University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 10 - Section 10.6 - Conics in Polar Coordinates - Exercises - Page 591: 40

Answer

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Work Step by Step

$\displaystyle \frac{4\div 2}{(2-2\cos\theta)\div 2}=\frac{2}{1-\cos\theta}$ $r=\displaystyle \frac{2}{1-\cos\theta}$ Compare with $r=\displaystyle \frac{ke}{1\pm e\cos\theta}$ $e=1, \qquad $ This is a parabola. $k=2$ $x=-2 \ $ is the directrix, so the parabola opens right. The vertex is halfway between the directrix and the focus, 1 unit left of the focus at the origin. The polar coordinates are: $(-1,0)$ or $(1, \pi)$
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