— -—’ 
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A MEMOIR ON CURVES OF THE THIRD ORDER. 
391 
13. The line EF joining a pair of conjugate poles of the cubic is a tangent of 
the Pippian 1 . 
In fact, the equations (A), by the elimination of a., /3, 7, give 
— I (| 3 + v 3 +£ 3 ) + ( — 1 + 4l 3 ) = 0, 
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. 
The equation of the pair of lines considered as the first or conic polar of the 
conjugate pole E, is 
X (x 2 + 2lyz) + Y (y 2 + 2Izx) + Z (z 2 + 2Ixy) = 0. 
Let one of the lines be 
Xx + y,y + vz = 0, 
then the other is 
and we find 
X Y Z A 
--x + -y + -z = 0; 
X fl ^ V 
2lXyv — Yv 2 — Zy 2 = 0, 
— Xv 2 + 2 lYvX — Z\ 2 = 0, 
— X/jl 2 — Fx 2 + 2 IZyv = 0, 
any two of which determine the ratios A,, y, v. 
The elimination of X, F, Z gives 
2lyv , — v 2 , — /i 2 =0, 
— v 2 , 2lv\, — X 2 
— fx 2 , — X 2 , 2 IXy 
which is equivalent to 
Xyv { — l (A, 3 + y? -f v 3 ) + ( — 1 + 4<l 3 ) Xyv) = 0 ; 
or, omitting a factor, to 
— I (A 3 + [F + v 3 ) + ( — 1 + 4>l 3 ) X/xv = 0, 
which shows that the line in question is a tangent of the Pip2Dian. 
15. To find the equation of the pair of lines through 0. 
The equation of the pair of lines through E is in like manner 
X' (x 2 + 2 lyz) + Y' (y 2 + 2Izx) + Z' (z 2 + 2Ixy) = 0; 
1 Steiner’s curve JR 0 , in the particular case of a cubic basis-curve, is according to definition the envelope 
of the line EF, that is, the curve B 0 in the particular case in question is the Pippian.