Answer
Vertices: $(0,±12)$
Eccentricity: $e=\frac{\sqrt {23}}{12}$
Work Step by Step
$\frac{x^2}{121}+\frac{y^2}{144}=1$
$\frac{x^2}{11^2}+\frac{y^2}{12^2}=1$
The major axis is vertical.
Standard form when major axis is vertical:
$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$
So: $a=12$ and $b=11$
$a^2=b^2+c^2$
$c^2=a^2-b^2=144-121=23$
$c=\sqrt {23}$
$e=\frac{c}{a}=\frac{\sqrt {23}}{12}$
Vertices when major axis is horizontal:
$(0,±a)=(0,±12)$
