Find the coordinates of the foci and the vertices, the eccentricity,

Solution:

The given equation is $49 y^{2}-16 x^{2}=784$.

It can be written as

$49 y^{2}-16 x^{2}=784$

Or, $\frac{y^{2}}{16}-\frac{x^{2}}{49}=1$

Or, $\frac{y^{2}}{4^{2}}-\frac{x^{2}}{7^{2}}=1$ $\ldots(1)$

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 $a=4$ and $b=7$.

We know that $a^{2}+b^{2}=c^{2}$.

$\therefore c^{2}=16+49=65$

$\Rightarrow c=\sqrt{65}$

Therefore,

The coordinates of the foci are $(0, \pm \sqrt{65})$.

The coordinates of the vertices are $(0, \pm 4)$.

Eccentricity, $e=\frac{c}{a}=\frac{\sqrt{65}}{4}$

Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 49}{4}=\frac{49}{2}$