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: 39


See graph and explanations.

Work Step by Step

Step 1. Rewrite the given equation as $\frac{(x+3)^2}{16}+\frac{(y-2)^2}{4}=1$; we have $a^2=16, b^2=4$ and $c=\sqrt {a^2-b^2}=2\sqrt 3$. The ellipse is centered at $(-3,2)$ with a horizontal major axis. Step 2. We can graph the equation as shown in the figure with foci at $(-3\pm2\sqrt 3,2)$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.