Answer
$y^{2}=8(x+2)$
Work Step by Step
Multiply $1-\cos\theta$ to both sides of the equation to obtain:
\begin{align*}
(1-\cos\theta)(r)&=\frac{4}{1-\cos\theta} \cdot (1-\cos\theta)\\
r-r\cos\theta&=4\end{align*}
Knowing that $r=\sqrt {x^{2}+y^{2}}$ and $r\cos\theta=x$, then the equation above becomes:
\begin{align*}
\sqrt {x^{2}+y^{2}}-x&=4\\
\sqrt {x^{2}+y^{2}}&=x+4\\
\left(\sqrt {x^{2}+y^{2}}\right)^2&=(x+4)^2\\
x^2+y^2&=x^2+8x+16\\
y^2&=x^2+8x+16-x^2\\
y^2&=8x+16\\
y^2&=8(x+2)\end{align*}
