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

(N/A) The given equation is $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$,which can be written as $\frac{x^{2}}{4^{2}}-\frac{y^{2}}{3^{2}}=1$.
Comparing this with the standard equation $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$,we get $a=4$ and $b=3$.
We know that $c^{2}=a^{2}+b^{2}$.
Substituting the values,$c^{2}=4^{2}+3^{2}=16+9=25$,so $c=5$.
$1$. The coordinates of the foci are $(\pm c, 0) = (\pm 5, 0)$.
$2$. The coordinates of the vertices are $(\pm a, 0) = (\pm 4, 0)$.
$3$. The eccentricity $e = \frac{c}{a} = \frac{5}{4}$.
$4$. The length of the latus rectum $= \frac{2b^{2}}{a} = \frac{2 \times 9}{4} = \frac{9}{2} = 4.5$.