# Dupin Cyclide: Cartesian coordinates to parametric coordinates

I have been given points in Cartesian coordinates that lie on Dupin's cyclide. I am simply trying to extract the corresponding parametric coordinates. Given two parameters $$u,v \in [0,2\pi]$$, the cyclide is defined parametrically as: \begin{align} x &= \frac{d(c - a\cos u \cos v) + b^2 \cos u}{a - c \cos u \cos v}, \\ y &= \frac{b\sin u (a-d\cos v)}{a - c \cos u \cos v},\\ z &= \frac{b\sin v (c\cos u - d)}{a - c \cos u \cos v}. \end{align} I've been trying to solve for $$u,v$$ with no luck. I'd appreciate it if someone could give me analytic expressions for $$u$$ and $$v$$, like one would get for spherical coordinates (for instance). Note that for my problem, $$c^2 = \sqrt{a^2 - b^2}$$,$$d = 1$$, $$a = 2$$, $$b=1.9$$. Thanks!

Eliminate $$\cos v$$ from $$x$$ and $$y$$:
$$\frac{y}{x-\dfrac{cd}{a}}=\frac{a}{b}\tan u$$
Eliminate $$\cos u$$ from $$x$$ and $$z$$:
$$\frac{z}{x-\dfrac{ad}{c}}=\frac{c}{b}\sin v$$
Standard texts give the following implicit equations: $$(x^2+y^2+z^2-d^2+b^2)^2-4(ax-cd)^2-4b^2y^2=0$$ or equivalently $$(x^2+y^2+z^2-d^2-b^2)^2-4(cx-ad)^2+4b^2z^2=0$$