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

Answer

See graph and explanations.
1584984459

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 $(0,\pm\frac{\sqrt 3}{2})$
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