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: 40

Answer

See graph and explanations.

Work Step by Step

Step 1. Rewrite the given equation as $\frac{(x-3)^2}{18}+\frac{(y+2)^2}{2}=1$; we have $a^2=18, b^2=2$ and $c=\sqrt {a^2-b^2}=4$. 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 $(7,-2)$ and $(-1,-2)$
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