Answer
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:
$(±a,0)=(±2,0)$
