Chapter 9.2, Problem 73AYU

To determine

To find: The graph of the polar equation $r=\frac{1}{3-2\text{cos}\theta }\left(ellipse\right)$.

Answer to Problem 73AYU

Solution:

Graph of the polar equation $r=\frac{1}{3-2\text{cos}\theta }\left(ellipse\right)$ is

Given:

It is asked to find the graph of the polar equation $r=\frac{1}{3-2\text{cos}\theta }\left(hyperbola\right)$:

Explanation of Solution

Check for symmetry test:

Polar axis:

Replace $\theta$ by $-\theta$. The result is $r=\frac{1}{3-2\text{cos}\left(-\theta \right)}=\frac{1}{3-2\text{cos}\theta }$.

The test satisfied. So the graph is symmetric with respect to the polar axis.

The line $\theta =\frac{\pi }{2}$:

Replace $\theta$ by $\pi -\theta$. The result is

$r=\frac{1}{3-2\text{cos}\left(\pi -\theta \right)}=\frac{1}{3-2\left(\text{cos}\pi \text{cos}\theta +\text{sin}\pi \text{sin}\theta \right)}=\frac{1}{3-2\left(\left(-1\right) \text{cos}\theta +\left(0\right) \text{sin}\theta \right)}$

$=\frac{1}{3+2\text{cos}\theta }$

The test fails, so the graph may or may not symmetric with respect to the line $\theta =\frac{\pi }{2}$.

The Pole:

Replace $r$ by $-r$. Then the result is $-r=\frac{1}{3-2\text{cos}\theta }$, so $r=\frac{-1}{3-2\text{cos}\theta }$. This test fails. Replace $\theta$ by $\theta +\pi$. The result is

$r=\frac{1}{3-2\text{cos}\left(\theta +\pi \right)}=\frac{1}{3-2\left(\text{cos}\pi \text{cos}\theta -\text{sin}\pi \text{sin}\theta \right)}=\frac{1}{3-2\left(-1\right)\text{cos}\theta }=\frac{1}{3+2\text{cos}\theta }.$

This test also fails, so the graph may or may not be symmetric with respect to the pole.

Let’s identify points on the graph by assigning values to the angle $\theta$ and calculating the corresponding values of $r$.

As the polar equation is symmetric about the polar axis, it is enough to find the values from 0 to $\pi$.

$\theta$	$r=\frac{1}{3-2\text{cos}\theta }$
0	$\frac{1}{3-2\left(1\right)}=1$
$\frac{\pi }{6}$	$\frac{1}{3-2\times \frac{\sqrt{3}}{2}}=\frac{1}{3-\sqrt{3}}$
$\frac{\pi }{4}$	$\frac{1}{3-2\times \frac{1}{\sqrt{2}}}=\frac{1}{3-\sqrt{2}}$
$\frac{\pi }{3}$	$\frac{1}{3-2\times \frac{1}{2}}=\frac{1}{2}$
$\frac{\pi }{2}$	$\frac{1}{3-2\times \left(0\right)}=\frac{1}{3}$
$\frac{2\pi }{3}$	$\frac{1}{3+2\times \frac{1}{2}}=\frac{1}{4}$
$\frac{3\pi }{4}$	$\frac{1}{3+2\frac{1}{\sqrt{2}}}=\frac{1}{3+\sqrt{2}}$
$\frac{5\pi }{6}$	$\frac{1}{3+2\times \frac{\sqrt{3}}{2}}=\frac{1}{3+\sqrt{3}}$
$\pi$	$\frac{1}{3-2\times \left(-1\right)}=\frac{1}{5}$

To sketch:

$r=\frac{1}{3-2\text{cos}\theta }$