Precalculus (6th Edition) Blitzer

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

Chapter 9 - Review Exercises - Page 1035: 23


$-\frac{x^2}{12}+\frac{y^2}{4}=1$ or $\frac{y^2}{4}-\frac{x^2}{12}=1$

Work Step by Step

Step 1. From the given conditions, we have $c=4, a=2$; thus $b=\sqrt {4^2-2^2}=2\sqrt 3$ with center at $(0,0)$ and foci on $x=0$ Step 2. We can write the equation as $-\frac{x^2}{12}+\frac{y^2}{4}=1$ or $\frac{y^2}{4}-\frac{x^2}{12}=1$
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