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: 40

Answer

Graph: .

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}\Rightarrow\quad $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 and the parabola opens right. The vertex is halfway between the directrix and focus, 1 unit left of the focus at the origin, with polar coordinates: $(-1,0)$ or $(1, \pi)$.
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