Answer
$x^{2}-3y^{2}+16y-16=0$
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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$r=\displaystyle \frac{4}{1+2\sin\theta}\qquad/\times(1+2\sin\theta)$
$ r+2r\sin\theta=4\qquad$... subsitute: $r\sin\theta=y$
$r+2y=4\qquad / -2y$
$ r=4-2y\qquad$... square both sides
$ r^{2}=(4-2y)^{2}\qquad$...substitute $r^{2}$
$x^{2}+y^{2}=16-16y+4y^{2}$
... move all from RHS to LHS
$x^{2}-3y^{2}+16y-16=0$
