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 983: 64


$\frac{x^2}{9}-\frac{4y^2}{9}=1$ or $x^2-4y^2=9$; right branch.

Work Step by Step

Step 1. Use the figure given and rewrite the equation as $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. We have $a=3$ and $\frac{b}{a}=\frac{1}{2}$ (slope of the asymptote) with center at $(0,0)$ and a horizontal transverse axis. Step 2. Thus, we have $b=\frac{3}{2}$ and the equation becomes $\frac{x^2}{9}-\frac{4y^2}{9}=1$ or $x^2-4y^2=9$ (right branch).
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