Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 9 - Section 9.6 - Conic Sections in Polar Coordinates - Exercise Set - Page 1031: 44

Answer

See graph and explanations.

Work Step by Step

Step 1. Graph both $r=\frac{3}{2+6cos\theta}$ (red) and $r=\frac{3}{2+6cos(\theta+\frac{\pi}{3})}$ (blue) as shown in the figure. Step 2. We can see that the blue curve is the result of the red curve rotated $\frac{\pi}{3}$ clockwise around the pole.
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