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 981: 12



Work Step by Step

Step 1. From the given conditions, the hyperbola has a center at $(-2,1)$ and a vertical transverse axis; thus the equation is in the form of $\frac{(y-1)^2}{a^2}-\frac{(x+2)^2}{b^2}=1$ Step 2. From the given coordinates, we have $a=4-1=3, c=6-1=5$ Thus $b^2=c^2-a^2=16$ Step 3. The equation is $\frac{(y-1)^2}{9}-\frac{(x+2)^2}{16}=1$
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