If $y=mx+4$ ($n>0$ is a typo,likely $m>0$ or irrelevant) is a tangent to the hyperbola $\frac{x^2}{25}-\frac{y^2}{9}=1$,then the point of contact of this tangent is

  • A
    $\left(-\frac{25}{4}, -\frac{9}{4}\right)$
  • B
    $\left(\frac{25}{4}, \frac{9}{4}\right)$
  • C
    $(1, 5)$
  • D
    $\left(-\frac{1}{2}, \frac{7}{2}\right)$

Explore More

Similar Questions

The equation $\frac{x^2}{29 - p} + \frac{y^2}{4 - p} = 1$ $(p \neq 4, 29)$ represents:

$A$ rectangular hyperbola of latus rectum $2$ units passes through $(0, 0)$ and has $(1, 0)$ as its one focus. The other focus lies on the curve -

The equation of the tangent to the conic ${x^2} - {y^2} - 8x + 2y + 11 = 0$ at $(2, 1)$ is

If $3x + 2\sqrt{2}y + k = 0$ is a normal to the hyperbola $4x^2 - 9y^2 - 36 = 0$ making positive intercepts on both the axes,then $k=$ (in $\sqrt{2}$)

If a circle intersects the hyperbola $xy = 1$ at four points $(x_r, y_r)$ for $r = 1, 2, 3, 4$,then:

Difficult
View Solution

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