## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Chapter 4 - 4.3 - Conics - 4.3 Exercises - Page 338: 52

#### Answer

Vertices: $(±1,0)$ Eccentricity: $e=\frac{2\sqrt 2}{3}$

#### Work Step by Step

$x^2+9y^2=1$ $\frac{x^2}{1}+\frac{y^2}{\frac{1}{9}}=1$ $\frac{x^2}{1^2}+\frac{y^2}{(\frac{1}{3})^2}=1$ The major axis is horizontal. Standard form when major axis is horizontal: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ So: $a=1$ and $b=\frac{1}{3}$ $a^2=b^2+c^2$ $c^2=a^2-b^2=1-\frac{1}{9}=\frac{8}{9}$ $c=\frac{2\sqrt 2}{3}$ $e=\frac{c}{a}=\frac{\frac{2\sqrt 2}{3}}{1}=\frac{2\sqrt 2}{3}$ Vertices when major axis is horizontal: $(±a,0)=(±1,0)$

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