Find the coordinates of the foci,the vertices,the length of the major axis,the minor axis,the eccentricity,and the latus rectum of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$.

(N/A) Given the equation of the ellipse is $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$.
Since the denominator of $\frac{x^{2}}{25}$ is larger than the denominator of $\frac{y^{2}}{9}$,the major axis is along the $x$-axis.
Comparing the given equation with $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,we get $a^{2} = 25$ and $b^{2} = 9$,which implies $a = 5$ and $b = 3$.
We calculate $c = \sqrt{a^{2} - b^{2}} = \sqrt{25 - 9} = \sqrt{16} = 4$.
$1$. The coordinates of the foci are $(\pm c, 0)$,which are $(-4, 0)$ and $(4, 0)$.
$2$. The coordinates of the vertices are $(\pm a, 0)$,which are $(-5, 0)$ and $(5, 0)$.
$3$. The length of the major axis is $2a = 2 \times 5 = 10$ units.
$4$. The length of the minor axis is $2b = 2 \times 3 = 6$ units.
$5$. The eccentricity $e$ is $\frac{c}{a} = \frac{4}{5} = 0.8$.
$6$. The length of the latus rectum is $\frac{2b^{2}}{a} = \frac{2 \times 9}{5} = \frac{18}{5} = 3.6$ units.