Chapter 8.5, Problem 38E

Solution

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

The Cartesian equation for the given polar equation is $\frac{{\left(x-4\right)}^2}{4}-\frac{y^2}{12}=1$ .

Given information:

The given equation is $r=\frac{6}{1+2\cos \theta }$ .

Formula used:

The formula to calculate the value of $c$ is given by,

  $c=ea$   ..... (1)

The formula to calculate the value of $b$ is given by,

  $\begin{array}{c}{a}^2=b^2+c^2\\ b=\sqrt{a^2-c^2}\end{array}$   ..... (2)

Calculation:

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

  $\begin{array}{c}e=2\\ >1\left(\text{hyperbola}\right)\end{array}$

The focal axis is the polar axis (or x -axis) since $\cos \theta$ is present in the polar equation so find the endpoints of the transverse axis using $\theta =0$ and $\theta =\pi$ .

When $\theta =0$ ,

  $\begin{array}{c}r=\frac{6}{1+2\cos 0}\\ =\frac{6}{1+2}\\ =2\end{array}$

When $\theta =\pi$ ,

  $\begin{array}{c}r=\frac{6}{1+2\cos \pi }\\ =\frac{6}{1-2}\\ =-6\end{array}$

The endpoints of the transverse axis are $\left(2,0\right)$ and $\left(-6,\pi \right)$ so the major axis length is as follows.

  $\begin{array}{c}2a=6-2\\ 2a=4\\ a=2\end{array}$

Substitute $2$ for $e$ and $2$ for $a$ in equation (1) to find $c$ .

  $\begin{array}{c}c=\left(2\right)\left(2\right)\\ =4\end{array}$

Substitute $2$ for $a$ and $2$ for $c$ in equation (2) to find $b$ .

  $\begin{array}{c}b=\sqrt{4^2-2^2}\\ =\sqrt{16-4}\\ =\sqrt{12}\end{array}$

In the Cartesian plane, the focal axis is the x -axis and the endpoints of the transverse axis are $\left(2,0\right)$ and $\left(6,0\right)$ so the center is at:

  $\begin{array}{c}\left(h,k\right)=\left(\frac{2+6}{2},\frac{0+0}{2}\right)\\ =\left(4,0\right)\end{array}$

Using the standard form of a horizontal hyperbola, the equation is as follows.

  $\begin{array}{c}\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1\\ \frac{{\left(x-4\right)}^2}{2^2}-\frac{{\left(y-0\right)}^2}{{\left(\sqrt{12}\right)}^2}=1\\ \frac{{\left(x-4\right)}^2}{4}-\frac{y^2}{12}=1\end{array}$

Therefore, the Cartesian equation for the given polar equation is $\frac{{\left(x-4\right)}^2}{4}-\frac{y^2}{12}=1$ .