<s>[INST] Parameterizing the intersection between $4x^2+y^2=z^2$ and $xy-1=z$ I would like to parameterize the intersection between the surfaces
\begin{align}
4x^2+y^2&=z^2\\
xy&=z+1
\end{align}
I started by noting that $z=xy-1$ and then substituting that into the first equation to get \begin{align*}
4x^2+y^2&=(xy-1)^2\\
&=x^2y^2-2xy+1
\end{align*}
Then, I introduced the substitution $u=xy$ and $v=2x+y$ to obtain
\begin{align*}
v^2-4u&=u^2-2u+1\\
\implies v^2=u^2+2u+1\implies v^2&=(u+1)^2\\
\therefore v=u+1
\end{align*}
Clearly, $z=u-1$.
Where do I go from here to obtain a parameterization of the curve of intersection?
 [/INST] You have that
$$
4x^2  + y^2  = \left( {xy - 1} \right)^2 
$$
thus
$$
\left( {2x + y} \right)^2  = \left( {xy + 1} \right)^2 
$$
This means
$$
2x + y =  \pm \left( {xy + 1} \right)
$$
So you have two cases: the first
$$
\left\{ \begin{array}{l}
 2x + y = xy + 1 \\ 
 z = xy - 1 \\ 
 \end{array} \right.
$$
and the second
$$
\left\{ \begin{array}{l}
 2x + y =  - xy - 1 \\ 
 z = xy - 1 \\ 
 \end{array} \right.
$$
Let us consider the first. You have
$$
y\left( {x - 1} \right) = 2x - 1
$$
From this you have
$$
\left\{ \begin{array}{l}
 x = t \\ 
 y = \frac{{2t - 1}}{{t - 1}} \\ 
 z = t\frac{{2t - 1}}{{t - 1}} - 1 \\ 
 \end{array} \right.\,\,\,\,\,\,t \ne 1
$$
The second case is similar.
</s>