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: 55

Answer

$\frac{(x+2)^2}{16}+\frac{(y-3)^2}{64}=1$, foci: $(-2, 3\pm4\sqrt 3)$.
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Work Step by Step

Step 1. Rewrite the given equation as $4(x^2+4x+4)+(y^2-6y+9)=39+16+9=64$ or $\frac{(x+2)^2}{16}+\frac{(y-3)^2}{64}=1$, Step 2. We have $a^2=64, b^2=16$ and $c=\sqrt {a^2-b^2}=4\sqrt 3$. The ellipse is centered at $(-2,3)$ with a vertical major axis. Step 3. We can graph the equation as shown in the figure with foci at $(-2, 3\pm4\sqrt 3)$.
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