Differentiation 367the conicEX^2 + 2F XY + GY^2 = 1 (6.16-2)
Since EG - F^2 = (uxVy - uyvx)^2 = J2, this conic is an ellipse if J # 0,
and two parallel lines, namely,VEX+ (sgnF)../GY = ±1
if J = o.
To express (6.16-1) in Cartesian coordinates, let x = pcosB, y = psinB.
We obtain(x^2 + y^2 )^2 = Ex^2 + 2Fxy + Gy^2 (6.16-3)
The inversion transformation formulas are
x y
x = x2 + y2 ' Y = x2 + y2
When the new y,ariables (X, Y) are introduced in (6.16-3), equation (6.16-2)
results..
In the CCJ;~e J = p, equation (6.16-3) becomes
x^2 + y^2 = ±[VEx + (sgnF)../Gy]
which represents two tangent circles at the origin, both with radius
~vfE+G ..
Equa~!9!1 ( 6.16-2) in polar coordinates ( r, B) reads
1
2 = E cos
(^2) B + 2F cos B sin B + G sin (^2) B
r
(6.16-4)
Clearly, p = 1/r.
Ii:i·1~'ig. 6.11 we have sketched for the case J # 0 the graph of (6.16-2)
or (6.16-4) in a light line, and that of (6.16-1) or (6.16-3) in a heavy line.
In Fig. 6.12 the corresponding graphs are shown for the case J = 0. Those
graphs may be laid off from the fixed point z, if desired, so as to show the
behavior of the magnification ratio at that point.
For F = 0 and E = G equation (6.16-1) reduces to
p= VE
so that the magnification ratio is constant, and from Theorem 6.17 it follows
that f is either monogenic or conjugate monogenic at z.
In order to determine in the general case the maximum and minimum
values of p we find, from (6.16-1),
:o (p^2 ) = -2E sin B cos B + 2F( cos^2 B -sin^2 B) + 2G sin B cos B