Answer
$y^{2}-4x-4=0$
Work Step by Step
Formulas relating rectangular $(x,y)$ and polar $(r,\theta)$ coordinates:
$ x=r\cos\theta$ and $ y=r\sin\theta$
$r^{2}=x^{2}+y^{2}$ and $\displaystyle \tan\theta=\frac{y}{x}$.
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$r=\displaystyle \frac{2}{1-\cos\theta}\qquad/\times(1-\cos\theta)$
$r-r\cos\theta=2\qquad $... subsitute: $r\cos\theta=x$
$r-x=2 \qquad/+x$
$ r=2+x\qquad$... square both sides
$ r^{2}=4+4x+x^{2}\qquad$...substitute $r^{2}$
$x^{2}+y^{2}=4+4x+x^{2}$
... move all from RHS to LHS
$y^{2}-4x-4=0$