The given equation is $\frac{x^{2}}{100}+\frac{y^{2}}{400}=1$ or $\frac{x^{2}}{10^{2}}+\frac{y^{2}}{20^{2}}=1$

Here, the denominator of is greater than the denominator of $\frac{x^{2}}{100}$. 

Therefore, the major axis is along the $y-$ axis, while the minor axis is along the $x-$ axis.

On comparing the given equation with, $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$ we obtain $b=10$ and $a=20$

$\therefore c=\sqrt{a^{2}-b^{2}}=\sqrt{400-100}=\sqrt{300}=10 \sqrt{3}$

Therefore,

The coordinates of the foci are $(0,\, \pm 10 \sqrt{3})$

The coordinates of the vertices are $(0,\,±20)$ 

Length of major axis $=2 a =40$

Length of minor axis $=2 b =20$

Eccentricity,  $e=\frac{c}{a}=\frac{10 \sqrt{3}}{20}=\frac{\sqrt{3}}{2}$

Length of latus rectum  $=\frac{2 b^{2}}{a}=\frac{2 \times 100}{20}=10$