Answer
$\dfrac{x^2}{100}+\dfrac{y^2}{94.24}=1$
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: $(\pm 10,0)$
Now, $c=(10)(0.24)=2.4$ and $n^2=m^2-c^2=100-5.76=94.24
$
Here, the equation of the ellipse is
$\dfrac{x^2}{100}+\dfrac{y^2}{94.24}=1$