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 of hyperbola is, $49y^2-16x^2=784$
$\frac{49y^2}{784}-\frac{16x^2}{784}=1\Rightarrow \frac{y^2}{16}-\frac{y^2}{16}-\frac{x^2}{49}=1\text{Which is of the form}\frac{y^2}{a^2}-\frac{x^2}{a^2}=1$
The foci and vertices of the hyperbola lie on y- axis.
$\therefore a^2=16\Rightarrow a=4\text{and}{b}^2=49\Rightarrow b=7\text{Now}{c}^2=a^2+b^2=16+49=65\Rightarrow c=\sqrt{65}\therefore \text{Coordinates of foci are}(0,\pm c)i.e.(0,\pm \sqrt{65})\text{Coordinates of vertices are}(0,\pm a)i.e.(0,\pm 4)\text{Eccentricity}\left(e\right)=\frac{c}{a}=\frac{\sqrt{65}}{4}\text{Length of latus rectum}=\frac{2b^2}{a}=\frac{2\times 49}{4}=\frac{49}{2}.$