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: 7



Work Step by Step

Foci: $(-4,0), (4,0)$ and Vertices: $(-3,0)$ and $(3,0)$ We are given that $ c=4; a=3$` Since $ c^2=a^2+b^2$ so, $ b^2=c^2-a^2=16-9=7$ Thus, the equation for the hyperbola is: $\dfrac{x^2}{9}-\dfrac{y^2}{7}=1$
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