Answer
$x=\frac{\sqrt {4+4y^{2}+z^{2}}}{2},y=y,z=z$
Work Step by Step
Given:The part of the hyperboloid $4x^{2}-4y^{2}-z^{2}= 4$ that lies in front of the $yz$ plane.
We need to use two parameters for the surface when we parametrize it. Since, it lies in the front of the $yz$ plane, we can let $y$ and $z$ be the independent variables and express $x$ in terms of $y$ and $z$.
$4x^{2}-4y^{2}-z^{2}= 4$
$4x^{2}=4+4y^{2}+z^{2}$
$x^{2}=\frac{4+4y^{2}+z^{2}}{4}$
$x=\sqrt {\frac{4+4y^{2}+z^{2}}{4}}$
or
$x=\frac{\sqrt {4+4y^{2}+z^{2}}}{2}$
We choose the positive square root (rather than negative square root) because it lies in the front of the $yz$ plane.
Hence, the parametric representation of the hyperboloid is
$x=\frac{\sqrt {4+4y^{2}+z^{2}}}{2},y=y,z=z$
