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

(N/A) The given equation is $9 y^{2}-4 x^{2}=36$.
Dividing both sides by $36$,we get:
$\frac{9 y^{2}}{36} - \frac{4 x^{2}}{36} = 1$
$\frac{y^{2}}{4} - \frac{x^{2}}{9} = 1$
This is the standard form of a hyperbola $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$,where $a^{2} = 4$ and $b^{2} = 9$.
Thus,$a = 2$ and $b = 3$.
For a hyperbola,$c^{2} = a^{2} + b^{2} = 4 + 9 = 13$,so $c = \sqrt{13}$.
$1$. The coordinates of the foci are $(0, \pm c) = (0, \pm \sqrt{13})$.
$2$. The coordinates of the vertices are $(0, \pm a) = (0, \pm 2)$.
$3$. The eccentricity $e = \frac{c}{a} = \frac{\sqrt{13}}{2}$.
$4$. The length of the latus rectum $= \frac{2 b^{2}}{a} = \frac{2 \times 9}{2} = 9$.