Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 9 - Mid-Chapter Check Point - Page 1000: 33


a. $\frac{400x^2}{441}-\frac{400y^2}{3159}=1$ b. See graph.

Work Step by Step

a. Based on the given conditions, we can write the general equation of the hyperbola as $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ with $2c=6, c=3$ and $2a=(6sec)(0.35km/sec)=2.1, a=\frac{21}{20}, a^2=\frac{441}{400}$ Thus $b^2=c^2-a^2=\frac{3159}{400}$ and the equation is $\frac{400x^2}{441}-\frac{400y^2}{3159}=1$ b. See graph.
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