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: 59


$\frac{(x-3)^2}{9}+\frac{(y)^2}{36}=1$, foci: $(3,\pm3\sqrt {3})$.

Work Step by Step

Step 1. Rewrite the given equation as $36(x^2-6x+9)+9(y^2)=324$ or $\frac{(x-3)^2}{9}+\frac{(y)^2}{36}=1$, Step 2. We have $a^2=36, b^2=9$ and $c=\sqrt {a^2-b^2}=3\sqrt {3}$. The ellipse is centered at $(3,0)$ with a vertical major axis. Step 3. We can graph the equation as shown in the figure with foci at $(3,\pm3\sqrt {3})$.
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