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: 28

Answer

$\dfrac{x^2}{9}-\dfrac{y^2}{4}=1$

Work Step by Step

Center: $(0,0)$ and vertices: $(-3,0); (3,0)$ This gives: $ a=3$ and Asymptote: $ y=\dfrac{2}{3} x \implies b=2$ And we have a horizontal traverse axis. Standard Equation for a hyperbola is : $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ We have $ a^2=9; b^2 =4$ So, the equation for the hyperbola is: $\dfrac{x^2}{9}-\dfrac{y^2}{4}=1$
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