Answer
$\dfrac{y^2}{16}-\dfrac{x^2}{9}=1$
Work Step by Step
The eccentricity of the ellipse $\dfrac{x^2}{m^2}+\dfrac{y^2}{n^2}=1$ when $m \gt n$ is given by:
$e=\dfrac{\sqrt {m^2-n^2}}{m}$
The foci of the ellipse are: $(\pm me,0)$ and the directrices are given as: $x=\pm \dfrac{m}{e}$
Given: Vertices: $(0, \pm 5)$ and $e=1.25$
so, $e=\dfrac{c}{a}=1.25$ or, $5=\dfrac{5}{4}(m)$
and $m=4$
Also, $q^2=c^2-p^2=25-16=9$
so, the equation of the ellipse is:
$\dfrac{y^2}{16}-\dfrac{x^2}{9}=1$
