Precalculus (6th Edition) Blitzer

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

Chapter 9 - Section 9.2 - The Hyperbola - Exercise Set - Page 982: 37

Answer

asymptotes: $y=\pm\frac{1}{2}(x-1)-2$ foci: $(1,-2\pm2\sqrt {5})$
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Work Step by Step

Step 1. From the given equation $\frac{(y+2)^2}{4}-\frac{(x-1)^2}{16}=1$, we have $a=2, b=4, c=\sqrt {a^2+b^2}=2\sqrt {5}$ centered at $(1,-2)$ with a vertical transverse axis. Step 2. We can find the vertices as $(1,-4),(1,0)$ and asymptotes as $y+2=\pm\frac{a}{b}(x-1)$ or $y=\pm\frac{1}{2}(x-1)-2$ Step 3. We can graph the equation as shown in the figure with foci at $(1,-2\pm2\sqrt {5})$
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