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: 30



Work Step by Step

Center: $(0,0)$ and vertices: $(0,-5); (0,5)$ This yields: $ a=2$ and Asymptote: $ y=\dfrac{5}{3} x \implies b=3$ And we have a vertical traverse axis. Standard Equation for a hyperbola is : $\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1$ We have $ a^2=25; b^2 =9$ So, the equation for the hyperbola is: $\dfrac{y^2}{25}-\dfrac{x^2}{9}=1$
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