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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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    The derivation is just writing down the relevant identities, $\cos^2 u+\sin^2 u=1$ and $\cosh^2 v-sinh^2 v=1$. We also need to check we do obtain the whole hyperboloid, which involves the range of $\sinh$.2012-06-14
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    I know they work, using these identities. I guess I'd like a geometric reason. I can see easily how to derive spherical coordinates, which I find similar to these. Can they be derived by considering the hyperboloid as a surface of revolution?2012-06-14
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    Surface of revolution works fine, $\cosh and \sinh$ are to the hyperbola $x^2-y^2=1$ as $\cos$ and $\sin$ are to the circle $x^2+y^2=1$. That gets you to the parametrization of the hyperboloid of revolution $x^2+y^2-z^2=1$. The multiplications by $a$, $b$, $c$ are at the end, they are scalings in the $x$, $y$, and $z$ directions.2012-06-14
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    Oh! Thank you. I just realized this whole time I was misreading the cos and sin terms in the parameterizations that come from surface of revolution as a second pair of cosh and sinh. If you leave your comment as an answer I will accept.2012-06-14
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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, $x = (a\cos \theta) \color{green}{\cosh(s)} \text{ and } y = (b\sin \theta) \color{green}{\cosh(s)} \text{ and } \color{green}{z = c\sinh(s)}$.