Question

Let $S$ and $T$ be the foci of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$ and $B$ be an extreme point of the minor axis. If $SBT$ is an equilateral triangle, then the value of $4e$ is (where $e$ is the eccentricity of the ellipse)

Answer 1

$S\equiv (ae,0),T\equiv (-ae,0)$ and $B\equiv (0,b)$
Since $STB$ is an equilateral triangle
$\Rightarrow S{T}^2=T{B}^2\Rightarrow 4a^2{e}^2=a^2{e}^2+b^2$
$\Rightarrow 3a^2{e}^2=a^2(1-e^2)$
$\Rightarrow 4a^2{e}^2=a^2$
$\Rightarrow e=\frac{1}{2}\therefore 4e=2$