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: 9

Answer

$\frac{y^2}{36}-\frac{x^2}{9}=1$

Work Step by Step

Step 1. From the given conditions, we have $a=6$ with a vertical transverse axis; thus the equation is in the form of $\frac{y^2}{36}-\frac{x^2}{b^2}=1$ Step 2. The asymptotes can be found as $y=\pm\frac{a}{b}x=\pm\frac{6}{b}x$. Comparing with the given asymptote, we have $\frac{6}{b}=2$, which gives $b=3$ Step 3. The equation is $\frac{y^2}{36}-\frac{x^2}{9}=1$
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