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ON THE QUARTIC SURFACES (*^U, V, W) 2 = 0.
[491
conic, perpendicular to MN, and QR perpendicular to 00', and join PR: the point
P of the conic describes a circle radius RP, = /^(RQ 2 + QP 2 ). Hence if Z.0MN=cl,
and if MQ = x, QP — y are the coordinates in the plane of the conic of the point P ;
and if the coordinates x, y, z are measured, 2 upwards from M in the direction MO,
and x, y in the plane at right angles to the axis 00': we have
z = x cos a, ^/(¿c 2 + y 2 ) = V(x 2 sin 2 cl + y 2 );
or, what is the same thing,
x = z sec cl, y = \J(oc 2 + y 2 — z 2 tan 2 a).
Hence the equation of the conic being F(x, y) = 0, that of the torus is
F (z sec a, \Zipc 2 + y 2 — z 2 tan 2 a)) = 0.
Thus taking the equation of the conic to be
(a, b, c, f, g, hjx, y, l) 2 = 0;
-or, as this may be written,
(ax 2 4- 2 gx + c + by 2 ) 2 = 4y 2 (hx+f ) 2 ,
we have at once the equation of the torus in the form
[az 2 sec 2 a + 2gz sec a + c + b (x 2 + y 2 — z 2 tan 2 a)} 2 = 4 (x 2 + y 2 — z 2 tan 2 a) (hz sec a 4- f) 2 ,
which is of the form V 2 — 4<UW = 0; or, as it is better to write it, V 2 — 4 UL 2 = 0, where
V = az 2 sec 2 a + 2gz sec a + c + b (x 2 -f y 2 — z 2 tan 2 a),
JJ — x 2 + y 2 — z 2 tan 2 a,
L = hz sec cl +/, W = L 2 .
There is thus a nodal circle V = 0, L = 0, that is
/
Z — — y OOS CL,
h
bh 2 (x 2 + y 2 ) — bf 2 sin 2 a + af 2 — Zgfh + cli 2 = 0.
But the origin of this nodal circle is better seen geometrically. For observe that the
radius of the circle described by the point P of the conic depends only on the
square of the ordinate PQ: hence if we have on the conic two points S, S' situate
symmetrically in regard to the line MN, these points S, S' will describe one and the
same circle, which will be a nodal circle on the surface. And there is in fact one
such pair of points S, S'; for (see fig. 2) considering in the plane of the conic the
equal conic situate symmetrically thereto on the other side of the line MN, the two
conics intersect in two points T, T' (real or imaginary) on the line MN, and in two
