Parameterizing a surface The question I was asked goes like this:

The part of the hyperboloid $5x^2 − 5y^2 − z^2 = 5$ that lies in front of the yz-plane. Let x, y, and z be in terms of u and/or v. Find a parametric representation for the surface.

So, by fixing $x$ as a constant $u$, I find that the slice is an ellipse which can be parameterized as $y = \sqrt{\frac{5u^2-5}{5}}\cos{v}$ and $z = \sqrt{5u^2-5}\sin{v}$, where $u$ ranges from $1$ to infinity, and $v$ ranges from $0$ to $2\pi$.
However, I was told that this is not the correct parameterization. The correct answer is
$$
\begin{align*}
x&=\sqrt{1+u^2+\frac{1}{5}v^2} \\
y&= u \\
z&= v
\end{align*}
$$
I believe that this is just another way of parameterizing the same surface, and I don't see why my answer is wrong. Any idea?

I interpret being 'in front of the $yz$-plane' as the first coordinate being non-negative, thus the surface $S$ in question $\{(x,y,z)\in \mathbb R^3\colon x\ge 0\land 5x^2 − 5y^2 − z^2 = 5\}$.
It can easily be proved that this set equals $\left\{\left(\sqrt{1+y^2+\dfrac{z^2}5}, y,z\right)\colon y,z\in \mathbb R\right\}$, therefore $S=\vec r\left[\mathbb R^2\right]$, where and $\vec r\colon \mathbb R^2\to \mathbb R^3, (u,v)\mapsto \left(\sqrt{1+u^2+\dfrac{v^2}5}, u,v\right)$. So this $\vec r$ is certainly a suitable candidate for parametrization (you need to check that it satisfies the other properties for being one). In my mind this is the most natural option. 
You suggest taking $\vec s\colon D\to \mathbb R^3, (\rho,\theta)\mapsto \left(\rho, \sqrt{\rho ^2-1}\cos(\theta), \sqrt 5\sqrt{\rho^2-1}\sin(\theta)\right)$, where $D=[1,+\infty[\times [0,2\pi[$. 
I don't see how to make any variation of this parametrization become injective (I think it's not possible) and that might be why yours was marked wrong.
Anyway, I'll prove that $S=\vec s[D]$. The inclusion $\supseteq$ is the easy one. I'll help out with the other one.
Let $(x,y,z)\in S$. Due to $x^2=1+y^2+\dfrac{z^2}5$ it's easy to conclude that $x\ge 1$.
One wishes to find $(\rho, \theta)\in [1,+\infty[\times[0,2\pi[$ such that $(x,y,z)= \left(\rho, \sqrt{\rho ^2-1}\cos(\theta), \sqrt 5\sqrt{\rho^2-1}\sin(\theta)\right)$.
If $x=1$, then $y=0=z$ and $(x,y,z)=\vec s(1,\theta)$, for any $\theta$. 
Suppose $x>1$. If such $(\rho, \theta)$ exists, necessarily $\rho =x$ and $y=\sqrt{\rho ^2-1}\cos(\theta)$. This ensues the equality $\cos(\theta)=\dfrac{y}{\sqrt{x^2-1}}$. So it's necessary to check that $-1\leq \dfrac{y}{x^2-1} \leq 1$ or equivalently that$\dfrac{y^2}{x^2-1}\leq 1$, and this in fact holds because $x^2=1+y^2+\dfrac{z^2}5$.
So take $\theta=\arccos\left(\dfrac y{\sqrt{x^2-1}}\right)\lor \theta=2\pi-\arccos\left(\dfrac y{\sqrt{x^2-1}}\right)$, depending on whether you're in $[0,\pi]$ or $[\pi, 2\pi]$.
All that is left is to check that this choice of $(\rho, \theta)$ works. I'll check the first one.
In fact 
$$\require{cancel}
\begin{align} \vec s(\rho, \theta)&=\left(\rho, \sqrt{\rho ^2-1}\cos(\theta), \sqrt 5\sqrt{\rho^2-1}\sin(\theta)\right)\\
&=\left(x, \sqrt{x ^2-1}\cos\left(\arccos\left(\dfrac y{\sqrt{x^2-1}}\right)\right), \sqrt 5\sqrt{x^2-1}\sin\left(\arccos\left(\dfrac y{\sqrt{x^2-1}}\right)\right)\right)\\
&=\left(x,y,\sqrt 5\sqrt{x^2-1}\sqrt{1-\left(\dfrac y{\sqrt{x^2-1}}\right)^2}\right)\\
&=\left(x,y,\sqrt 5\cancel{\sqrt{x^2-1}}\sqrt{\dfrac{x^2-1}{\cancel{x^2-1}}-\dfrac{y^2}{\cancel{x^2-1}}}\right)\\
&=\left(x,y,\sqrt 5\sqrt{y^2+\frac{z^2}5-y^2}\right)\\
&=(x,y,z).
\end{align}
$$
To ensure that you get an appropriate $z$ just choose one of the two possible values of $\theta$.
