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: 10


$\dfrac{x^2}{16}-\dfrac{y^2}{64}=1$ With horizontal traverse axis.

Work Step by Step

Endpoints: $(-4,0), (4, 0)$ and Asymptote: $ y=2x $ We are given that $ a=4$ Since $\dfrac{b}{a}=2 \implies b=8$ so, $ b^2=64$ Thus, the equation for hyperbola is: $\dfrac{x^2}{16}-\dfrac{y^2}{64}=1$ And the traverse axis is horizontal.
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