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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!

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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 $$(x^2+y^2+z^2-d^2-b^2)^2-4(cx-ad)^2+4b^2z^2=0$$

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  • $\begingroup$ Thanks! This is very helpful. It's just a single further step to solve for u and v explicitly also. Cheers! $\endgroup$ – vergere6 Apr 19 at 18:40

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