528
ON POLYZOMAL CURVES.
[414
In this case the pencils of tangents are = kdz, by — kdz, where d is determined by
a quartic equation, or taking the corresponding lines (which by their intersections
determine the foci A, B, G, D) to be (a^ — kd x z, by — kdf), &c., these four points lie in
the line — by = 0, which is a line through the centre of the curve, or point £ = 0,
7) = 0: the formulae just obtained belong therefore to the symmetrical case of the
circular cubic. Passing to rectangular coordinates, writing z — 1, and taking y = 0 for
the equation of the axis, it is easy to see that the equation may be written
(x 2 + y 2 ) (x — a) + k (x — b) = 0 ;
or, changing the origin and constants,
xy 2 + (x — a) (x — b) (x — c) = 0.
Article Nos. 145 to 149. Analytical Theory for the Bicircular Quartic.
145. The equation for the bicircular quartic may be taken to be
k (£' — a 2 2 2 ) (t — + ezfn + z 3 (a£ + by) 4- C2 4 = 0,
viz. (£, y, z) being any coordinates whatever, this is the equation of a quartic curve
having a node at each of the points (£ = 0, z = 0) and (y = 0, z = 0): the equations of
the two tangents at the one node are £ — az = 0, £ + clz = 0; and those of the two
tangents at the other node are y — /3z = 0, y + fiz = 0; £ = 0 is thus the harmonic of
the line z = 0 in regard to the tangents at (£ = 0, 2 = 0), and y = 0 is the harmonic
of the same line 2 = 0 in regard to the tangents at (?7 = 0, 2=0). If (£, y, z= 1) be
circular coordinates, then we have the general equation of the bicircular quartic having
the lines f + az = 0, £ — az = 0 for one pair, and the lines y — ftz — 0, y + /32 = 0 for the
other pair of parallel asymptotes; and therefore the point £ = 0, ?? = 0 for centre, and
the lines — ay = 0, /3£ + ay = 0 for nodal axes. In order that the curve may be real
we must have (a, ¡3), (a, b) conjugate imaginarles, k, e, c real. The points (£ = 0, 2 = 0)
and (y = 0, 2=0) are as before the points /, J. If a = 0, the node at I becomes a
cusp, and so if /3 = 0, the node at J becomes a cusp; the form thus includes the case
of a bicuspidal or Cartesian curve.
146. To find the tangents from /, writing in the equation of the curve i;=6xz
we have
ko. 2 ( 6- — 1) (rf — ¡3 2 z 2 ) + eadyz + z {aadz + by) + cz 2 — 0 ;
that is
V 2 . kar(6' 2 — 1),
+ yz . ead + b,
+ 2 2 . — ka 2 /3~ (d 2 — 1) + aad 4- c = 0,
and the condition of tangency is
4tk (d 2 — 1) {ka 2 /3- (d 2 — 1) — a ad — c} +