<s>[INST] How can I convert the following parametric equation to cartesian equation? \begin{align}
x&=\left(1 + \frac{1}{\,\sqrt{\,2t^{2} - 4t + 4\,}\,}\right)t\ -\ 2
\\[3mm]
y&=\left(1 - \frac{1}{\,\sqrt{\,2t^{2} - 4t + 4\,}\,}\right)t\
+\ \frac{2}{\,\sqrt{\,2t^{2} - 4t + 4\,}\,}
\end{align}
 [/INST] Let $u=1/\sqrt{2t^{2}-4t+4}$. Then
\begin{equation*}
\left\{ 
\begin{array}{c}
x=t+ut-2 \\ 
y=t+u\left( 2-t\right), 
\end{array}
\right. \iff t=\frac{x+2}{1+u}=\frac{y-2u}{1-u}.
\end{equation*}
Square and add both equations
\begin{eqnarray*}
x^{2}+y^{2} &=&\left( t+ut-2\right) ^{2}+\left( t+u\left( 2-t\right) \right)
^{2} \\
&=&2t^{2}-4t+4+\left( 2t^{2}-4t+4\right) u^{2} \\
&=&2t^{2}-4t+5.
\end{eqnarray*}
So
\begin{equation*}
u=\frac{1}{\sqrt{x^{2}+y^{2}-1}}.
\end{equation*}
Subtract both equations
\begin{eqnarray*}
x-y &=&2\left( u\left( t-1\right) -1\right)  \\[2ex]
&=&2\left( u\left( \frac{x+2}{1+u}-1\right) -1\right)  \\[2ex]
&=&2\frac{\sqrt{x^{2}+y^{2}-1}\ x-x^{2}-y^{2}}{\sqrt{x^{2}+y^{2}-1}\left( 
\sqrt{x^{2}+y^{2}-1}+1\right) }.
\end{eqnarray*}
Consequently,
\begin{equation*}
\left( x-y\right) \left( \sqrt{x^{2}+y^{2}-1}+1\right) \sqrt{x^{2}+y^{2}-1}
-2\left( \sqrt{x^{2}+y^{2}-1}\ x-x^{2}-y^{2}\right) =0
\end{equation*}
or
\begin{equation*}
x^{3}-y^{3}-x^{2}y+xy^{2}+2x^{2}+2y^{2}-x+y-\left( x+y\right) \sqrt{
x^{2}+y^{2}-1}=0.
\end{equation*}
</s>