The area enclosed by the ellipse $4x^2 + 9y^2 = 1$ is . . . . . . sq. units.

  • A
    $\frac{\pi}{6}$
  • B
    $\frac{\pi}{36}$
  • C
    $6\pi$
  • D
    $\frac{\pi}{12}$

Explore More

Similar Questions

An ellipse passes through the foci of the hyperbola $9x^2 - 4y^2 = 36$,and its major and minor axes lie along the transverse and conjugate axes of the hyperbola,respectively. If the product of the eccentricities of the two conics is $\frac{1}{2}$,then which of the following points does not lie on the ellipse?

An ellipse,with foci at $(0, 2)$ and $(0, -2)$ and minor axis of length $4$,passes through which of the following points?

The curve represented by $x = 3(\cos t + \sin t)$ and $y = 4(\cos t - \sin t)$ is

If $\beta$ is one of the angles between the normals to the ellipse $x^2 + 3y^2 = 9$ at the points $(3\cos \theta, \sqrt{3} \sin \theta)$ and $(-3\sin \theta, \sqrt{3} \cos \theta)$,where $\theta \in (0, \pi/2)$,then $\frac{2 \cot \beta}{\sin 2\theta}$ is equal to

Let a line $L$ pass through the point of intersection of the lines $bx + 10y - 8 = 0$ and $2x - 3y = 0$,where $b \in R - \{\frac{4}{3}\}$. If the line $L$ also passes through the point $(1, 1)$ and touches the circle $17(x^2 + y^2) = 16$,then the eccentricity of the ellipse $\frac{x^2}{5} + \frac{y^2}{b^2} = 1$ 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