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: 51

Answer

$\frac{(x-2)^2}{25}+\frac{(y+1)^2}{9}=1$, see graph, foci $(6,-1)$ and $(-2,-1)$.
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Work Step by Step

Step 1. Rewrite the given equation as $9(x^2-4x+4)+25(y^2+2y+1)=164+36+25=225$ or $\frac{(x-2)^2}{25}+\frac{(y+1)^2}{9}=1$, Step 2. We have $a^2=25, b^2=9$ and $c=\sqrt {a^2-b^2}=4$. The ellipse is centered at $(2,-1)$ with a horizontal major axis. Step 3. We can graph the equation as shown in the figure with foci at $(6,-1)$ and $(-2,-1)$.
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