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: 41

Answer

asymptotes: $y=\pm(x-1)+2$ foci: $(1\pm\sqrt {6},2)$
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Work Step by Step

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