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

Solution:

The given equation is $9 y^{2}-4 x^{2}=36$.

It can be written as

$9 y^{2}-4 x^{2}=36$

Or, $\frac{y^{2}}{4}-\frac{x^{2}}{9}=1$

$\mathrm{Or}, \frac{y^{2}}{2^{2}}-\frac{x^{2}}{3^{2}}=1$ .(1)

Or, $\frac{y^{2}}{2^{2}}-\frac{x^{2}}{3^{2}}=1$ (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 $a=2$ and $b=3$.

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

$\therefore c^{2}=4+9=13$

$\Rightarrow c=\sqrt{13}$

Therefore,

The coordinates of the foci are $(0, \pm \sqrt{13})$.

The coordinates of the vertices are $(0, \pm 2)$.

Eccentricity, $e=\frac{c}{a}=\frac{\sqrt{13}}{2}$

Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 9}{2}=9$