Answer
Vertices: $(0,±1)$
Eccentricity: $e=\frac{\sqrt 3}{2}$
Work Step by Step
$4x^2+y^2=1$
$\frac{x^2}{\frac{1}{4}}+\frac{y^2}{1}=1$
$\frac{x^2}{(\frac{1}{2})^2}+\frac{y^2}{1^2}=1$
The major axis is vertical.
Standard form when major axis is vertical:
$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$
So: $a=1$ and $b=\frac{1}{2}$
$a^2=b^2+c^2$
$c^2=a^2-b^2=1-\frac{1}{4}=\frac{3}{4}$
$c=\frac{\sqrt 3}{2}$
$e=\frac{c}{a}=\frac{\frac{\sqrt 3}{2}}{1}=\frac{\sqrt 3}{2}$
Vertices when major axis is horizontal:
$(0,±a)=(0,±1)$
