108 
ADDITION TO MU WALTON S PAPER 
[576 
same sign ; that is, the point on the qnartic curve must lie within the triangle. 
Hence, when in the quartic curve the oval becomes a conjugate point, the octic curve 
has no real branch, but it consists wholly of conjugate points ; viz. it consists of the 
points A, B, C as conjugate points; two imaginary conjugate points answering to the 
point a of the figure, two other imaginary conjugate points answering to the point y ; 
and two conjugate points answering to the point /3, these last being not ordinary 
conjugate points, but conjugate tacnodal points, or points of contact of two imaginary 
branches of the curve. 
The case in question, ¡3 a conjugate point on the quartic curve, answers to 
46 2 + ay 
Mr Walton’s critical value of sec 2 0, viz. in the present notation sec 2 0 = . To 
ay 
show this I consider the intersection of the curve by the line yz — ax = 0; and I 
write for convenience 7z — ojx — 7au, that is, x = 7u, z — au. Substituting these values, 
the equation divides by yu, or omitting this factor it is 
a 3 7 3 /3 2 sec 2 0 . u [y + (a + 7) u) 
= a 3 {yu (be — a- — aa) + a(3y}- 
+ ¡3-ay. uy (ca — b 2 ) 2 
+ y 3 {au (ab - c- + cy) - efiy] 3 , 
or observing that we have a + 7 = - /3, be - a 2 - aa = ftS, ab -c 2 +cy = - 8/3, this becomes 
a 3 7 3 sec 2 0u (y — /3u) 
= a 3 (7£u + ay)- 
+ ay (ca — 6 2 ) 2 uy 
+ 7 3 (aSu + cy)-, 
viz. this is 
u 2 (a :! 7 2 £" 2 + a 2 7 3 8 2 + a A y 3 sec 2 0 . ¡3] 
-f uy {2a 3 a7^ + 27 3 ca8 + ay (ca — 6 2 ) 2 — oty 5 sec 2 0} 
+ y 2 (a 2 a 3 + c 2 7 3 ) = 0.