Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 9 - Review Exercises - Page 1035: 22

Answer

See graph: foci $(1\pm\sqrt {2},-1)$, asymptotes $y=\pm(x-1)-1$
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Work Step by Step

Step 1. Rewrite the given equation as $(x^2-2x+1)-(y^2+2y+1)=1+1-1$ or $(x-1)^2-(y+1)^2=1$. We have $a=1, b=1$. Thus $c=\sqrt {1^2+1^2}=\sqrt {2}$ and the center is at $(1,-1)$ with foci on $y=-1$. Step 2. See graph; we can locate the foci at $(1\pm\sqrt {2},-1)$ Step 3. The asymptotes can be found as $y+1=\pm\frac{1}{1}(x-1)$ or $y=\pm(x-1)-1$
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