Find the equation of the hyperbola with foci $(0,\,\pm 3)$ and vertices $(0,\,\pm \frac {\sqrt {11}}{2})$.
Find the equation of the hyperbola with foci $(0,\,\pm 3)$ and vertices $(0,\,\pm \frac {\sqrt {11}}{2})$.
Solution since the foci is on $y-$ axis, the equation of the hyperbola is of the form $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$
since vertices are $\left(0,\,\pm \frac{\sqrt{11}}{2}\right)$ , $a=\frac{\sqrt{11}}{2}$
Also, since foci are $(0,\,±3)$; $c=3$ and $b^{2}=c^{2}-a^{2}=\frac{25}{4}$
Therefore, the equation of the hyperbola is
$\frac{y^{2}}{\left(\frac{11}{4}\right)}$ $-\frac{x^{2}}{\left(\frac{25}{4}\right)}=1$, i.e., $100 y^{2}-44 x^{2}=275$
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- [JEE MAIN 2019]