Find the coordinates of the foci and the vertices,the eccentricity,and the length of the latus rectum of the hyperbola $5y^{2} - 9x^{2} = 36$.

The given equation is $5y^{2} - 9x^{2} = 36$.
Dividing by $36$,we get $\frac{y^{2}}{(36/5)} - \frac{x^{2}}{4} = 1$.
Comparing this with the standard equation $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$,we have $a^{2} = \frac{36}{5}$ and $b^{2} = 4$.
Thus,$a = \frac{6}{\sqrt{5}}$ and $b = 2$.
For a hyperbola,$c^{2} = a^{2} + b^{2} = \frac{36}{5} + 4 = \frac{56}{5}$.
So,$c = \sqrt{\frac{56}{5}} = \frac{2\sqrt{14}}{\sqrt{5}}$.
Coordinates of the foci are $(0, \pm c) = (0, \pm \frac{2\sqrt{14}}{\sqrt{5}})$.
Coordinates of the vertices are $(0, \pm a) = (0, \pm \frac{6}{\sqrt{5}})$.
Eccentricity $e = \frac{c}{a} = \frac{2\sqrt{14}/\sqrt{5}}{6/\sqrt{5}} = \frac{\sqrt{14}}{3}$.
Length of the latus rectum $= \frac{2b^{2}}{a} = \frac{2 \times 4}{6/\sqrt{5}} = \frac{4\sqrt{5}}{3}$.