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 966: 39

Answer

See graph and explanations.
1584988691

Work Step by Step

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