Answer
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:
$(±a,0)=(±\frac{5}{3},0)$
