$A$ curve is represented by the equations $x = \sec^2 t$ and $y = \cot t$,where $t$ is a parameter. If the tangent at the point $P$ on the curve where $t = \pi / 4$ meets the curve again at the point $Q$,then $|PQ|$ is equal to

  • A
    $\frac{5\sqrt{3}}{2}$
  • B
    $\frac{5\sqrt{5}}{2}$
  • C
    $\frac{2\sqrt{5}}{2}$
  • D
    $\frac{3\sqrt{5}}{2}$

Explore More

Similar Questions

The function $f(x) = \begin{cases} e^{2x} - 1, & x \le 0 \\ ax + \frac{bx^2}{2} - 1, & x > 0 \end{cases}$ is continuous and differentiable for

The equation of a tangent to the curve $y \cot x = y^3 \tan x$ at the point where the abscissa is $\frac{\pi}{4}$ is:

The value of $f(4)-f(3)$ is

Let $y(x) = (1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})$. Then $y'(x) - y''(x)$ at $x = -1$ is equal to:

If $f(x) = \begin{cases} -x-\frac{\pi}{2}, & x \leq-\frac{\pi}{2} \\ -\cos x, & -\frac{\pi}{2} < x \leq 0 \\ x-1, & 0 < x \leq 1 \\ \ln x, & x > 1 \end{cases}$,then which of the following statements are true?
$(A)$ $f(x)$ is continuous at $x=-\frac{\pi}{2}$
$(B)$ $f(x)$ is not differentiable at $x=0$
$(C)$ $f(x)$ is differentiable at $x=1$
$(D)$ $f(x)$ is differentiable at $x=-\frac{3}{2}$

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