Answer
Center at $(0,0).$
Vertices at $(0,\pm 5).$
Foci at $(0,\pm 4)$
Work Step by Step
$\displaystyle \frac{x^{2}}{3^{2}}+\frac{y^{2}}{5^{2}}=1\qquad $.... Center at $(0,0)$.
The greater denominator is $a^{2}=5^{2}$,
it is below $y^{2}$ so the major axis is along the y-axis. Apply the theorem:
An equation of the ellipse with center at $(0,0),$
foci at $(0,-c)$ and $(0,c),$ and vertices at $(0,-a)$ and $(0,a)$ is
$\displaystyle \frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\quad $ where $a \gt b \gt 0$ and $b^{2}=a^{2}-c^{2}$
$a=5,\quad b=3$
Center at $(0,0).$
Vertices at $(0,\pm 5).$
Find $c$ from $b^{2}=a^{2}-c^{2}$
$c^{2}=a^{2}-b^{2}$
$c^{2}=25-9$
$c^{2}=16$
$c=4$
Foci at $(0,\pm 4)$
