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: 8

Answer

See graph and explanations.
1584720971

Work Step by Step

Step 1. Graph the equation as shown in the figure. Step 2. Rewrite the equation as $\frac{y^2}{16}-\frac{x^2}{4}=1$; 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 $(0,0)$ and the major axis is vertical, the foci are $(0,\pm2\sqrt {5})$ and the asymptotes are $y=\pm \frac{4}{2}x=2x$ as shown in the figure.
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