## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 9 - Section 9.6 - Conic Sections in Polar Coordinates - Exercise Set - Page 1032: 51

#### Answer

$r=\frac{12}{2+cos\theta}$ or $r=\frac{4}{2-cos\theta}$

#### Work Step by Step

Step 1. Given an ellipse with focus at the pole and vertex $(4,0)$, we can see that the directrix will be perpendicular to the polar axis and the equation will be one of the forms: $r=\frac{ep}{1+e\ cos\theta}$ or $r=\frac{ep}{1-e\ cos\theta}$. Step 2. As $e=\frac{1}{2}$, we have $r=\frac{\frac{1}{2}p}{1+\frac{1}{2} cos\theta}=\frac{p}{2+cos\theta}$ or $r=\frac{\frac{1}{2}p}{1-\frac{1}{2}cos\theta}=\frac{p}{2-cos\theta}$ Step 3. The first equation gives $r(0)=\frac{p}{3}$. Let $\frac{p}{3}=4$ (vertex). We have $p=12$ and the equation becomes $r=\frac{12}{2+cos\theta}$ Step 4. The second equation gives $r(0)=p$. Let $p=4$ (vertex). The equation becomes $r=\frac{4}{2-cos\theta}$

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