# Question #02b85

Aug 4, 2015

$x = \frac{1}{8} {y}^{2} - 2$.

#### Explanation:

One thing you can do is start by multiplying both sides of the equation $r = \frac{4}{1 - \cos \left(\theta\right)}$ by $1 - \cos \left(\theta\right)$ to get $r - r \cos \left(\theta\right) = 4$.

Next, rearrange this to get $r = 4 + r \cos \left(\theta\right)$.

Now square both sides to get ${r}^{2} = 16 + 8 r \cos \left(\theta\right) + {r}^{2} {\cos}^{2} \left(\theta\right)$.

The reason this was a good idea is that you can now substitute rectangular coordinates $\left(x , y\right)$ pretty quickly using the facts that ${r}^{2} = {x}^{2} + {y}^{2}$ and $r \cos \left(\theta\right) = x$ to get:

${x}^{2} + {y}^{2} = 16 + 8 x + {x}^{2}$
${y}^{2} = 16 + 8 x$.

Solving this equation for $x$ as a function of $y$ gives

$x = \left(\frac{1}{8}\right) \left({y}^{2} - 16\right) = \frac{1}{8} {y}^{2} - 2$.

The graph of $r = \frac{4}{1 - \cos \left(\theta\right)}$, as $\theta$ varies over the open interval $\left(0 , 2 \pi\right)$, is the sideways parabola shown below. 