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 $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$.

(N/A) The given equation is $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$,which can be written as $\frac{x^{2}}{4^{2}} + \frac{y^{2}}{3^{2}} = 1$.
Since the denominator of $\frac{x^{2}}{16}$ is greater than the denominator of $\frac{y^{2}}{9}$,the major axis is along the $x$-axis.
Comparing this with the standard form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,we get $a = 4$ and $b = 3$.
We calculate $c = \sqrt{a^{2} - b^{2}} = \sqrt{16 - 9} = \sqrt{7}$.
$1$. The coordinates of the foci are $(\pm \sqrt{7}, 0)$.
$2$. The coordinates of the vertices are $(\pm 4, 0)$.
$3$. The length of the major axis is $2a = 2 \times 4 = 8$.
$4$. The length of the minor axis is $2b = 2 \times 3 = 6$.
$5$. The eccentricity $e = \frac{c}{a} = \frac{\sqrt{7}}{4}$.
$6$. The length of the latus rectum is $\frac{2b^{2}}{a} = \frac{2 \times 9}{4} = \frac{9}{2} = 4.5$.