Answer Solution: Graph of the polar equation r = 1 3 − 2 cos θ ( e l l i p s e ) is Given: It is asked to find the graph of the polar equation r = 1 3 − 2 cos θ ( h y p e r b o l a ) : Explanation Check for symmetry test: Polar axis: Replace θ by − θ . The result is r = 1 3 − 2 cos ( − θ ) = 1 3 − 2 cos θ . The test satisfied. So the graph is symmetric with respect to the polar axis. The line θ = π 2 : Replace θ by π − θ . The result is r = 1 3 − 2 cos ( π − θ ) = 1 3 − 2 ( cos π cos θ + sin π sin θ ) = 1 3 − 2 ( ( − 1 ) cos θ + ( 0 ) sin θ ) = 1 3 + 2 cos θ The test fails, so the graph may or may not symmetric with respect to the line θ = π 2 . The Pole: Replace r by − r . Then the result is − r = 1 3 − 2 cos θ , so r = − 1 3 − 2 cos θ . This test fails. Replace θ by θ + π . The result is r = 1 3 − 2 cos ( θ + π ) = 1 3 − 2 ( cos π cos θ − sin π sin θ ) = 1 3 − 2 ( − 1 ) cos θ = 1 3 + 2 cos θ . 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 θ 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 π . θ r = 1 3 − 2 cos θ 0 1 3 − 2 ( 1 ) = 1 π 6 1 3 − 2 × 3 2 = 1 3 − 3 π 4 1 3 − 2 × 1 2 = 1 3 − 2 π 3 1 3 − 2 × 1 2 = 1 2 π 2 1 3 − 2 × ( 0 ) = 1 3 2 π 3 1 3 + 2 × 1 2 = 1 4 3 π 4 1 3 + 2 1 2 = 1 3 + 2 5 π 6 1 3 + 2 × 3 2 = 1 3 + 3 π 1 3 − 2 × ( − 1 ) = 1 5 To sketch: r = 1 3 − 2 cos θ

Precalculus Enhanced with Graphing Utilities

6th Edition

ISBN: 9780321795465

Author: Michael Sullivan, Michael III Sullivan

Publisher: PEARSON

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Question

Chapter 9.2, Problem 73AYU

To determine

To find: The graph of the polar equation $r = \frac{1}{3 - 2 cos θ} (e l l i p s e)$ .

Expert Solution

Answer to Problem 73AYU

Solution:

Graph of the polar equation $r = \frac{1}{3 - 2 cos θ} (e l l i p s e)$ is

Precalculus Enhanced with Graphing Utilities, Chapter 9.2, Problem 73AYU , additional homework tip 1

Given:

It is asked to find the graph of the polar equation $r = \frac{1}{3 - 2 cos θ} (h y p e r b o l a)$ :

Explanation of Solution

Check for symmetry test:

Polar axis:

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

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

The line $θ = \frac{π}{2}$ :

Replace $θ$ by $π - θ$ . The result is

$r = \frac{1}{3 - 2 cos (π - θ)} = \frac{1}{3 - 2 (cos π cos θ + sin π sin θ)} = \frac{1}{3 - 2 ((- 1) cos θ + (0) sin θ)}$

$= \frac{1}{3 + 2 cos θ}$

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

The Pole:

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

$r = \frac{1}{3 - 2 cos (θ + π)} = \frac{1}{3 - 2 (cos π cos θ - sin π sin θ)} = \frac{1}{3 - 2 (- 1) cos θ} = \frac{1}{3 + 2 cos θ} .$

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 $θ$ 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 $π$ .

$θ$	$r = \frac{1}{3 - 2 cos θ}$
0	$\frac{1}{3 - 2 (1)} = 1$
$\frac{π}{6}$	$\frac{1}{3 - 2 \times \frac{\sqrt{3}}{2}} = \frac{1}{3 - \sqrt{3}}$
$\frac{π}{4}$	$\frac{1}{3 - 2 \times \frac{1}{\sqrt{2}}} = \frac{1}{3 - \sqrt{2}}$
$\frac{π}{3}$	$\frac{1}{3 - 2 \times \frac{1}{2}} = \frac{1}{2}$
$\frac{π}{2}$	$\frac{1}{3 - 2 \times (0)} = \frac{1}{3}$
$\frac{2 π}{3}$	$\frac{1}{3 + 2 \times \frac{1}{2}} = \frac{1}{4}$
$\frac{3 π}{4}$	$\frac{1}{3 + 2 \frac{1}{\sqrt{2}}} = \frac{1}{3 + \sqrt{2}}$
$\frac{5 π}{6}$	$\frac{1}{3 + 2 \times \frac{\sqrt{3}}{2}} = \frac{1}{3 + \sqrt{3}}$
$π$	$\frac{1}{3 - 2 \times (- 1)} = \frac{1}{5}$