Answer
The foci are ${F_1} = \left( {\sqrt 5 ,0} \right)$ and ${F_2} = \left( { - \sqrt 5 ,0} \right)$.
The focal vertices are $\left( { \pm a,0} \right) = \left( { \pm 3,0} \right)$ and the minor vertices are $\left( {0, \pm b} \right) = \left( {0, \pm 2} \right)$.
Work Step by Step
By Theorem 1 of Section 12.5, this is the equation of an ellipse in standard position, where $a=3$ and $b=2$. Since $a>b>0$, we have
$c = \sqrt {{a^2} - {b^2}} = \sqrt {9 - 4} = \sqrt 5 $
The foci are ${F_1} = \left( {\sqrt 5 ,0} \right)$ and ${F_2} = \left( { - \sqrt 5 ,0} \right)$.
The focal vertices are $\left( { \pm a,0} \right) = \left( { \pm 3,0} \right)$ and the minor vertices are $\left( {0, \pm b} \right) = \left( {0, \pm 2} \right)$.
