Let $F_{1}$ and $F_{2}$ be the foci of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ and $(0, b)$ be an end point of the minor axis. If triangle $B F_{1} F_{2}$ is equilateral, $e$ is the eccentricity of the ellipse and $∆$ is the area of the triangle $B F_{1} F_{2}$ , then

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$∆ = \frac{\sqrt{3}}{4} a^{2}$

$Δ = \sqrt{3} a^{2}$

$e = \frac{1}{2}$

$e = \frac{1}{3}$

answer is A, C.

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Detailed Solution

Coordinates of $F_{1}$ are $(a e, 0)$ and of $F_{2}$ are $(- a e, 0)$

Also $B F_{1} = B F_{2} = F_{1} F_{2}$

$\begin{array}{l} \Rightarrow a^{2} e^{2} + b^{2} = 4 a^{2} e^{2} \\ \Rightarrow a^{2} e^{2} + a^{2} (1 - e^{2}) = 4 a^{2} e^{2} \\ \Rightarrow e = \frac{1}{2} \Rightarrow F_{1} F_{2} = a \end{array}$

and the area of the

$Δ B F_{1} F_{2} = \frac{\sqrt{3}}{4} a^{2}$

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