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: 57

Answer

$\frac{(x-3)^2}{4}+\frac{(y+4)^2}{25}=1$, foci: $(3,-4\pm\sqrt {21})$.
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Work Step by Step

Step 1. Rewrite the given equation as $25(x^2-6x+9)+4(y^2+8y+16)=225+64-189=100$ or $\frac{(x-3)^2}{4}+\frac{(y+4)^2}{25}=1$, Step 2. We have $a^2=25, b^2=4$ and $c=\sqrt {a^2-b^2}=\sqrt {21}$. The ellipse is centered at $(3,-4)$ with a vertical major axis. Step 3. We can graph the equation as shown in the figure with foci at $(3,-4\pm\sqrt {21})$.
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