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: 42

Answer

asymptotes: $y=\pm(x+3)+2$ foci: $(-3,2\pm\sqrt {10})$
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Work Step by Step

Step 1. Rewrite the equation as $\frac{(y-2)^2}{5}-\frac{(x+3)^2}{5}=1$, we have $a=b=\sqrt 5, c=\sqrt {a^2+b^2}=\sqrt {10}$ centered at $(-3,2)$ with a vertical transverse axis. Step 2. We can find the vertices as $(-3,2\pm\sqrt 5)$ and asymptotes as $y-2=\pm\frac{a}{b}(x+3)$ or $y=\pm(x+3)+2$ Step 3. We can graph the equation as shown in the figure with foci at $(-3,2\pm\sqrt {10})$
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