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: 48

Answer

asymptotes: $y=\pm\frac{2}{3}(x+1)-1$ foci: $(-1\pm\frac{\sqrt {13}}{6},-1)$
1585148173

Work Step by Step

Step 1. Rewriting the equation as $4(x^2+2x+1)-9(y^2+2y+1)=6+4-9=1$ or $\frac{(x+1)^2}{1/4}-\frac{(y+1)^2}{1/9}=1$ we have $a=\frac{1}{2},b=\frac{1}{3}, c=\sqrt {a^2+b^2}=\frac{\sqrt {13}}{6}$ centered at $(-1,-1)$ with a horizontal transverse axis. Step 2. We can find the vertices as $(-\frac{3}{2},-1),(-\frac{1}{2},-1)$ and asymptotes as $y+1=\pm\frac{b}{a}(x+1)$ or $y=\pm\frac{2}{3}(x+1)-1$ Step 3. We can graph the equation as shown in the figure with foci at $(-1\pm\frac{\sqrt {13}}{6},-1)$
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