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: 34

Answer

$r=\dfrac{2}{4-\cos \theta}$

Work Step by Step

The equation of conic with eccentricity $e$ and directrix $x=k$ leads to focus can be written as: $r=\dfrac{ke}{1+e \cos \theta}$ ....(1) Given: $e=\dfrac{1}{4},k=2$ Then $x=-2$ Thus, the equation (1) can be written as: $r=\dfrac{ke}{1+e \cos \theta}=\dfrac{(\dfrac{1}{2})}{1-(\dfrac{1}{4})\cos \theta}$ or, $r=\dfrac{2}{4-\cos \theta}$
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