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: 60

Answer

$\frac{(x)^2}{25}+\frac{(y-6)^2}{16}=1$, foci: $(\pm3,6)$.
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Work Step by Step

Step 1. Rewrite the given equation as $16(x^2)+25(y^2-12y+36)=25(36)-500=400$ or $\frac{(x)^2}{25}+\frac{(y-6)^2}{16}=1$, Step 2. We have $a^2=25, b^2=16$ and $c=\sqrt {a^2-b^2}=3$. The ellipse is centered at $(0,6)$ with a horizontal major axis. Step 3. We can graph the equation as shown in the figure with foci at $(\pm3,6)$.
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