Given equation of hyperbola is $\frac{x^2}{16}-\frac{y^2}{9}$ = 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, $a^2$ = 16 ⇒ a = 4 and b2 = 9 ⇒ b = 3 Also, $c^2$ = $a^2 + b^2$ = 16 + 9 = 25 ⇒ c = 5 ∴ Coordinates of foci are (± c, 0) i.e. (± 5, 0) Coordinates of vertices are (± a, 0) i.e. (± 4, 0) Eccentricity (e) = $\frac{c}{a}$ = $\frac{5}{4}$ Length of latus rectum = $\frac{2b^2}{a}$ = $\frac{2 \times 9}{4}$ = $\frac{9}{2}$