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: 43

Answer

See graph and explanations.

Work Step by Step

Step 1. Given $\frac{(x)^2}{25}+\frac{(y-2)^2}{36}=1$, we have $a^2=36, b^2=25$ and $c=\sqrt {a^2-b^2}=\sqrt {11}$. The ellipse is centered at $(0,2)$ with a vertical major axis. Step 2. We can graph the equation as shown in the figure with foci at $(0,2\pm\sqrt {11})$.
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