Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $49 y^{2}-16 x^{2}=784$
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, $\frac{y^{2}}{4^{2}}-\frac{x^{2}}{7^{2}}=1$ ......... $(1)$
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 $(0,\,±4)$
Eccentricity, $e=\frac{c}{a}=\frac{\sqrt{65}}{4}$
Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 49}{4}=\frac{49}{2}$
Curve $xy = {c^2}$ is said to be
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