How to make a Trigonometric Parametrization of $3x^2+y^2-4z^2=1$, if $y=z^2-1$, $z\geq 0$. To find $\theta$ such that $r(\theta)=(\sqrt{3},2,\sqrt{2})$
Attempt Replacing the second equation on the first we arrived to $$3x^2+y^2-4y-4=1 $$ which is an ellipse but, we know that the parametrization in such case $r=h+a\cos(t)$, $y=k+b\sin(t)$ here $(h,k)=(0,2)$ how to find $a,b$?
HINT
Add 8 on both sides of equation and simplify $$\frac{x^2}{(\sqrt3)^2} + \frac{(y-2)^2}{3^2} =1 $$ $$ \cos^2 t + \sin^2 t =1 $$
You want to put it into the form:
$x=h+a\cos(t)$, $y=k+b\sin(t)$ here $(h,k)=(0,2)$
Can you take it from here?