Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 9 - Section 9.6 - Conic Sections in Polar Coordinates - Exercise Set - Page 1032: 54


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Work Step by Step

Step 1. Using the figure given in the exercise, we can write a general form for the ellipse as $r=\frac{ep}{1-e\ cos\theta}$ Step 2. At $\theta=0$, we have $r(0)=\frac{ep}{1-e}$ Step 3. At $\theta=\pi$, we have $r(\pi)=\frac{ep}{1+e}$ Step 4. As $r(0)+r(\pi)=2a$, we have $\frac{ep}{1-e}+\frac{ep}{1+e}=2a$ Step 5. To solve for $p$, multiply both sides with $(1+e)(1-e)$. We have $(e+e^2)p+(e-e^2)p=2a(1-e^2)$ or $2ep=2a(1-e^2)$, thus $p=\frac{(1-e^2)a}{e}$ Step 6. The equation from Step-1 becomes $r=\frac{(1-e^2)a}{1-e\ cos\theta}$, which is the equation given in the exercise.
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