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I know there is a parameterization of a hyperboloid $\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1$ in terms of $\cosh$ and $\sinh$, but I don't see how these equations are derived. I would appreciate it if either someone could explain to me how such a parameterization is derived or recommend a reference.

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    There was$a$partial answer by @Argon, later deleted. Argon can revive and amplify her/his answer. Or, perhaps better, now that you have figured it out you can answer, and, after some time has passed, accept.2012-06-14

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This derivation has been done by André Nicolas!

The parametrization of the hyperbola is

$x(t)=\cosh t$ $y(t)=\pm \sinh t$

A circle of radius $r$ is parametrized as:

$x(t)=\cos t$ $y(t)=\sin t$

Rotating the hyperbola above around a circle of radius $\cosh$ (distance of a regular hyperbola from y axis):

$x(u, v)=\cosh v \cos u$ $y(u, v)=\cosh v \sin u$ $z(u, v)=\sinh v$

It is easy to imaging the hyperboloid from two ways - from the top and from the side. This helped me understand the derivation.

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By virtue of user bondesan, here's a picture:

enter image description here

The $(P_u, P_z)$ might confuse, so I'd rewrite it as follows.

$\color{green}{\text{On the $uv$-plane, any hyperbola is given by: } u = \cosh(s) \text{ and } z = c\sinh(s) \, \,\forall -1 \leq s \leq 1}$.

Then by definition of polar coordinates, $x = (a\cos \theta) \color{green}{u} \text{ and } y = (b\sin \theta) \color{green}{u} \text{ and } \color{green}{z = z}$.

Altogether, $x = (a\cos \theta) \color{green}{\cosh(s)} \text{ and } y = (b\sin \theta) \color{green}{\cosh(s)} \text{ and } \color{green}{z = c\sinh(s)}$.