If $\mathrm{B}$ is end point of minor axis of the ellipse $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}(a>b)$ and…
If $\mathrm{B}$ is end point of minor axis of the ellipse $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}(a>b)$ and $\mathrm{S}$ and
S' are foci of ellipse such that $\Delta \mathrm{SBS}^{\prime}$ is an equilateral triangle, then eccentricity $\mathrm{e}=$
$\frac{1}{2}$
$\frac{1}{3}$
$\frac{3}{5}$
$\frac{4}{5}$
Solution
Correct option is A
Given $\mathrm{S}(-\mathrm{ae}, 0), \mathrm{T}(\mathrm{ae}, 0), \mathrm{B}(0, \mathrm{~b})$
As STB is an equilateral triangle
In $\triangle \mathrm{TOB}$
$\tan 60^{\circ}=\frac{\mathrm{OB}}{\mathrm{OS}}$
$\sqrt{3}=\frac{\mathrm{OB}}{\mathrm{OS}}=\frac{\mathrm{b}}{\mathrm{ae}}$
$(\sqrt{3} a e)^{2}=b^{2}$
$3 a^{2}\left(1-\frac{b^{2}}{a^{2}}\right)=b^{2}$
$3\left(1-\frac{b^{2}}{a^{2}}\right)=\frac{b^{2}}{a^{2}}$
$\frac{3}{4}=\frac{b^{2}}{a^{2}}$
$e=\sqrt{1-\frac{3}{4}}=\frac{1}{2}$
