From any point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,tangents are drawn to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 2$. The area of the figure formed by the chord of contact of that point and the asymptotes is

  • A
    $\frac{ab}{2}$
  • B
    $ab$
  • C
    $2ab$
  • D
    $4ab$

Explore More

Similar Questions

$A$ square $ABCD$ has all its vertices on the curve $x^{2}y^{2}=1$. The midpoints of its sides also lie on the same curve. Then,the square of the area of $ABCD$ is

If the angle between the asymptotes of a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $2 \tan^{-1}\left(\frac{b}{a}\right) = 2 \tan^{-1}\left(\frac{2}{3}\right)$ and $a^2-b^2=45$,then $ab=$

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

The midpoint of the chord $4x - 3y = 5$ of the hyperbola $2x^2 - 3y^2 = 12$ is

Let $P(a \sec \theta, b \tan \theta)$ and $Q(a \sec \phi, b \tan \phi)$ be two points such that $\theta+\phi=\frac{\pi}{2}$ on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. If $(h, k)$ is the point of intersection of the normals at $P$ and $Q$,then $k=$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo