Find the coordinates of the foci and the vertices,the eccentricity,and the length of the latus rectum of the hyperbola $49 y^{2}-16 x^{2}=784$.

(N/A) The given equation is $49 y^{2}-16 x^{2}=784$.
Dividing both sides by $784$,we get:
$\frac{49 y^{2}}{784} - \frac{16 x^{2}}{784} = 1$
$\frac{y^{2}}{16} - \frac{x^{2}}{49} = 1$
$\frac{y^{2}}{4^{2}} - \frac{x^{2}}{7^{2}} = 1$ $(1)$
Comparing equation $(1)$ with the standard equation of the hyperbola $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$,we get $a = 4$ and $b = 7$.
We know that $c^{2} = a^{2} + b^{2}$.
$c^{2} = 16 + 49 = 65$
$c = \sqrt{65}$
Coordinates of the foci are $(0, \pm \sqrt{65})$.
Coordinates of the vertices are $(0, \pm 4)$.
Eccentricity $e = \frac{c}{a} = \frac{\sqrt{65}}{4}$.
Length of the latus rectum $= \frac{2 b^{2}}{a} = \frac{2 \times 49}{4} = \frac{49}{2} = 24.5$.