Explanation Check for symmetry test: Polar axis: Replace θ by − θ . The result is r = 1 1 − cos ( − θ ) = 1 1 − 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 1 − cos ( π − θ ) = 1 1 − ( cos π cos θ + sin π sin θ ) = 1 1 − ( ( − 1 ) cos θ + ( 0 ) sin θ ) = 1 1 + 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 1 − cos θ , so r = − 1 1 − cos θ . This test fails. Replace θ by θ + π . The result is r = 1 1 − cos ( θ + π ) = 1 1 − ( cos π cos θ − sin π sin θ ) = 1 1 − ( − 1 ) cos θ = 1 1 + 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 ...

Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (4th Edition)

4th Edition

ISBN: 9780134686974

Author: Michael Sullivan, Michael Sullivan III

Publisher: PEARSON

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Chapter 8.2, Problem 76AYU

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To find: The graph of the polar equation $r = \frac{1}{1 - cos θ} (p a r a b o l a)$ .

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Chapter 8.2, Problem 75AYU

Chapter 8.2, Problem 77AYU

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Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (4th Edition)

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