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: 40

Answer

asymptotes: $y=\pm\frac{1}{3}(x+3)+4$ foci: $(-3\pm\sqrt {10},4)$
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Work Step by Step

Step 1. Rewriting the equation as $\frac{(x+3)^2}{9}-\frac{(y-4)^2}{1}=1$, we have $a=3, b=1, c=\sqrt {a^2+b^2}=\sqrt {10}$ centered at $(-3,4)$ with a horizontal transverse axis. Step 2. We can find the vertices as $(-6,4),(0,4)$ and asymptotes as $y-4=\pm\frac{b}{a}(x+3)$ or $y=\pm\frac{1}{3}(x+3)+4$ Step 3. We can graph the equation as shown in the figure with foci at $(-3\pm\sqrt {10},4)$
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