Find the coordinates of the foci,the vertices,the length of the major axis,the minor axis,the eccentricity,and the length of the latus rectum of the ellipse $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$.

The given equation is $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$.
Here,the denominator of $\frac{x^{2}}{36}$ is greater than the denominator of $\frac{y^{2}}{16}$.
Therefore,the major axis is along the $x$-axis,while the minor axis is along the $y$-axis.
On comparing the given equation with $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,we obtain $a=6$ and $b=4$.
$\therefore c=\sqrt{a^{2}-b^{2}}=\sqrt{36-16}=\sqrt{20}=2\sqrt{5}$.
Therefore:
The coordinates of the foci are $(\pm 2\sqrt{5}, 0)$.
The coordinates of the vertices are $(\pm 6, 0)$.
Length of the major axis $= 2a = 12$.
Length of the minor axis $= 2b = 8$.
Eccentricity,$e = \frac{c}{a} = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3}$.
Length of the latus rectum $= \frac{2b^{2}}{a} = \frac{2 \times 16}{6} = \frac{16}{3}$.