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

Solution: Given equation is,
$49y^2-16x^2=784$
We will convert this equation into standard form.
$\Rightarrow \frac{49}{784}{y}^2-\frac{16}{784}{x}^2=1$
$\Rightarrow \frac{y^2}{16}-\frac{x^2}{49}=1$
So, this is our standard equation of hyperbola with,
$a=4,b=7$
$c=\sqrt{a^2+b^2}=\sqrt{16+49}=\sqrt{65}$
Here, as $y^2$ is positive, major-axis will be $Y-$axis.
Now, foci will be $\left(0,\pm c\right)=\left(0,\pm \sqrt{65}\right).$
Vertices will be $\left(0,\pm a\right)=\left(0,\pm 4\right).$
Eccentricity $=\frac{c}{a}=\frac{\sqrt{65}}{4}.$
Length of latus rectum $=2\frac{b^2}{a}=2\ast \frac{49}{4}=\frac{49}{2}.$