Question

$S$ and $T$ are the foci of the ellipse $\left(\frac{x^2}{a^2}\right)+\left(\frac{y^2}{b^2}\right)=1$ and $B$ is an end of the minor axis. If $STB$ is an equilateral triangle, then eccentricity of the ellipse is:

• A. $\frac{1}{4}$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $\sqrt{\frac{3}{2}}$

Answer 1

The correct option is C $\frac{1}{2}$

$S=(ae,0),T=(-ae,0)$ and $B=(0,b)$

Since triangle is equilateral,

$ST=TB\Rightarrow S{T}^2=T{B}^2$

$\Rightarrow (2ae{)}^2=(ae{)}^2+b^2$

$\Rightarrow 3a^2{e}^2=b^2$

$\Rightarrow \frac{b^2}{a^2}=3e^2$.......(1)

Thus, eccentricity of the ellipse is,

$e^2=1-\frac{b^2}{a^2}=1-3e^2$

$\Rightarrow 4e^2=1$

$\Rightarrow e^2=\frac{1}{4}$

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