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: 46

Answer

See graph and explanations.
1585007205

Work Step by Step

Step 1. Given $\frac{(x+2)^2}{16}+\frac{(y-3)^2}{1}=1$, we have $a^2=16, b^2=1$ and $c=\sqrt {a^2-b^2}=\sqrt {15}$. The ellipse is centered at $(-2,3)$ with a horizontal major axis. Step 2. We can graph the equation as shown in the figure with foci at $(-2\pm\sqrt {15},3)$.
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