Given equation of hyperbola is $16x^2 - 9y^2$ = 576 i.e., $\frac{16x^2}{576} - \frac{9y^2}{576}$ = 1 ⇒ $\frac{x^2}{36} - \frac{y^2}{64}$ = 1 which is of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2}$ = 1. The foci and vertices of the hyperbola lie on x-axis. Now, $a^2$ = 36 ⇒ a = 6 and $b^2$ = 64 ⇒ b = 8 Also, $c^2$ = $a^2 + b^2$ = 36 + 64 = 100 ⇒ c = 10 ∴ Coordinates of foci are (± c, 0) i.e, (±10, 0) ∴ Coordinates of vertices are (± a, 0) i.e. (± 6, 0) Eccentricity (e) = $\frac{c}{a}$ = $\frac{10}{6}$ = $\frac{5}{3}$ Length of latus rectum = $\frac{2b^2}{a}$ = $\frac{2 \times 64}{6}$ = $\frac{64}{3}$.