Simplest way of parameterize the surface of a two-sheeted hyperboloid? I want to parametrize the surface of a given Two-Sheeted Hyperboloid expression:$$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=-1 $$I have tried with the parametrization that Wolfram|Mathworld give for the two-sheeted hyperboloid (with hiperbolic functions) but I would like to use the simplest way to do it. Maybe I just need a correct parametrization for the ellipse on the xy plane. 
 A: We can combine the identities $\cos^2\phi+\sin^2\phi=1$ and $\sec^2\theta-\tan^2\theta=1$ to get:
$$x=a\tan \theta\cdot\cos \phi$$
$$y=b\tan \theta\cdot\sin \phi$$
$$z=c\sec \theta $$
This directly gives 
$$\frac{z^2}{c^2}-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)=1$$
There are, of course, many other ways. You may consider this way simplest, compared to using hyperbolic functions.
I believe the parameter intervals are:
$$0\le\theta<\frac{\pi}2,\qquad \frac{\pi}2<\theta\le\pi$$
$$0\le\phi<2\pi$$
The two intervals for $\theta$ correspond to the two sheets.
And here is a parametrization for the ellipse on the $xy$ plane:
$$x=a\cos\phi$$
$$y=b\sin\phi$$
For $0\le\phi<2\pi$, this gives the Cartesian equation
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
A: Please note:
$$x=a\sin \theta\cdot\cos \phi,y=b\sin \theta\cdot\sin \phi, z= c\cos \theta $$
parametrizes a sphere. Now, plug in as follows to change an  argument and coefficient to convert to its corresponding hyperbolic form:
$$ \theta  \rightarrow  i\theta,a \rightarrow ia, b \rightarrow ib. $$
Taking positive terms you get:
$$x=a\sinh \theta\cdot\cos \phi, y=b\sinh \theta\cdot\sin \phi, z= \pm c\cosh \theta $$
which parametrizes a two sheeted hyperboloid.
The hyperbolic functions are as much easy and natural to hyperbolic geometry surfaces as are the circular trigonometric functions to elliptic geometry surfaces.
EDIT1:
z has  $ \pm $ double sign valid, one sign for each of two nappes. 
