(N/A) The given equation is $16 x^{2}-9 y^{2}=576$.
It can be written as:
$\frac{x^{2}}{36}-\frac{y^{2}}{64}=1$
$\Rightarrow \frac{x^{2}}{6^{2}}-\frac{y^{2}}{8^{2}}=1$ $(1)$
On comparing equation $(1)$ with the standard equation of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$,we obtain $a=6$ and $b=8$.
We know that $c^{2}=a^{2}+b^{2}$.
$\therefore c^{2}=36+64=100$
$\Rightarrow c=10$.
Therefore:
$1$. The coordinates of the foci are $(\pm 10, 0)$.
$2$. The coordinates of the vertices are $(\pm 6, 0)$.
$3$. Eccentricity,$e = \frac{c}{a} = \frac{10}{6} = \frac{5}{3}$.
$4$. Length of the latus rectum $= \frac{2b^{2}}{a} = \frac{2 \times 64}{6} = \frac{64}{3}$.