Question

Find the coordinates of the foci, the vertices, the eccentricity and the length of the latus rectum of the hyperbola,

$9y^2-4x^2=36$

Answer 1

The given equation of hyperbola is,
$9y^2-4x^2=36$
$i.e.\frac{9y^2}{36}-\frac{4x^2}{36}=1\Rightarrow \frac{y^2}{4}-\frac{x^2}{9}=1\text{Which is of the form}\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
The foci and vertices of the hyperbola lie on y - axis.
$\therefore a^2=4\Rightarrow a=2\text{and}{b}^2=9\Rightarrow b=3\text{Now}{c}^2=a^2+b^{2}4+9=13\Rightarrow c=\sqrt{13}\therefore \text{Coordinates of foci are}(0,\pm c)i.e.(0,\pm \sqrt{13})\text{Coordinats of vertices are}(0,\pm a)i.e.(0,\pm 2)\text{Eccentricity}\left(e\right)=\frac{c}{a}=\frac{\sqrt{13}}{2}\text{Length fo latus rectum}=\frac{2b^2}{a}=\frac{2\times 9}{2}=9.$