Find the coordinates of the foci, the vertices the eccentricity and the length of latus rectum of the hyperbola $9y^2-4x^2=36$

The given equation is $9y^2-4x^2=36$
It can be written as
$9y^2-4x^2=36$
$\frac{y^2}{4}-\frac{x^2}{9}=1$
$\frac{y^2}{2^2}-\frac{x^2}{3^2}=\mathrm{1...}(1)$
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 $a^2+b^2=c^2,$ where $c=ae$
$\therefore c^2=4+9=13$
$\Rightarrow c=\sqrt{13}$
Therefore, the coordinates of the foci are $\left(0,\pm \sqrt{13}\right)$
The coordinates of the vertices are $\left(0,\pm 2\right)$
Eccentricity $e$ $=\frac{c}{a}=\frac{\sqrt{13}}{2}$
Length of latus rectum $=$ $\frac{2b^2}{a}=\frac{2\times 9}{2}=9$