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: 42

Answer

See graph and explanations.
1585006316

Work Step by Step

Step 1. Given $\frac{(x-3)^2}{9}+\frac{(y+1)^2}{16}=1$, we have $a^2=16, b^2=9$ and $c=\sqrt {a^2-b^2}=\sqrt 7$. The ellipse is centered at $(3,-1)$ with a vertical major axis. Step 2. We can graph the equation as shown in the figure with foci at $(3,-1\pm\sqrt 7)$.
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