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

$\dfrac{x^2}{1}-\dfrac{y^2}{8}=1$
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 the vertices are: $(\pm 3,0)$ and $e=3$ so, $e=\dfrac{c}{a}=3\implies c=3p=(3)(1)=3$ Now, $q^2=c^2-p^2=9-1=8$ Thus, the equation of the ellipse becomes: $\dfrac{x^2}{1}-\dfrac{y^2}{8}=1$