Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 11: Parametric Equations and Polar Coordinates - Section 11.7 - Conics in Polar Coordinates - Exercises 11.7 - Page 686: 44

Answer

Graph: .

Work Step by Step

Sine in the denominator $\Rightarrow\qquad$ the main axis is vertical. We want $1\pm...$ in the denominator, $\displaystyle \frac{4\div 2}{(2-\sin\theta)\div 2}$=$\displaystyle \frac{2}{1-\frac{1}{2}\sin\theta}$ $ r=\displaystyle \frac{4(\frac{1}{2})}{1-\frac{1}{2}\sin\theta},\qquad$ Compare with $r=\displaystyle \frac{ke}{1\pm e\sin\theta}$ $e=\displaystyle \frac{1}{2}, \qquad$ This is an ellipse with vertical major axis. $ k=4,\quad y=-4$ is a directrix $a(1-e^{2})=ke$ $a\displaystyle \cdot\frac{3}{4}=2$ $a=\displaystyle \frac{8}{3}$ $c=ae=\displaystyle \frac{4}{3}$ Center is at $\displaystyle \frac{4}{3}$ units above the focus at the origin; polar coordinates $(\displaystyle \frac{4}{3},\frac{\pi}{2}).$ The vertices are at $\displaystyle \frac{8}{3}$ units above and below the center, $(\displaystyle \frac{8}{3}+\frac{4}{3},\frac{\pi}{2})=(4,\frac{\pi}{2})$ $(\displaystyle \frac{4}{3}-\frac{8}{3},\frac{\pi}{2})=(-\frac{4}{3},\frac{\pi}{2})=(\frac{4}{3},\frac{3\pi}{2})$
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