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

  • A
    $3 x-y+2=0$
  • B
    $3 x+y-14=0$
  • C
    $x+y+3=0$
  • D
    $x-y+6=0$

Explore More

Similar Questions

The area of the triangle formed by the tangent to the hyperbola $xy = a^2$ and the coordinate axes is:

The length of the latus rectum and directrices of a hyperbola with eccentricity $e$ are $9$ and $x = \pm \frac{4}{\sqrt{13}}$,respectively. Let the line $y - \sqrt{3}x + \sqrt{3} = 0$ touch this hyperbola at $(x_0, y_0)$. If $m$ is the product of the focal distances of the point $(x_0, y_0)$,then $4e^2 + m$ is equal to ...........

Find the equation of the tangent to the hyperbola $x^2 - 4y^2 = 36$ which is perpendicular to the line $x - y + 4 = 0$.

Let $P(a \sec \theta, b \tan \theta)$ and $Q(a \sec \phi, b \tan \phi)$ where $\theta + \phi = \frac{\pi}{2}$ be two points 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 drawn at $P$ and $Q$,then $k=$

If the eccentricity of the hyperbola $\frac{x^2}{9} - \frac{y^2}{b^2} = 1$ passing through the point $(k, 2)$ is $\frac{\sqrt{13}}{3}$,then the value of $k^2$ 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