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: 21

Answer

See graph: foci $(1, 2\pm\sqrt {5})$, asymptotes $y=\pm2(x-1)+2$
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Work Step by Step

Step 1. Rewrite the given equation as $(y^2-4y+4)-4(x^2-2x+1)=4+4-4$ or $\frac{(y-2)^2}{4}-(x-1)^2=1$; we have $a=2, b=1$ and thus $c=\sqrt {2^2+1^2}=\sqrt {5}$ and the center is at $(1,2)$ with foci on $x=1$. Step 2. See graph; we can locate the foci at $(1, 2\pm\sqrt {5})$ Step 3. The asymptotes can be found as $y-2=\pm\frac{2}{1}(x-1)$ or $y=\pm2(x-1)+2$
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