अतिपरवलयों के शीर्षों, नाभियों के निर्देशांक, उत्केंद्रता और नाभिलंब जीवा की लंबाई ज्ञात कीजिए

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

The given equation is $9 y^{2}-4 x^{2}=36$

It can be written as

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

Or, $\frac{y^{2}}{4}-\frac{x^{2}}{9}=1$

Or, $\frac{y^{2}}{2^{2}}-\frac{x^{2}}{3^{2}}=1$          ........... $(1)$

On comparing equation $(1)$ with the standard equation of hyperbola i.e., $\frac{y^{2}}{a^{2}}-\frac{ x ^{2}}{b^{2}},$ we obtain $a=2$ and $b=3$

We know that $a^{2}+b^{2}=c^{2}$

$\therefore c^{2}=4+9=13$

$\Rightarrow c=\sqrt{13}$

Therefore,

The coordinates of the foci are $(0, \,\pm \sqrt{13})$

The coordinates of the vertices are $(0,\,±2)$

Eccentricity, $e=\frac{c}{a}=\frac{\sqrt{13}}{2}$

Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 9}{2}=9$