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: $$x=\sqrt{2pu} \cos v ,\ \ y=\sqrt{2qu} \sin v, \ \ z=u $$

For the first: $$\frac{x^2}{a^2}\color{red}{-}\left(\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)=1$$
\begin{eqnarray}
&x&=a\cosh\theta,\\
&y&=b\cos\phi\sinh\theta,\\
&z&=c\sin\phi\sinh\theta.\\
\end{eqnarray}
