The slope of a tangent drawn at the point $P(\alpha, \beta)$ lying on the curve $y=\frac{1}{2x-5}$ is $-2$. If $P$ lies in the fourth quadrant,then $\alpha-\beta=$

  • A
    $4$
  • B
    $3$
  • C
    $2$
  • D
    $1$

Explore More

Similar Questions

If the normal to the curve $y=f(x)$ at $(1,2)$ makes an angle $\frac{3 \pi}{4}$ with the positive $X$-axis, then $f^{\prime}(1)=$

The length of the subtangent at any point of the curve $x^my^n = a^{m+n}$ is proportional to

Difficult
View Solution

If the tangent of the curve $y=e^{x}$ at a point $(c, e^{c})$ and the normal to the parabola $y^{2}=4x$ at the point $(1,2)$ intersect at the same point on the $x$-axis,then the value of $c$ is

The normal to the curve $y=f(x)$ at the point $(3,4)$ makes an angle $\frac{3 \pi}{4}$ with the positive $X$-axis. Then $f^{\prime}(3)$ is equal to:

Show that the normal at any point $\theta$ to the curve $x=a \cos \theta+a \theta \sin \theta, y=a \sin \theta-a \theta \cos \theta$ is at a constant distance from the origin.

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