The collected mathematical papers of Arthur Cayley.

108

ADDITION TO MR WALTON S PAPER

[576

same sign; that is, the point on the quartic 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
C
a
point a of the figure, two other imaginary conjugate points answering to the point y;
and two conjugate points answering to the point 83, 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, 38 a conjugate point on the quartic curve, answers to
462 + dl T
Mr Walton's critical value of sec20, viz. in the present notation sec2 0= 4b2 + To
acy
show this I consider the intersection of the curve by the line yz -ax= 0; and I
write for convenience yz = ax = 7yau, that is, x = y7, z = au. Substituting these values,
the equation divides by yu, or omitting this factor it is
a373s2 sec2 0. u {y + (a + 7) u}
= y3 {u (be - a2- aa) + a/3y}2
+ 2aLy. ny (ca - b2)2
+ y3 {au (ab - c2 + cy) - c/3y}2,
or observing that we have a + y = - A, be - a2 - at = /3, ab - c2+ cy = - y 83, this becomes
a y3 sec2 tO (y - /u)
= a3 (7'u + ay)2
+ ay (ca -b2)2 uy
+ 73 (a8u + cy)2,
viz. this is
u2 {a3y2~2 + a0y382 + a373 sec2 0. /3}
+ uy {2a3ay7 + 27yca8 + ay (ca - b2)2 - a373 sec2 0}

+ y2 (a2a3 + C2y3) = 0.