Precalculus (6th Edition) Blitzer

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

Chapter 9 - Section 9.1 - The Ellipse - Exercise Set - Page 966: 20


$\frac{x^2}{16}+\frac{y^2}{4}=1$, foci $(\pm2\sqrt 3,0)$

Work Step by Step

Step 1. From the given graph, we can identify $a=4, b=2$; thus $c=\sqrt {a^2-b^2}=2\sqrt {3}$. The ellipse is centered at $(0,0)$ with a horizontal major axis. Step 2. We can write the equation as $\frac{x^2}{16}+\frac{y^2}{4}=1$ with foci at $(\pm2\sqrt 3,0)$
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