Question

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

$5y^2-9x^2=36$

Answer 1

The given equation of hyperbola is, $5y^2-9x^2=36$
$\frac{5y^2}{36}-\frac{9x^2}{36}=1\Rightarrow \frac{y^2}{\frac{36}{5}}-\frac{x^2}{4}=1\text{Which is of the form}\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
The foci and vertices of the hyperbola lie on y - axis
$\therefore a^2=\frac{36}{5}\Rightarrow a=\frac{6}{\sqrt{5}}\text{and}{b}^2=4\Rightarrow b=2\text{Now}{c}^2=a^2+b^2=\frac{36}{5}+4=\frac{56}{5}\Rightarrow c=\sqrt{\frac{56}{5}}\therefore \text{Coordinates of foci are}(0,\pm c)i.e.(0,\pm \sqrt{\frac{56}{5}})\text{Coordinates of vertices are}(0,\pm a)i.e.(0,\pm \frac{6}{\sqrt{5}})\text{Eccentricity}\left(e\right)=\frac{c}{a}=\frac{\sqrt{\frac{56}{5}}}{\frac{6}{\sqrt{5}}}=\frac{\sqrt{56}}{6}=\frac{2\sqrt{14}}{6}=\frac{\sqrt{14}}{3}\text{length of latus rectum}=\frac{2b^2}{a}=\frac{2\times 4}{\frac{6}{\sqrt{6}}}=\frac{2\times 4\times \sqrt{5}}{6}=\frac{4\sqrt{5}}{3}$