There are many parametrizations, so no such thing as the parametrization.
You have $x^2 + y^2 + 1 = z^2$. If you write $x^2 + y^2 = r^2$ (so you can take
$x = r \cos(v)$, $y = r\sin(v)$), then $r^2 + 1 = z^2$.
Now since $\cosh^2(u) - \sinh^2(u) = 1$, you could take $z = \cosh(u)$, $r = \sinh(u)$, and thus
$$ \eqalign{x &= \sinh(u) \cos(v)\cr
y &= \sinh(u) \sin(v)\cr
z &= \cosh(u)\cr}$$
However,this is not a good parametrization at $u=0$ (corresponding to the single point $(x,y,z) = (0,0,1)$.
Now let's try a different way. Write the equation as $t^2 - y^2 = 1$ where
$t^2 = z^2 - x^2$. We can take $y = \sinh(v)$ (note that this ranges over
$\mathbb R$ as $v$ does), and then $t^2 = \cosh^2(v)$. If we choose the
positive $t$ (and we can), we have $t = \cosh(v)$. Now
$(z/t)^2 - (x/t)^2 = 1$. For the sheet with $z > 0$ we can take
$z/t = \cosh(u)$ and $x/t = \sinh(u)$. As $u$ ranges over $\mathbb R$,
we get all possible $(x,z)$ pairs with $z > 0$. So we have
$$
\eqalign{
x &= \sinh(u) \cosh(v)\cr
y &= \sinh(v)\cr
z &= \cosh(u) \cosh(v)\cr}
$$
This parametrization is better because it maps ${\mathbb R}^2$ one-to-one onto one sheet.
Assume that the Gaussian curvature is positive. It means that the second fundamental form is positive (or negative, depends on the choice of orientation). It then follows, that the curvature of every curve on the surface is also positive, as it is at least equal to the second fundamental form of the derivative at each point. This contradicts, obviously, the existence of a straight line at every point.
Best Answer
I've never heard about vertex of a hyperboloid, but I suppose that, here, it means the points $(\pm a,0,0)$. In fact, these points belong to the hyperboloid, and if $(x,y,z)$ is such that $|x|<a$, then$$\frac{x^2}{a^2}-\frac{y^2}{b^2}-\frac{z^2}{c^2}\leqslant\frac{x^2}{a^2}<1$$and therefore $(x,y,z)$ does not belong to hyperbolod.