Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 10 - Parametric Equations and Polar Coordinates - 10.6 Conic Sections in Polar Coordinates - 10.6 Exercise - Page 728: 21


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

Work Step by Step

Consider the eccentricity which is defined as $e=\dfrac{|PF|}{|Pl|}$ Since, $|PF|=r$ and $|Pl|=d-r \cos \alpha$ But $\alpha =\pi-\theta$ This implies that, $|Pl|=d-r \cos (\alpha)=d-r\cos (\pi-\theta)$ $\implies |Pl|=d+r \cos \theta$ Further, we have $\dfrac{|PF|}{|Pl|}=e \implies \dfrac{r}{d+r \cos \theta}=e $ $\implies r=ed+er \cos \theta$ Thus, $r(1-e \cos \theta)=ed \implies r=\dfrac{ed}{1-e \cos \theta}$
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