Question

If the area of the auxiliary circle of the ellipse \[\frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{b}^2}=1(\mathrm{a}>\mathrm{b})\] is twice the area of the ellipse, then the eccentricity of the ellipse is______.

विकल्प

• \[\frac{1}{\sqrt{2}}\]

• \[\frac{\sqrt{3}}{2}\]

• \[\frac{1}{\sqrt{3}}\]

• \[\frac{1}{2}\]

Answer 1

If the area of the auxiliary circle of the ellipse \[\frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{b}^2}=1(\mathrm{a}>\mathrm{b})\] is twice the area of the ellipse, then the eccentricity of the ellipse is \[\frac{\sqrt{3}}{2}\].

Explanation:

• Area of auxiliary circle = \[\pi a^{2}\]

• Area of ellipse = \[\pi ab\]

Given condition: Area of auxiliary circle = 2 × Area of ellipse

                             \[\begin{aligned} \pi a^2 & =2\pi ab \\ \\ a=2b & \Longrightarrow b=\frac{a}{2} \end{aligned}\]

Now find eccentricity using \[c^2=a^2-b^2{:}\]

                                                 \[c^2=a^2-\frac{a^2}{4}=\frac{3a^2}{4}\]

                                                \[e=\frac{c}{a}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}\]