Find the coordinates of the foci and the vertices, the eccentricity,

The given equation is $5 y^{2}-9 x^{2}=36$.

$\Rightarrow \frac{y^{2}}{\left(\frac{36}{5}\right)}-\frac{x^{2}}{4}=1$

$\Rightarrow \frac{y^{2}}{\left(\frac{6}{\sqrt{5}}\right)^{2}}-\frac{x^{2}}{2^{2}}=1$ $\ldots(1)$

On comparing equation (1) with the standard equation of hyperbola i.e., $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$, we obtain $a=\frac{6}{\sqrt{5}}$ and $b=2$

We know that $a^{2}+b^{2}=c^{2}$.

$\therefore c^{2}=\frac{36}{5}+4=\frac{56}{5}$

$\Rightarrow c=\sqrt{\frac{56}{5}}=\frac{2 \sqrt{14}}{\sqrt{5}}$

Therefore, the coordinates of the foci are $\left(0, \pm \frac{2 \sqrt{14}}{\sqrt{5}}\right)$.

The coordinates of the vertices are $\left(0, \pm \frac{6}{\sqrt{5}}\right)$.

Eccentricity, $e=\frac{c}{a}=\frac{\left(\frac{2 \sqrt{14}}{\sqrt{5}}\right)}{\left(\frac{6}{\sqrt{5}}\right)}=\frac{\sqrt{14}}{3}$

Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 4}{\left(\frac{6}{\sqrt{5}}\right)}=\frac{4 \sqrt{5}}{3}$