Precalculus (6th Edition) Blitzer

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

Chapter 9 - Section 9.1 - The Ellipse - Exercise Set - Page 967: 52

Answer

$\frac{(x-4)^2}{9}+\frac{(y+2)^2}{4}=1$; foci at $(4\pm\sqrt 5,-2)$.
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Work Step by Step

Step 1. Rewrite the given equation as $4(x^2-8x+16)+9(y^2+4y+4)=64+36-64=36$ or $\frac{(x-4)^2}{9}+\frac{(y+2)^2}{4}=1$, Step 2. We have $a^2=9, b^2=4$ and $c=\sqrt {a^2-b^2}=\sqrt 5$. The ellipse is centered at $(4,-2)$ with a horizontal major axis. Step 3. We can graph the equation as shown in the figure with foci at $(4\pm\sqrt 5,-2)$.
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