Find the equation of the hyperbola with foci $(0,\,\pm 3)$ and vertices $(0,\,\pm \frac {\sqrt {11}}{2})$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store

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., $100 y^{2}-44 x^{2}=275$

Similar Questions

Let $\lambda x-2 y=\mu$ be a tangent to the hyperbola $a^{2} x^{2}-y^{2}=b^{2}$. Then $\left(\frac{\lambda}{a}\right)^{2}-\left(\frac{\mu}{b}\right)^{2}$ is equal to

  • [JEE MAIN 2022]

Let $H$ be the hyperbola, whose foci are $(1 \pm \sqrt{2}, 0)$ and eccentricity is $\sqrt{2}$. Then the length of its latus rectum is

  • [JEE MAIN 2023]

Let the foci of a hyperbola be $(1,14)$ and $(1,-12)$. If it passes through the point $(1,6)$, then the length of its latus-rectum is :

  • [JEE MAIN 2025]

Find the equation of the hyperbola satisfying the give conditions: Vertices $(0,\,\pm 5),$ foci $(0,\,±8)$

The equation of the hyperbola in the standard form (with transverse axis along the $x$ -  axis) having the length of the latus rectum = $9$ units and eccentricity = $5/4$ is