Precalculus (6th Edition) Blitzer

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

Chapter 9 - Review Exercises - Page 1035: 18

Answer

See graph. Foci $(0,\pm2\sqrt 5)$; asymptotes $y= \pm\frac{1}{2}x$
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Work Step by Step

Step 1. Rewrite the given equation as $-\frac{x^2}{16}+\frac{y^2}{4}=1$. We have $a=2, b=4$ and thus $c=\sqrt {4^2+2^2}=2\sqrt 5$. The foci are on the y-axis. Step 2. See graph; we can locate the foci at $(0,\pm2\sqrt 5)$ Step 3. The asymptotes can be found as $y=\pm\frac{2}{4}x=\pm\frac{1}{2}x$
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