392] 
AND TOUCH TWO GIVEN LINES. 
45 
or (what is the same thing) introducing the arbitrary coefficient k, we have 
kx + 7/4 — v \J (-mn) = 0, 
ky + yX — v — 0, 
kz-\sj (mn) - fi - ~ {V (m) - V (w)} 2 y = 0 ; 
the first two equations give 
that is 
k : 7 : — 1— v{/x — X V (mn)] : v [y \/(wm) — ¿c} : Xw; — /xy, 
^ - (dm)} _~ v [y V (® w ) — #} 
\x — /xy ’ ^ \x — /xy 
or, substituting this value of 7 in the third equation, 
( w (mB)1 + WMzlMÏ = 0, 
\x— ¡xy 1 n x - y V (tow) 2 
that is 
(\æ - /¿y) 2 .1 {V (to) - V (w)} 2 + [x - y V (tow)} (X« - yay) {/i + X V (mn)] 
+ z [x — y V (wiw)} v {fx — X V (??iw)} = 0, 
which is the equation of the curve, the locus of the pole of the line \x + /xy + vz = 0 
in regard to the conic 
(x — my) (x - ny) + 2 {« + y sj (mn)] 7z + r fz 1 = 0. 
In particular, if X = /x = v = 1, then for the coordinates of the centre of the conic, 
we have 
x : y : £ = — 7 d* V (wiw) : — 7 + 1 : V (wm) + 1 + ^ {V (to) — V (w)} 2 ; 
and for the locus of the centre, 
(x - y)\ I y(m) - v/(w)l 2 + (a - y) {x - y *J(mn)] {1 + >J(mn)] +z{x-y >J(mn)} {1 - V(tow)} = 0, 
so that the locus is a conic, and it is obvious that this conic is a hyperbola. Putting 
for greater simplicity 
x—y = X, 
x — y\l (mn) = F, 
2 = Z, 
the equation of the curve of centres is 
X 2 . \ (V (to) - V (w)} 2 + XY{ 1 + V (tow)} + YZ{ 1 - V (tow)} = 0, 
or, writing this under the form 
Y[X {1 + V(wiw)} + Z{ 1 — V(tow)}] + \ {V(to) — V(w)}- A r “ = 0,