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 968: 90

Answer

See explanations.

Work Step by Step

From $c^2=a^2-b^2$, divide by $a^2$ on both sides; we have $\frac{c^2}{a^2}=1-\frac{b^2}{a^2}$. As $\frac{c}{a}\to 0$, we have $\frac{b^2}{a^2}\to 1$, which means that the graph will approach to a circle.
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