The point at which the tangent line to the curve of $y=\frac{16}{x}-x^2$ is horizontal, is

  • A
    $2, 4$
  • B
    $-2, -12$
  • C
    $1, 15$
  • D
    $-1, -17$

Explore More

Similar Questions

The equation of the tangent to the curve $y = b e^{-x / a}$ at the point where it crosses the $Y$ axis is

The angle between the curves $y^2=4x+4$ and $y^2=36(9-x)$ is (in $^{\circ}$)

The subnormal at any point on the curve $xy^n = a^{n + 1}$ is constant for $n=$ ............

$y=f(x)$ and $x=g(y)$ are two curves and $P(x, y)$ is a common point of the two curves. If at $P$, on the curve $y=f(x)$, $\frac{dy}{dx}=Q(x)$ and at the same point $P$ on the curve $x=g(y)$, $\frac{dx}{dy}=-Q(x)$, then

If $y=4x-5$ is a tangent to the curve $y^2=px^3+q$ at $(2,3)$,then $p-q$ 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