Question

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

Answer 1

The given equation is $49y^2-16x^2=784$
It can be written as
$49y^2-16x^2=784$
$\frac{y^2}{16}-\frac{x^2}{49}=1$
$\frac{y^2}{4^2}-\frac{x^2}{7^2}=1$ ...(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 $a^2+b^2=c^2,$ where $c=ae$
$\therefore c^2=16+49=65$
$\Rightarrow c=\sqrt{65}$
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 $=$ $\frac{2b^2}{a}=\frac{2\times 49}{4}=\frac{49}{2}$