The eccentricity of the ellipse $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$ is

  • A
    $\frac{2 \sqrt{5}}{6}$
  • B
    $\frac{2 \sqrt{5}}{4}$
  • C
    $\frac{2 \sqrt{13}}{6}$
  • D
    $\frac{2 \sqrt{13}}{4}$

Explore More

Similar Questions

Equations of the latus rectum of the ellipse $9x^2+4y^2-18x-8y-23=0$ are:

If $a$ and $c$ are positive real numbers and the ellipse $\frac{x^2}{4c^2} + \frac{y^2}{c^2} = 1$ has four distinct points in common with the circle $x^2 + y^2 = 9a^2$,then

If the latus rectum of an ellipse subtends a right angle at the centre of that ellipse,then the eccentricity of that ellipse is

The eccentricity of the ellipse $9x^2 + 25y^2 = 225$ is

An ellipse $\frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2} = 1$ with $a > b$ is tangent to both the $x$ and $y$ axes and is located in the first quadrant. Let $F_1$ and $F_2$ be the two foci of the ellipse and $O$ be the origin such that $OF_1 < OF_2$. Suppose the triangle $OF_1F_2$ is an isosceles triangle with $\angle OF_1F_2 = 120^{\circ}$. Then the eccentricity of the ellipse 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