Area of the region enclosed between the curves $y^2=4(x+7)$ and $y^2=5(2-x)$ is

  • A
    $\frac{32 \sqrt{2}}{3}$
  • B
    $\frac{8}{3}$
  • C
    $\frac{1}{6}$
  • D
    $24 \sqrt{5}$

Explore More

Similar Questions

Let the area enclosed between the curves $|y|=1-x^2$ and $x^2+y^2=1$ be $\alpha$. If $9\alpha=\beta\pi+\gamma$,where $\beta$ and $\gamma$ are integers,then the value of $|\beta-\gamma|$ equals

The area enclosed by $y=\sqrt{5-x^{2}}$ and $y=|x-1|$ is

Area of the region bounded by two parabolas $y=x^{2}$ and $x=y^{2}$ is

The area (in sq. units) of the region $A = \{(x, y) : x^2 \le y \le x + 2\}$ is

Let the area of the region enclosed by the curve $y = \min \{\sin x, \cos x\}$ and the $x$-axis between $x = -\pi$ to $x = \pi$ be $A$. Then $A^2$ is equal to...........

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