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: 10

Answer

See graph and explanations.
1584978341

Work Step by Step

Step 1. Given $\frac{x^2}{81/4}+\frac{y^2}{25/16}=1$, we can identify $a^2=\frac{81}{4}, b^2=\frac{25}{16}$; thus $c=\sqrt {a^2-b^2}=\frac{\sqrt {299}}{4}$. The ellipse is centered at $(0,0)$ with a horizontal major axis. Step 2. We can graph the equation as shown in the figure, and the foci can be located at $(\pm\frac{\sqrt {299}}{4},0)$
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