Answer
$\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$
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: $e=0.24$ ; vertices: $(\sqrt 5,0)$
the directrix is given as: $x=\dfrac{9}{\sqrt 5}$
Now, $e=\dfrac{\sqrt 5}{3}$
So, the equation of the ellipse is:
$(x-\sqrt 5)^2+y^2=(\dfrac{5}{9})(x-\dfrac{9}{\sqrt 5})^2$
or, $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$