#### Answer

The vertices are $(\pm9,0)$ and $(0,\pm4)$. The foci are $(\pm \sqrt{65},0)$.

#### Work Step by Step

The equation
$$ \left(\frac{x}{9}\right)^{2}+\left(\frac{y}{4}\right)^{2}=1. $$
is an ellipse. We have $a=9$, $b=4$, so $a\gt b$, $c=\sqrt{a^2-b^2}=\sqrt{81-16}=\sqrt{65}$ and hence: The vertices are $(\pm9,0)$ and $(0,\pm4)$. The foci are $(\pm \sqrt{65},0)$.