Answer
$\dfrac{x^2}{27}+\dfrac{y^2}{36}=1$
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: $e=0.5$ ; foci: $(0,\pm 3)$
Now, $m=\dfrac{c}{e}=\dfrac{3}{0.5}=6$ and $q^2=m^2-c^2=36-9=27$
Here, the equation of the ellipse is:
$\dfrac{x^2}{27}+\dfrac{y^2}{36}=1$
