An elementary treatise on cubic and quartic curves, by A. B. Basset.

178

SPECIAL QUARTICS.

move off to infinity, whilst the inner oval becomes an ellipse
whose foci are F and F1 and whose major axis is a.
When m > 1 and a > me, the oval (4) becomes imaginary, but
the line F1Q now cuts PF produced so that F2R bisects the
external angle of the triangle F1PQ; hence
QR PR
F1Q= - = m,
also since
FQ + FR = QR = mFQ,
the locus of Q is the oval
r + a = mr.......................(11).
Writing (3) and (11) in the forms
r, - r/m = a/m J
it follows that if a > nc, these ovals are of the same species as the
pair of conjugate ovals we have previously been discussing, but
the foci F and Fx are interchanged. Also writing (11) in the
form r - mr = -a, it follows that (11) belongs to Case IV., in
which a and m are negative quantities which are numerically
greater than c and 1 respectively.
From (6) it follows that FF2 is negative, so that F2 lies on the
left of F, and its value is (a2 - mc2)/c (m2 - 1).
When m = a/c, FF2= 0, and F2 coincides with F; also both
ovals pass through F since (12) are satisfied by ri = c, r.= 0. The
polar equations of the ovals referred to the focus F are
r (a2 - c2) = 2ac (a cos 0 T c),
the upper and lower signs being used for the internal and external
ovals respectively. But when polar coordinates are employed
negative as well as positive values of r are admissible, whence
both ovals are included in the equation
r (a2 - c2) = 2ac (a cos 0 - c)...............(13),
which is a hyperbolic limafon whose node is at F, which is also a
double focus. The focus F, lies inside the internal loop.
Lastly let m > a/c; then from (6) c' is positive, and therefore
F2 lies on the right of F; but FF1 > FF2, so that F2 now becomes
the middle and F the external focus. To find the conjugate oval,