Answer
$\dfrac{y^2}{16}-\dfrac{x^2}{9}=1$
Work Step by Step
The eccentricity of the ellipse $\dfrac{x^2}{p^2}+\dfrac{y^2}{q^2}=1$ when $p \gt q$ is given by:
$e=\dfrac{\sqrt {p^2-q^2}}{p}$
The foci of the ellipse are: $(\pm pe,0)$ and the directrices are given as: $x=\pm \dfrac{p}{e}$
We are given that the vertices are: $(0, \pm 5)$ and $e=1.25$
so, $e=\dfrac{c}{a}=1.25\implies 5=\dfrac{5}{4}p \implies p=4$
Now, $q^2=c^2-p^2=25-16=9$
Thus, the equation of the ellipse becomes:
$\dfrac{y^2}{16}-\dfrac{x^2}{9}=1$
