At which point does the line $2x + \sqrt{6}y = 2$ touch the hyperbola $x^2 - 2y^2 = 4$?

  • A
    $(4, -\sqrt{6})$
  • B
    $(\sqrt{6}, 1)$
  • C
    $(\frac{1}{2}, \frac{1}{\sqrt{6}})$
  • D
    $(\frac{\pi}{6}, \pi)$

Explore More

Similar Questions

The angle between the asymptotes of the hyperbola $x^2-3y^2=3$ is

If $PQ$ is a double ordinate of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ such that $\Delta OPQ$ is equilateral,where $O$ is the centre,then the eccentricity $e$ satisfies:

If $x=9$ is a chord of contact of the hyperbola $x^2-y^2=9$,then the equation of the tangent at one of the points of contact is

Let $A(2 \sec \theta, 3 \tan \theta)$ and $B(2 \sec \phi, 3 \tan \phi)$ where $\theta+\phi=\frac{\pi}{2}$,be two points on the hyperbola $\frac{x^2}{4}-\frac{y^2}{9}=1$. If $(\alpha, \beta)$ is the point of intersection of normals to the hyperbola at $A$ and $B$,then $\beta$ is equal to

Equation of one of the tangents passing through $(2,8)$ to the hyperbola $5 x^2-y^2=5$ is

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