## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$y^{2}=8(x+2)$
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*}