S and T are the foci of an ellipse and B is t | vedclass

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(C) Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.Foci are $S(ae, 0)$ and $T(-ae, 0)$.$B(0, b)$ is the end point of t

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$\frac{1}{2}$

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(C) Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.Foci are $S(ae, 0)$ and $T(-ae, 0)$.$B(0, b)$ is the end point of the minor axis.Since $	riangle STB$ is an equilateral triangle, the distance $SB = ST = TB$.$ST = ae - (-ae) = 2ae$.$SB = \sqrt{(ae-0)^{2} + (0-b)^{2}} = \sqrt{a^{2}e^{2} + b^{2}}$.Since $SB = ST$, we have $SB^{2} = ST^{2}$.$a^{2}e^{2} + b^{2} = (2ae)^{2} = 4a^{2}e^{2}$.$b^{2} = 3a^{2}e^{2}$.Using the relation $b^{2} = a^{2}(1-e^{2})$, we get:$a^{2}(1-e^{2}) = 3a^{2}e^{2}$.$1 - e^{2} = 3e^{2}$.$4e^{2} = 1$.$e^{2} = \frac{1}{4}$.$e = \frac{1}{2}$ (since eccentricity $e > 0$).

$S$ and $T$ are the foci of an ellipse and $B$ is the end point of the minor axis. If $\triangle STB$ is an equilateral triangle, the eccentricity of the ellipse is

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