Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 9 - Mid-Chapter Check Point - Page 1000: 11


See graph and explanations.

Work Step by Step

Step 1. Graph the equation as shown in the figure. Step 2. Rewrite the equation as $4(x^2+2x+1)-(y^2-6y+9)=4-11-9$ or $-\frac{(x+1)^2}{4}+\frac{(y-3)^2}{16}=1$; we have $a=4, b=2$ and thus $c=\sqrt {4^2+2^2}=2\sqrt {5}$ Step 3. As the center of the ellipse is at $(-1,3)$ and the major axis is vertical, the foci are $(-1, 3\pm2\sqrt {5})$ and the asymptotes are $y-3=\pm \frac{4}{2}(x+1)$ or $y=\pm 2(x+1)+3$ as shown in the figure.
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