Find the equation of the hyperbola satisfying the give conditions: Foci $(0,\,\pm 13),$ the conjugate axis is of length $24.$

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

Foci $(0,\,\pm 13),$ the conjugate axis is of length $24$.

Here, the foci are on the $y-$ axis.

Therefore, the equation of the hyperbola is of the form $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$

since the foci are $(0,\,\pm 13)$,  $c=13$

since the length of the conjugate axis is $24$,  $2 b=24 \Rightarrow b=12$

We know that  $a^{2}+b^{2}=c^{2}$

$\therefore a^{2}+12^{2}=13^{2}$

$\Rightarrow a^{2}=169-144=25$

Thus, the equation of the hyperbola is $\frac{y^{2}}{25}-\frac{x^{2}}{144}=1$

Similar Questions

The equation of the tangent to the hyperbola $4{y^2} = {x^2} - 1$ at the point $(1, 0)$ is

Eccentricity of rectangular hyperbola is

The equation of the common tangent to the curves $y^2 = 8x$ and $xy = -1$ is

The equation of the tangent at the point $(a\sec \theta ,\;b\tan \theta )$ of the conic $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$, is

Let $\mathrm{P}$ be a point on the hyperbola $\mathrm{H}: \frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{4}=1$, in the first quadrant such that the area of triangle formed by $\mathrm{P}$ and the two foci of $\mathrm{H}$ is $2 \sqrt{13}$. Then, the square of the distance of $\mathrm{P}$ from the origin is

  • [JEE MAIN 2024]