If the area of the auxiliary circle of the el | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

Given ellipse is $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{b}^{2}}=1$ whose area is $\pi$ ab.

The auxiliary circle to th

Vedclass · Accepted Answer

$\frac{{\sqrt 3 }}{2}$

Vedclass · Answer

Given ellipse is $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{b}^{2}}=1$ whose area is $\pi$ ab.

The auxiliary circle to the given ellipse is $\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{a}^{2}$ whose area is $\pi \mathrm{a}^{2}.$

Given that, $\pi \mathrm{a}^{2}=2 \pi \mathrm{ab} \Rightarrow \mathrm{a}=2 \mathrm{b}$

Now, eccentricity of ellipse

$=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{b^{2}}{4 b^{2}}}=\frac{\sqrt{3}}{2}$

If the area of the auxiliary circle of the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\left( {a > b} \right)$ is twice the area of the ellipse, then the eccentricity of the ellipse is

Similar Questions

Which of the following points lies on the locus of the foot of perpendicular drawn upon any tangent to the ellipse, $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$ from any of its foci?

If the lines $x -2y = 12$ is tangent to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ at the point $\left( {3,\frac{-9}{2}} \right)$, then the length of the latus rectum of the ellipse is

The equation of the tangent at the point $(1/4, 1/4)$ of the ellipse $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{{12}} = 1$ is

If a tangent to the ellipse $x^{2}+4 y^{2}=4$ meets the tangents at the extremities of its major axis at $\mathrm{B}$ and $\mathrm{C}$, then the circle with $\mathrm{BC}$ as diameter passes through the point:

If the length of the latus rectum of an ellipse is $4\,units$ and the distance between a focus and its nearest vertex on the major axis is $\frac {3}{2}\,units$ , then its eccentricity is?