Given equation of hyperbola is $5y^2 - 9x^2$ = 36 i.e., $\frac{5y^2}{36} - \frac{9x^2}{36}$ = 1 ⇒ $\frac{y^2}{\frac{36}{5}} - \frac{x^2}{4}$ = 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, $a^2$ = $\frac{36}{5}$ ⇒ a = $\frac{6}{\sqrt{5}}$ and $b^2$ = 4 ⇒ b = 2 Also, $x^2$ = $a^2 + b^2$ = $\frac{36}{5} + 4$ = $\frac{56}{5}$ ⇒ c = $\sqrt{\frac{56}{5}}$ = $2 \sqrt{\frac{14}{5}}$ ∴ Coordinates of foci are (0, ± c) i.e. $\left(0, \pm \frac{2\sqrt{14}}{\sqrt{5}}\right)$ Coordinates of vertices are (0, ± a) i.e. $\left(0, \pm \frac{6}{\sqrt{5}}\right)$ Eccentricity (e) = $\frac{c}{a}$ = $\frac{2\sqrt{\frac{14}{5}}}{\frac{6}{\sqrt{5}}}$ = $\frac{\sqrt{14}}{3}$ Length of latus rectum = $\frac{2b^2}{a}$ = $\frac{2 \times 4}{\frac{6}{\sqrt{5}}}$ = $\frac{4\sqrt{5}}{3}$.