Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $9 y^{2}-4 x^{2}=36$
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 $=\frac{2 b^{2}}{a}=\frac{2 \times 9}{2}=9$