Find the coordinates of the foci and the vertices,the eccentricity,and the length of the latus rectum of the hyperbola $\frac{y^{2}}{9}-\frac{x^{2}}{27}=1$.

(N/A) The given equation is $\frac{y^{2}}{9}-\frac{x^{2}}{27}=1$,which can be written as $\frac{y^{2}}{3^{2}}-\frac{x^{2}}{(\sqrt{27})^{2}}=1$.
Comparing this with the standard equation of a vertical hyperbola $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$,we get $a=3$ and $b=\sqrt{27}$.
For a hyperbola,the relation between $a, b,$ and $c$ is $c^{2}=a^{2}+b^{2}$.
Substituting the values,$c^{2}=3^{2}+(\sqrt{27})^{2}=9+27=36$,so $c=6$.
The coordinates of the foci are $(0, \pm c) = (0, \pm 6)$.
The coordinates of the vertices are $(0, \pm a) = (0, \pm 3)$.
The eccentricity $e$ is given by $e=\frac{c}{a}=\frac{6}{3}=2$.
The length of the latus rectum is $\frac{2b^{2}}{a}=\frac{2 \times 27}{3}=18$.