PROBLEMS AND SOLUTIONS. 
583 
[383 
383] 
ie three 
mtioned 
rve, and 
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conics 
nj ugate 
(A', B), 
of the 
> given 
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given 
e two 
ce the 
find a 
I, BJ 
points. 
) may 
in a 
i that 
there exists a conic touching the tangents from A, B to a given conic ©, and besides 
passing through three given points. It is rather more convenient to consider the 
reciprocal question—though it is to be borne in mind that for any two reciprocal 
questions a solution of the one question by means of coordinates (x, y, z) regarded as 
point-coordinates is in fact a solution of the other question by means of the same 
coordinates (x, y, z) regarded as line-coordinates. The reciprocal question is: through 
a given point to draw a line A, and through another given point to draw a line B, 
such that there exists a conic passing through the intersections of these lines with a 
given conic ©, and besides touching three given lines. The given points may be taken 
to be (x = 0, z = 0), (y = 0, z = 0); this determines the line z = 0, but not the lines 
x = 0, y — 0, so that the point (x = 0, y = 0) may without loss of generality be supposed 
to lie on the conic ©; the equation of this conic will therefore be 
(a, b, 0, /, g, h\x, y, z)~ = 0. 
I take az + yx = 0 for the equation of the line A, ¡xy + vz = 0 for the equation of the 
line B (so that the quantities to be determined are the ratios a : y and ¡x : v); this 
being so, the required conic' passes through the intersections of these lines with the 
conic ©; its equation will therefore be 
{a, b, 0, f g, li$x, y, z) 2 + 2 (ax + yz) (/xy + vz) = 0 ; 
or what is the same thing 
(a, b, 2vy, f+ fxy, g + va, h + fic^x, y, zf = 0 ; 
where a, y, ¡x, v have to be determined in such manner that this conic may touch 
three given lines. It is to be observed that a, y, ¡x, v, enter into the equation 
through the combinations a/x, a : y, and ¡x : v, so that there are really only three 
disposable quantities. 
The condition in order that the conic may touch a line %x + rjy + £z = 0 is 
( 2bvy — (/+ fxy) 2 , 2avy — (g + va) 2 , ab — (h + fxu)i 2 , ] 
iff +va)(h +yaa)- a(/+fi y), 
(h - fxa) (/ + fxy) - b(g + va), 
, (/ + py) (9 + va )~ ( h + ^ a ) 
that is, putting for shortness G = ab — h 2 , F — gh—aj, G — hf—bg, and reversing the 
sign of the whole expression, 
{ / 2 r+ gw- -2Frt- 2g&- 2fgW 
+ 2fi { fy£ 2 + ha£ 2 + (ay - ga) v £- (hy +fa) ££ - gyfr] 
+ 2v {-by£ 2 -(ay - ga)r) 2 - hag^+ bag+ (hy-af) grj} 
+ A* {(7l - a 0 2 } + 2/m/ {ol v (yt; - ag] + v 2 {a 2 ?; 2 } = 0; 
v, d 2 =°> 
;