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: 45


Vertices: $(±\frac{5}{3},0)$ Eccentricity: $e=\frac{9}{25}$

Work Step by Step

$\frac{x^2}{\frac{25}{9}}+\frac{y^2}{\frac{16}{9}}=1$ $\frac{x^2}{(\frac{5}{3})^2}+\frac{y^2}{(\frac{4}{3})^2}=1$ The major axis is horizontal. Standard form when major axis is horizontal: $\frac{x^2}{a^2}+\frac{y^2}{a^2}=1$ So: $a=\frac{5}{3}$ and $b=\frac{4}{3}$ $a^2=b^2+c^2$ $c^2=a^2-b^2=\frac{25}{9}-\frac{16}{9}=\frac{9}{9}=1$ $c=1$ $e=\frac{c}{a}=\frac{1}{\frac{25}{9}}=\frac{9}{25}$ Vertices when major axis is horizontal: $(±a,0)=(±\frac{5}{3},0)$
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