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: 44

Answer

asymptotes: $y=\pm2(x+4)+3$ foci: $(-4\pm2\sqrt {5},3)$
1585145546

Work Step by Step

Step 1. Rewriting the equation as $4(x^2+8x+16)-(y^2-6y+9)=64-9-39=16$ or $\frac{(x+4)^2}{4}-\frac{(y-3)^2}{16}=1$, we have $a=2,b=4, c=\sqrt {a^2+b^2}=2\sqrt {5}$ centered at $(-4,3)$ with a horizontal transverse axis. Step 2. We can find the vertices as $(-8,3),(0,3)$ and asymptotes as $y-3=\pm\frac{b}{a}(x+4)$ or $y=\pm2(x+4)+3$ Step 3. We can graph the equation as shown in the figure with foci at $(-4\pm2\sqrt {5},3)$
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