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: 54

Answer

$\frac{(x+5)^2}{16}+\frac{(y-1)^2}{4}=1$, see graph, foci $(-5\pm2\sqrt 3,1)$.
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Work Step by Step

Step 1. Rewrite the given equation as $(x^2+10x+25)+4(y^2-2y+1)=25+4-13=16$ or $\frac{(x+5)^2}{16}+\frac{(y-1)^2}{4}=1$, Step 2. We have $a^2=16, b^2=4$ and $c=\sqrt {a^2-b^2}=2\sqrt 3$. The ellipse is centered at $(-5,1)$ with a horizontal major axis. Step 3. We can graph the equation as shown in the figure with foci at $(-5\pm2\sqrt 3,1)$.
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