Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 10 - Section 10.6 - Conic Sections in Polar Coordinates - 10.6 Exercises - Page 688: 23

Answer

$r=\dfrac{ed}{1-e \sin \theta}$

Work Step by Step

The eccentricity is defined as: $\dfrac{|PF|}{|Pl|}=e$ ...(1) Here, we have $|PF|=r$ and $|Pl|=d-r \sin \alpha$ and $\alpha =2\pi-\theta$ Thus, $|Pl|=d-r \sin \alpha=d-r\cos (2\pi-\theta)$ This implies that $|Pl|=d+r \sin \theta$ Now, the equation (1) becomes: $\dfrac{|PF|}{|Pl|}=e \implies \dfrac{r}{d+r \sin \theta}=e$ or, $r=ed+er \sin \theta$ This gives: $r(1-e \sin \theta)=ed$ Hence, $r=\dfrac{ed}{1-e \sin \theta}$
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