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: 4

Answer

See graph and explanations.
1584976954

Work Step by Step

Step 1. Given $\frac{x^2}{16}+\frac{y^2}{49}=1$, we can identify $a^2=49, b^2=16$; thus $c=\sqrt {a^2-b^2}=\sqrt {33}$. The ellipse is centered at $(0,0)$ with a vertical major axis. Step 2. We can graph the equation as shown in the figure, and the foci can be located at $(0, \pm\sqrt {33})$
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