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$\mathrm{S}$ and $\mathrm{T}$ are the foci of an ellipse and $\mathrm{B}$ is an end of the minor axis. If STB is an equilateral triangle, then the eccentricity of the ellipse is
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$\frac{1}{2}$
Let eq. of ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, $\mathrm{S}$ is $(-\mathrm{ae}, 0), \mathrm{T}$ is $(\mathrm{ae}, 0)$ and $\mathrm{B}$ is $(0, \mathrm{~b})$
$\Rightarrow \mathrm{SB}=\sqrt{(0+\mathrm{ae})^{2}+\mathrm{b}^{2}}$
Also $\mathrm{SB}^{2}=\mathrm{ST}^{2} \Rightarrow 4 \mathrm{a}^{2} \mathrm{e}^{2}=\mathrm{a}^{2} \mathrm{e}^{2}+\mathrm{b}^{2}$
$\Rightarrow 3 a^{2} e^{2}=a^{2}\left(1-e^{2}\right)=a^{2}-a^{2} e^{2}$
$\Rightarrow 4 a^{2} e^{2}=a^{2} \Rightarrow e^{2}=\frac{1}{4} \Rightarrow e=\frac{1}{2}$
$\Rightarrow \mathrm{SB}=\sqrt{(0+\mathrm{ae})^{2}+\mathrm{b}^{2}}$
Also $\mathrm{SB}^{2}=\mathrm{ST}^{2} \Rightarrow 4 \mathrm{a}^{2} \mathrm{e}^{2}=\mathrm{a}^{2} \mathrm{e}^{2}+\mathrm{b}^{2}$
$\Rightarrow 3 a^{2} e^{2}=a^{2}\left(1-e^{2}\right)=a^{2}-a^{2} e^{2}$
$\Rightarrow 4 a^{2} e^{2}=a^{2} \Rightarrow e^{2}=\frac{1}{4} \Rightarrow e=\frac{1}{2}$
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