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: 25

Answer

$\frac{x^2}{25}+\frac{y^2}{9}=1$

Work Step by Step

Step 1. From the given conditions, we have $c=4, a=5$; thus $b=\sqrt {5^2-4^2}=3$ Step 2. We can write the equation as $\frac{x^2}{25}+\frac{y^2}{9}=1$
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