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: 12


See graph and explanations.

Work Step by Step

Step 1. Rewrite the equation as $4x^2+y^2=1$; we can identify $a^2=1, b^2=\frac{1}{4}$. Thus $c=\sqrt {a^2-b^2}=\frac{\sqrt 3}{2}$. 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 $(0,\pm\frac{\sqrt 3}{2})$
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.