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 $\frac{x^{2}}{100}+\frac{y^{2}}{400}=1$.

The given equation is $\frac{x^{2}}{100}+\frac{y^{2}}{400}=1$,which can be written as $\frac{x^{2}}{10^{2}}+\frac{y^{2}}{20^{2}}=1$.
Since the denominator of the $y^{2}$ term is greater than the denominator of the $x^{2}$ term,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 $b=10$ and $a=20$.
We calculate $c$ using the relation $c = \sqrt{a^{2}-b^{2}} = \sqrt{400-100} = \sqrt{300} = 10\sqrt{3}$.
$1$. The coordinates of the foci are $(0, \pm 10\sqrt{3})$.
$2$. The coordinates of the vertices are $(0, \pm 20)$.
$3$. The length of the major axis is $2a = 2 \times 20 = 40$.
$4$. The length of the minor axis is $2b = 2 \times 10 = 20$.
$5$. The eccentricity $e = \frac{c}{a} = \frac{10\sqrt{3}}{20} = \frac{\sqrt{3}}{2}$.
$6$. The length of the latus rectum is $\frac{2b^{2}}{a} = \frac{2 \times 100}{20} = 10$.