<s>[INST] Weiestrass Form  How to convert this to weiestrass form?
$x^{2}y^{2}-2\left( 1+2\rho \right) xy^{2}+y^{2}-x^{2}-2\left( 1+2\rho
\right) x-1=0$
 [/INST] You can rewrite the form as
\begin{equation*}
y^2=\frac{x^2+2(2\rho+1)x+1}{x^2-2(2\rho+1)x+1}
\end{equation*}
so, for rational solutions (which I presume you want), there exists $z \in \mathbb{Q}$ with
\begin{equation*}
z^2=(x^2+2(2\rho+1)x+1)(x^2-2(\rho+1)x+1)=x^4-2(8\rho^2+8\rho+1)x^2+1
\end{equation*}
This quartic can be transformed to an equivalent elliptic curve using the method described by Mordell in his book Diophantine Equations. We get (after some fiddling)
\begin{equation*}
v^2=u(u+4\rho^2+4\rho)(u+4\rho^2+4\rho+1)
\end{equation*}
with $x=v/u$.
For example, $\rho=11$ gives a rank $1$ elliptic curve with point $(147,8190)$, giving $x=8190/147$ and $y=\pm 527/163$.
This elliptic curve could be transformed to Weierstrass form if you want, but the above form is probably more useful.
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