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: 35


$\dfrac{10}{5-\cos \theta}$

Work Step by Step

The polar equation of a conic with eccentricity $e$ and directrix $y=-k$ is defined as: $r=\dfrac{ke}{1-e \sin \theta}$ ...(1) We are given that the vertices are: $e=\dfrac{1}{5}$ Thus, equation (1), becomes $r=\dfrac{2}{1-(\dfrac{1}{5})\sin \theta}=\dfrac{10}{5-\cos \theta}$
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