How can I convert the following parametric equation to cartesian equation?

Question

\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}

Answer 1

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*}