## Calculus 8th Edition

Published by Cengage

# Chapter 10 - Parametric Equations and Polar Coordinates - 10.6 Conic Sections in Polar Coordinates - 10.6 Exercise - Page 728: 22

#### Answer

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

#### Work Step by Step

Consider the eccentricity which is defined as $e=\dfrac{|PF|}{|Pl|}$ and $|PF|=r$ and $|Pl|=d-r \sin \theta$ Thus, we have $e=\dfrac{|PF|}{|Pl|}=e \implies e=\dfrac{r}{d-r \sin (\theta)}$ After simplifications, we get $r=ed-er \sin (\theta)$ $\implies r(1+e \sin \theta)=ed$ Thus, $r=\dfrac{ed}{1+e \sin \theta}$

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