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PROBLEMS AND SOLUTIONS. 
[485 
10. To find where the line in question meets the cubic 
(A 2 4- y 2 4- 1) (Px + Qy) + lx 2 4- 2 mxy 4- oiy 2 = 0, 
writing in this equation 
x = — A + DD, y = — B 4- ED, 
we have for the determination of il the equation 
[A 2 + B 2 +1 - 2 (AD + BE) fl + (D 2 + E 2 ) fl 2 } x 
{- AP-BQ+ (DP + EQ) H} + (l, m, n\- A + DD, -B + ED) 2 = 0, 
or observing that we have AP + BQ = 0, this equation becomes 
(D 2 + E 2 ) (DP + EQ) D 3 
+ {- 2 (AD 4- BE) (DP + EQ) + ID 2 + 2mDE + nE 2 } D 2 
+ { (A 2 + B 2 + 1) (DP + EQ) -2lAD-2m(AE+BD)-2nBE}D 
4- { lA 2 4- 2mAB + nB 2 } = 0. 
11. Substituting for A, B, D, E their values in terms of (P, Q, a, 0), we find 
DP + EQ= -6 2 (nP 2 - 2mPQ + IQ 2 ), 
IA 2 + 2mAB + nB 2 = (nP 2 - 2mPQ + IQ 2 ), 
IAD + m (AE + BD) + nBE = -1«<9 2 (nP 2 - 2mPQ + IQ 2 ), 
ID 2 + 2mDE + nE' 2 = ((nl — m 2 ) O' 1 + a 2 0 2 ) (nP 2 — 2mPQ + IQ 2 ), 
and substituting these values in the equation for D, the whole equation divides by 
0 2 (nP 2 — 2mPQ + IQ 2 ), and it then becomes 
4 (D 2 + E 2 ) D 3 + 4 ; {-2(AD + BE) - (oil - on 2 ) 0 2 - a 2 } O 2 + 4 {A 2 + B 2 + 1 - a} D - 1 = 0, 
or, putting for shortness 
C'=C-A 2 -B 2 , = a _l-A 2 -B\ 
F' = F- 2 (AD + BE), =-a 2 -6 2 (oil - on 2 ) - 2 (AD + BE), 
the equation in O is 
4 (D 2 + E 2 ) D 3 + 4F'D 2 - 4 CD -1 = 0, 
so that, il satisfying this equation, the intersections of the axis with the cubic are given 
by x = — A + DD, y — — B + ED. 
12. The equation of the Cartesian, writing therein x+A=u and y + B — v, and 
attending to the values of G' and F', is 
(u 2 4- v 2 + CJ 4- 2Du 4- 2Ev 4- F' = 0. 
And to find the foci, writing in this equation u + p, v + ip in place of u, v, we find 
[u 2 4- v 2 4- G' 4- 2 (u 4- vi) p} 2 + 2 (D + Ei) p 4- 2Du 4- 2Ev + F' = 0,