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10-2. Parabola, Ellipse, Hyperbola
normal
If the area of the auxiliary circle of the ellipse $\frac{{{x^2}}}{{{a^2}}}\, + \,\frac{{{y^2}}}{{{b^2}}}\, = \,1(a\, > \,b)$ is twice the area of the ellipse, then the eccentricity of the ellipse is
A
$\frac{1}{\sqrt2}$
B
$\frac{\sqrt3}{2}$
C
$\frac{1}{\sqrt3}$
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 $\pi \mathrm{\,ab}.$
The auxiliary circle to the given ellipse is $x^{2}+y^{2}=a^{2}$ whose area is $\pi a^{2}.$
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
