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?

Usually, conic sections are parametrized best with some kind of cylindrical or spherical coordinate transformation.  Given $$5x^2 - 5y^2 - z^2 = 5, \quad x > 0,$$ this suggests a cylindrical parametrization with $$(y,z) = (u \cos v, \sqrt{5} u \sin v),$$ which easily gives $$5x^2 = 5 + 5y^2 + z^2 = 5 + 5u^2 (\cos^2 v + \sin^2 v) = 5(u^2+1),$$ or $$x = \sqrt{u^2 + 1},$$ where we take the positive square root since we were asked to provide the portion of the surface "in front of" the $yz$-plane (which I assume to mean $x > 0$).  Thus one parametrization is $$\begin{align*} x &= \sqrt{u^2 + 1}, \\ y &= u \cos v, \\ z &= \sqrt{5}u \sin v, \end{align*}$$ for $u \in [0,\infty)$ and $v \in [0, 2\pi)$.  Now, the above parametrization works just fine, but we can make it a little nicer and consider $$\begin{align*} x &= \sec u, \\ y &= \tan u \cos v, \\ z &= \sqrt{5} \tan u \sin v, \end{align*}$$ for $u \in [0,\pi/2)$, $v \in [0,2\pi)$, which gives the same result:  this was motivated by the familiar trigonometric identity $$\tan^2 \theta + 1 = \sec^2 \theta.$$  Clearly, we can see that parametrizations are not unique, and that in general, infinitely many legitimate parametrizations of the same surface are possible.
