## University Calculus: Early Transcendentals (3rd Edition)

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