Given equation of hyperbola is $49y^2 - 16x^2$ = 784 i.e., $\frac{49y^2}{784} - \frac{16x^2}{784}$ = 1 ⇒ $\frac{y^2}{16} - \frac{x^2}{49}$ = 1 which is of the form $\frac{y^2}{a^2} - \frac{x^2}{b^2}$ = 1. The foci and vertices of the hyperbola lie on y-axis. Now, $a^2$ = 16 ⇒ a = 4 and $b^2$ = 49 ⇒ b = 7 Also, $c^2$ = $a^2 + b^2$ = 16 + 49 = 65 ⇒ c = 65 ∴ Coordinates of foci are (0, ±c) i.e. $(0, \pm \sqrt{64})$ Coordinates of vertices are (0, ±a) i.e. (0, ±4) Eccentricity (e) = $\frac{c}{a}$ = $\frac{\sqrt{65}}{4}$ Length of latus rectum = $\frac{2b^2}{a}$ = $\frac{2 \times 49}{4}$ = $\frac{49}{2}$.