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 $\frac{x^{2}}{4}+\frac{y^{2}}{25}=1$.

(N/A) The given equation is $\frac{x^{2}}{4}+\frac{y^{2}}{25}=1$,which can be written as $\frac{x^{2}}{2^{2}}+\frac{y^{2}}{5^{2}}=1$.
Since the denominator of $\frac{y^{2}}{25}$ is greater than the denominator of $\frac{x^{2}}{4}$,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 $a=5$ and $b=2$.
We calculate $c$ using the relation $c = \sqrt{a^{2}-b^{2}} = \sqrt{25-4} = \sqrt{21}$.
The coordinates of the foci are $(0, \sqrt{21})$ and $(0, -\sqrt{21})$.
The coordinates of the vertices are $(0, 5)$ and $(0, -5)$.
The length of the major axis is $2a = 2 \times 5 = 10$.
The length of the minor axis is $2b = 2 \times 2 = 4$.
The eccentricity is $e = \frac{c}{a} = \frac{\sqrt{21}}{5}$.
The length of the latus rectum is $\frac{2b^{2}}{a} = \frac{2 \times 4}{5} = \frac{8}{5}$.