If the maximum distance of the normal to the ellipse $\frac{x^2}{4} + \frac{y^2}{b^2} = 1$,where $b < 2$,from the origin is $1$,then the eccentricity of the ellipse is:

  • A
    $\frac{1}{\sqrt{2}}$
  • B
    $\frac{\sqrt{3}}{2}$
  • C
    $\frac{1}{2}$
  • D
    $\frac{\sqrt{3}}{4}$

Explore More

Similar Questions

If the midpoint of a chord of the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ is $(\sqrt{2}, 4/3)$,and the length of the chord is $\frac{2 \sqrt{\alpha}}{3}$,then $\alpha$ is :

The locus of a variable point whose distance from $(-2, 0)$ is $\frac{2}{3}$ times its distance from the line $x = -\frac{9}{2}$ is

Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ be an ellipse with foci $F_1$ and $F_2$. Let $AO$ be its semi-minor axis,where $O$ is the centre of the ellipse. The lines $AF_1$ and $AF_2$,when extended,cut the ellipse again at points $B$ and $C$ respectively. Suppose that the $\triangle ABC$ is equilateral. Then,the eccentricity of the ellipse is

Let $E$ be an ellipse whose axes are parallel to the coordinate axes,having its center at $(3, -4)$,one focus at $(4, -4)$,and one vertex at $(5, -4)$. If $mx - y = 4$ with $m > 0$ is a tangent to the ellipse $E$,then the value of $5m^{2}$ is equal to .....

If a tangent having slope $m = \frac{1}{3}$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ is a normal to the circle $(x + 1)^2 + (y + 1)^2 = 1$,then $a^2$ lies in the interval:

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