Chapter 8.5, Problem 40E

Solution

To find: The Cartesian equation for the given polar equation.

The Cartesian equation for the given polar equation is $y^2=-8\left(x-2\right)$ .

Given information:

The given equation is $r=\frac{12}{3+3\sin \theta }$ .

Formula used:

The $\cos \theta$ is present in the polar equation, the focal axis is the polar axis ( xx -axis) so use the following equation.

  $y^2=4p\left(x-h\right)$   ..... (1)

Calculation:

Divide the numerator and denominator by $3$ .

  $\begin{array}{c}r=\frac{4}{1+\sin \theta }\\ r=\frac{4\left(1\right)}{1+\left(1\right)\sin \theta }\end{array}$

Compare the above equation with $r=\frac{ke}{1+e\sin \theta }$ .

  $k=4$

It is given that $k=2p$ . Then,

  $\begin{array}{c}2p=4\\ \left|p\right|=2\end{array}$

Since the directrix is to the right of the pole, the vertex is to the right of the pole, when $\theta =\frac{\pi }{2}$ .

  $\begin{array}{c}r=\frac{12}{3+3\sin \frac{\pi }{2}}\\ =\frac{12}{6}\\ =2\end{array}$

So, the vertex is at $\left(r,\theta \right)=\left(2,\frac{\pi }{2}\right)$ or at $\left(x,y\right)=\left(0,2\right)$ .

The parabola opens to the left so take $p=-2$ .

Substitute $-2$ for $p$ and $2$ for $h$ in equation (1) to find the equation.

  $\begin{array}{c}{y}^2=4\left(-2\right)\left(x-2\right)\\ y^2=-8\left(x-2\right)\end{array}$

Therefore, the Cartesian equation for the given polar equation is $y^2=-8\left(x-2\right)$ .