ME. A. CAYLEY’S MEMOIE ON CEEVES OE THE THIED OEDEE. 
423 
But by the first of the equations (A) and the preceding value thus 
becomes ( — l~7^®+8Z®)a^. Hence throwing out the constant factor the coordinates of 
the point O are found to be 
7^. 
12. The points G, O are conjugate poles of the cubic. 
Take «, c for the coordinates of G, and b\ c' for the coordinates of O, we have 
a, b, (5'^—ayi, 
a',b',d~a,l , , yt. 
These values give aa! ^l{bd c) 
~al{yn~^l)^-l{^ri{^l—a7i)-\-yl{(x,l—yl)] 
— 4- + '/i\ — /a/3 ) + r( lay)-\-'^t{ — ci^ — ly'^) ; 
or substituting for their values in terms of a, (3, y, this is 
( ~ 7 y' “ 72 (ay + //3") 
+ (^~47a^( — Za/3) 
+ — 4a/3^(/ay) 
+ ( - y /3" — y ay) ( - a/3 — ly^), 
which is identically equal to zero. Hence, completing the system, we find 
ad-\-l{bd -\-b'c)—^, 
bb' 4-/(m'4-c'«)=0, 
cd +/(«/»'+«'/>)= 0, 
equations which show that O (as well as G) is a point of the Hessian, and that the 
points G, 0 are corresponding poles of the cubic. 
13. The line EF joining a pair of conjugate poles of the cubic is a tangent of the 
Pippian*. 
In fact, the equations (A), by the elimination of a, j3, y, give 
-?(P+^^+r)+(-i+4/^)?^2:=o, 
which proves the theorem. 
14. To find the equation of the pair of lines through F, and to show that these lines 
are tangents of the Pippian. 
ITe equation of the pair of lines considered as the first or conic polar of the conjugate 
pole E, is 
X(^+2/^2)+Y(?/^-}-2fo^)+Z(z^+2/^)=0. 
* Steiner’s curve Eq, in the particular case of a cubic basis-curve, is according to definition the envelope 
of the line EF, that is, the curve E^, in the particular case in question is the Pippian. 
