University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 10 - Practice Exercises - Page 594: 61


$\dfrac{2}{2+\sin \theta}$; an ellipse

Work Step by Step

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