^li/3 + ^2j/3— ^33/2} = 0« (7-) 



Unless the three points (^'1,3/1), ('^2 > 3/2)5 (^3 » 3/3) are in 

 the same straight line 



^1+^2 + ^3+ f -2X+ (3/1+3/2+3/3-2 y) cos a; 3/ = 0. (8.) 



This equation is general, but simplifications result if 3/, = ; 

 (which supposition does not limit the generality of the solu- 

 tion of the problem proposed;) and in this case by p. 101, line 

 18, ^2 + ^'3 = 0, a\ = — 0^3, equation (8.) becomes 



^1 + |--2:t + (3/2 + 3/3-2 Y)cos^3^ = 0. (9.) 



and by equation (7.)> 



((^i-|-) (3/2 + 3/3) + i' ( Y'+ X cos .r3/) = 0. (10.) 



If Ax^ + Bxy + C3/"- + Dx ^Ey + F = 0, 

 is the equation to any curve of the second order, the free re- 

 sult generally from the intersection of the lines whose equa- 

 tions are, 



{B^-A^A C) {x'^-f) + {2B E-4> D C) x— (2 BD-^EA) y 



+ E^-iCF-D^ + 4^AF = 0. 



{B^-iACi {x^-y^) + {BD^2AC)x + {BD-2 DC)y 



+ 2BF-ED- {{B'-iACjy^ + {2BD-4>AE)y 



+ D- — iAE} cosxy = 0. (See Phil. Mag. Aug. 1831.) 



If the equation to the curve is y- = 2p x, in which case 

 A = 0, B = 0, C = I, D = - 2 p, £ = 0, 2^ = 0, it is 

 easy to deduce from the equations given above, or to prove 

 otherwise, that the focus results from the intersection of the 



straight lines whose equations are .r = ^, and 3/ = — p cos xy. 



Let ^'4,3/4 denote the coordinates of the intersection of the 

 circumscribing circle passing through (a-j , 3/1), {x^, 3/3), (^35^3) 

 with the line „ „ 



In order to prove that the circle passes through the focus 

 of the parabola it is sufficient to show that y^= — p cos xy. 

 By the equation to the circle 



(f - X)+{y,-Yf+2[^ -X) (3/4- Y) 008^3/ = B} 

 {x,-XY+ Y^-2{x,-X) Yco%xy =72* 



