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: 33

Answer

asymptotes: $y=\pm\frac{4}{3}(x+4)-3$ foci: $(-9,-3)$ and $(1,-3)$
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Work Step by Step

Step 1. From the given equation $\frac{(x+4)^2}{9}-\frac{(y+3)^2}{16}=1$, we have $a=3, b=4, c=\sqrt {a^2+b^2}=5$ centered at $(-4,-3)$ with a horizontal transverse axis. Step 2. We can find the vertices as $(-7,-3), (-1,-3), (-4,-7),(-4,1)$ and asymptotes as $y+3=\frac{b}{a}(x+4)$ or $y=\pm\frac{4}{3}(x+4)-3$ Step 3. We can graph the equation as shown in the figure with foci at $(-4\pm5,-3)$ or $(-9,-3)$ and $(1,-3)$
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