Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 4 - 4.3 - Conics - 4.3 Exercises - Page 338: 46


Vertices: $(±2,0)$ Eccentricity: $e=\frac{\sqrt {15}}{4}$

Work Step by Step

$\frac{x^2}{4}+\frac{y^2}{\frac{1}{4}}=1$ $\frac{x^2}{2^2}+\frac{y^2}{(\frac{1}{2})^2}=1$ The major axis is horizontal. Standard form when major axis is horizontal: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ So: $a=2$ and $b=\frac{1}{2}$ $a^2=b^2+c^2$ $c^2=a^2-b^2=4-\frac{1}{4}=\frac{15}{4}$ $c=\frac{\sqrt {15}}{2}$ $e=\frac{c}{a}=\frac{\frac{\sqrt {15}}{2}}{2}=\frac{\sqrt {15}}{4}$ Vertices when major axis is horizontal: $(±a,0)=(±2,0)$
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