<s>[INST] Parametrization of two surfaces $\frac{x^2}{a^2}-\frac{y^2}{b^2}-\frac{z^2}{c^2}=1$ and $\frac{x^2}{p}+\frac{y^2}{q}=2z$. Can someone please help me to parametrize the following surfaces in terms of  hyperbolic(for second it might not be possible but i need some more convenient set of parametric equation than mine ) and trigonometric functions
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}-\frac{z^2}{c^2}=1 $$ and  $$\frac{x^2}{p}+\frac{y^2}{q}=2z$$
I have tried to do but the set of parametric equations I got were too complicated as I have to use those in some further calculation which makes the result very ugly. 
For first equation the set of parametric equations is: $$x=a\sqrt{1+\frac{u^2}{c^2}}\cos v, \ \ y=b\sqrt{1+\frac{u^2}{c^2}}\sin v \ \ z=u$$
and for second: $$x=\sqrt{2pu} \cos v ,\ \ y=\sqrt{2qu} \sin v, \ \ z=u $$
 [/INST] 
$$
\frac{x^2}{a^2} - \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1
$$

Try
\begin{align}
x &=& a\cosh u \\
y &=& b\sinh u \cos v \\ 
z &=& c\sinh u \sin v
\end{align}
Then
\begin{eqnarray}
\frac{x^2}{a^2} - \frac{y^2}{b^2} - \frac{z^2}{c^2} &=& \cosh^2 u -(\sinh^2u\cos^2v + \sinh^2u\sin^2v) \\
&=& \cosh^2 u - \sinh^2 u \\
&=& 1
\end{eqnarray}

$$
\frac{x^2}{p} + \frac{\color{red}{y}^2}{q} = 2z
$$

Try
\begin{align}
x &=& \sqrt{2up}\cos v \\
y &=& \sqrt{2uq}\sin v \\ 
z &=& u
\end{align}
In this case
\begin{eqnarray}
\frac{x^2}{p} + \frac{y^2}{q} &=& \frac{2 u p}{p}\cos^2v + \frac{2 u q}{q}\sin^2 u = 2u \\
&=& 2z 
\end{eqnarray}
</s>