Question

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

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

Answer 1

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

Let the equation of ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$
Coordinates of $S$ and $T$ are $(ae,0)$ and $(-ae,0)$ respectively.
Length $ST=2ae$

Area of equilateral $△STB$ is,
$\frac{\sqrt{3}}{4}(\text{side}{)}^2=\frac{\sqrt{3}}{4}(2ae{)}^2$
Also, area $=\frac{1}{2}\times \text{base}\times \text{height}$
$=\frac{1}{2}\times \left(2ae\right)\times b$

Equating both areas, we get
$b=\frac{\sqrt{3}}{2}\left(2ae\right)$
Squaring both sides
$\Rightarrow b^2=3(ae{)}^2$
$\Rightarrow b^2=3(a^2-b^2)[\because e^2=1-\frac{b^2}{a^2}]$
$\Rightarrow 4b^2=3a^2$
$\Rightarrow \frac{b^2}{a^2}=\frac{3}{4}$

$\therefore e^2=1-\frac{b^2}{a^2}=1-\frac{3}{4}=\frac{1}{4}$
$\Rightarrow e=\frac{1}{2}$