48 5 J 
PROBLEMS AND SOLUTIONS. 
561 
that is 
(u 2 + V 2 + O') 2 4- 2Du + 2Ev + F' + {2 {u + vi) (u 2 + v 2 + O') + D + Ei) 2p 4- 4 (u + m) 2 p 2 = 0. 
Expressing that this equation in p has two equal roots, we find 
•1 (u + vi) 2 {('u 2 + v 2 + C') 2 + 2Du + 2Ev + F'} — {2 (u 4- vi) {u 2 + v 2 + C") + i) + i^'} 2 = 0, 
that is 
4 (2Dw 4- 2Ev 4- F') (u + vi) 2 — 4 (u 2 + v 2 + C) (u + vi) (D + Ei) — (D — Ei) 2 = 0, 
which equation is in fact the equation of the three tangents from one of the circular 
points at infinity. Writing it under the form U + Vi = 0, the nine foci of the 
Cartesian are given as the intersections of the two cubics TJ = 0, V = 0. But of these 
nine points, three, the foci that we are concerned with, lie on the axis, or line 
Eu — Bv = 0 ; in fact, we have 
U = 4 (u 2 — v 2 ) (2Du 4- 2Ev 4- F) 
— 4 (uD — vE) (u 2 4- v 2 4- O') 
- (D 2 - E 2 ), 
and hence 
V = 8uv (2Du+ 2Ev 4- F') 
— 4 (uE 4- vD) (u 2 + v 2 4- O') 
-2 DE; 
2DEU — (D 2 — E 2 ) V—{Eu — Dv) {8 {Du 4- Ev) (2Du 4- 2Ev + F') — 4 {D 2 + E 2 ) {u 2 + v 2 4- 0')} = 0, 
which shows that the nine points lie three of them on the line Eu — Dv = 0, and the 
remaining six on the conic 
2 {Du + Ev) (2Du + 2 Ev + F') — {D 2 4- E 2 ) {u 2 4- v 2 4- O') = 0. 
13. We have thus the three foci given as the intersections of the axis Eu — Dv = 0, 
with the cubic 
U =4 {u 2 — v 2 ) (2Du 4- 2Ev 4- F') — 4 {uD — vE) {u 2 4- v 2 4- O') — {D 2 — E 2 ) = 0; 
or, writing in this last equation u = Dll, v = ECl, that is x = - A 4- DZl, y = — B 4- EOl, 
we have 
u 2 — v 2 = {D 2 — E 2 ) il 2 , uD — vE = {D 2 — E 2 ) fl. 
The whole equation divides by {D 2 — E 2 ), and omitting this factor, it is 
4H 2 [2 {D 2 + E 2 ) O 4- F'} - 4il \{D 2 + E 2 ) H 2 + O') - 1 = 0, 
that is 
4 (D 2 + E 2 ) n 3 + 4^'fl 2 - 4C'il -1 = 0, 
the same equation as the equation in O before obtained; that is the intersections of 
the cubic with the axis are the three foci of the Cartesian. 
[Vol. vi. pp. 57—59.] 
1949. (Proposed by Professor Cayley.)—Find the conic of five-pointic intersection 
at any point of the cuspidal cubic y z = x 2 z. 
C. V1T. 
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