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: 44

Answer

Vertices: $(0,±12)$ Eccentricity: $e=\frac{\sqrt {23}}{12}$
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Work Step by Step

$\frac{x^2}{121}+\frac{y^2}{144}=1$ $\frac{x^2}{11^2}+\frac{y^2}{12^2}=1$ The major axis is vertical. Standard form when major axis is vertical: $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ So: $a=12$ and $b=11$ $a^2=b^2+c^2$ $c^2=a^2-b^2=144-121=23$ $c=\sqrt {23}$ $e=\frac{c}{a}=\frac{\sqrt {23}}{12}$ Vertices when major axis is horizontal: $(0,±a)=(0,±12)$
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