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10-2. Parabola, Ellipse, Hyperbola
normal
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
A
$\frac{1}{{\sqrt 2 }}$
B
$\frac{{\sqrt 3 }}{2}$
C
$\frac{1}{{\sqrt 3 }}$
D
$\frac{1}{2}$
Solution
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
$=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{b^{2}}{4 b^{2}}}=\frac{\sqrt{3}}{2}$
Standard 11
Mathematics
