Find the coordinates of the foci and the vertices,the eccentricity,and the length of the latus rectum of the hyperbola: $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$.

(N/A) Comparing the given equation $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$ with the standard equation $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$,we get:
$a^{2} = 9 \implies a = 3$
$b^{2} = 16 \implies b = 4$
For a hyperbola,$c = \sqrt{a^{2} + b^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5$.
$1$. The coordinates of the foci are $(\pm c, 0) = (\pm 5, 0)$.
$2$. The coordinates of the vertices are $(\pm a, 0) = (\pm 3, 0)$.
$3$. The eccentricity $e = \frac{c}{a} = \frac{5}{3}$.
$4$. The length of the latus rectum is $\frac{2b^{2}}{a} = \frac{2 \times 16}{3} = \frac{32}{3}$.