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