Answer
$x=x,y=-\sqrt {\frac{1-x^{2}-3z^{2}}{2}},z=z$
Work Step by Step
Given:The part of the ellipsoid $x^{2}+2y^{2}+3z^{2}= 1$ that lies to the left of the $xz$ plane.
We need to use two parameters for the surface when we parametrize it. Since, it lies in the front of the $xz$ plane, we can let $x$ and $z$ be the independent variables and express $y$ in terms of $x$ and $z$.
$x^{2}+2y^{2}+3z^{2}= 1$
$2y^{2}=1-x^{2}-3z^{2}$
$y^{2}=\frac{1-x^{2}-3z^{2}}{2}$
$y=-\sqrt {\frac{1-x^{2}-3z^{2}}{2}}$
We choose the negative square root (rather than positive square root) because it lies to the left of the $xz$ plane.
Hence, the parametric representation of the hyperboloid is
$x=x,y=-\sqrt {\frac{1-x^{2}-3z^{2}}{2}},z=z$
