52
ON THE FOC ALS OF A QUADRIC SURFACE.
[902
902]
ч
In connexion with the corresponding question in solid geometry, I obtain this
equation in a different manner. We investigate the envelope of the line gx + yy + £z = 0,
considering £, y, £ as parameters connected by the foregoing two equations; by the
ordinary process of indeterminate multipliers, we have
x + (Aft + /m)| = 0, y + (Aò + fib') y = 0, z + (Ac + ¡xc) £= 0,
and thence, eliminating £, y, £, we obtain
X 2 y 2 z 2
q. V- 4- = 0
Aft -f- ¡xa' Ab -(- /xb Ac -f- /xc
aar
+
by 2
+
cz 2
(A ft + /ха') 2 (A b + yh') 2 (Ac + /хс'У
ax 2
+
b'y 2
+
cz*
(A ft + /Aft') 2 (AÒ + /xb') 2 (Ac + ftc') !
= 0,
= 0,
equations equivalent to two equations, from which A and /x are to be eliminated. The
second and third equations are the derived functions of the first equation in regard
to A, /x respectively; and hence, expressing the first equation in an integral form,
the result is
Disct. [x 2 (Ab + fxb') (Ac + /xc') + &c.} = 0 ;
viz. this is
{(be' + b'c) x 2 + (cci' + c'a) y 2 + (ab' + a'b) z 2 } 2 — 4 (ihex 2 + cay 2 + abz 2 ) (b'cx 2 + c'a'y 2 + a'b'z 2 ) = 0,
or developing and reducing, we have the foregoing result, the coefficients entering
through the combinations
f g, h = be' — b'c, ca — ca, ab' — a'b.
Writing c = — 1, also «'= 6' = 1, c' = 0: the equation of the first conic is
- + %-* = 0,
ft b
and that of the second may be replaced by the two equations x 2 + y 2 = 0, z = 0, viz.
these give the circular points at infinity: we have f g, h — 1, —1, a — b, and the
equation of the line-system is
x 4, + 2x 2 y 2 + у 4 — 2Л (x 2 — y 2 ) z 2 + h 2 z 4 = 0.
If finally, z — 1, then for the tangents from the circular points at infinity to the
quadric
the equation is
-+f-l=0,
ft b
x 4 + 2 x 2 y 2 + y 4 —2h (x 2 — y 2 ) + h 2 = 0,
where, as before, h = a — 6 ; the four tangents intersect in pairs in the two circular
points at infinity, and in four other points which are the foci of the quadric.
and
equal
By v
VIZ.
entei
arise
pp.
d' = 0
and ti
f > g-
we hi
