Answer
$\displaystyle \frac{x^{2}}{4}+\frac{y^{2}}{9}=1 \quad$ (ellipse)
Foci: $(0, \pm\sqrt{5})$
Vertices: $(0, \pm 3)$
Work Step by Step
Ellipse, major axis is vertical.
Foci on the y-axis: $\quad \displaystyle \frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1 \quad (a\gt b)$
Center-to-focus distance: $\quad c=\sqrt{a^{2}-b^{2}}$
Foci: $(0, \pm c)$
Vertices: $(0, \pm a)$
Equation offered: $\quad \displaystyle \frac{x^{2}}{4}+\frac{y^{2}}{9}=1 \quad a=3, b=2$
Center-to-focus distance: $c=\sqrt{a^{2}-b^{2}}=\sqrt{9-4}=\sqrt{5}$
Foci: $(0, \pm\sqrt{5})$
Vertices: $(0, \pm 3)$
