Answer
$\dfrac{x^2}{100}+\dfrac{y^2}{94.24}=1$
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: $(\pm 10,0)$
Now, $c=ae=(10)(0.24)=2.4$ and $q^2=p^2-c^2=100-5.76=94.24
$
Thus, the equation of the ellipse becomes:
$\dfrac{x^2}{100}+\dfrac{y^2}{94.24}=1$
