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 $\frac{x^{2}}{25}+\frac{y^{2}}{100}=1$.

The given equation is $\frac{x^{2}}{25}+\frac{y^{2}}{100}=1$,which can be written as $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{10^{2}}=1$.
Since the denominator of $\frac{y^{2}}{100}$ is greater than the denominator of $\frac{x^{2}}{25}$,the major axis is along the $y$-axis.
Comparing this with the standard form $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$,we get $a=10$ and $b=5$.
We calculate $c = \sqrt{a^{2}-b^{2}} = \sqrt{100-25} = \sqrt{75} = 5\sqrt{3}$.
$1$. The coordinates of the foci are $(0, \pm 5\sqrt{3})$.
$2$. The coordinates of the vertices are $(0, \pm 10)$.
$3$. The length of the major axis is $2a = 2(10) = 20$.
$4$. The length of the minor axis is $2b = 2(5) = 10$.
$5$. The eccentricity $e = \frac{c}{a} = \frac{5\sqrt{3}}{10} = \frac{\sqrt{3}}{2}$.
$6$. The length of the latus rectum is $\frac{2b^{2}}{a} = \frac{2(25)}{10} = 5$.