Q80 / 155 · WBJEE 2019 · Mathematics
Sand $T$ are the foci of an ellipse and $B$ is the end point of the minor axis. If $S T B$ is equilateral triangle, the eccentricity of the ellipse is
- A$\frac{1}{4}$
- B$\frac{1}{3}$
- Correct answerC$\frac{1}{2}$
- D$\frac{2}{3}$
Solution
Equation of the ellipse is
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1
$$
Foci are $S(a e, 0), T(-a e, 0)$
$B(0, b)$ is the end of the minor axis. STB is an equilateral triangle.
$$
\begin{aligned}
S B &=S T \\
S B^{2} &=S T^{2}
\end{aligned}
$$
$\Rightarrow \quad a^{2} e^{2}+b^{2}=4 a^{2} e^{2}$
$\Rightarrow \quad b^{2}=3 a^{2} e^{2}$
$\Rightarrow \quad a^{2}\left(1-e^{2}\right)=3 a^{2} e^{2}$
$\Rightarrow \quad 1-e^{2}=3 e^{2}$
$\Rightarrow \quad 4 e^{2}=1$
$\Rightarrow \quad e^{2}=\frac{1}{4}$
$\Rightarrow \quad e=\frac{1}{2}$
Eccentricity of the ellipse, $e=\frac{1}{2}$
Foci are $S(a e, 0), T(-a e, 0)$
$B(0, b)$ is the end of the minor axis. STB is an equilateral triangle.
$$
\begin{aligned}
S B &=S T \\
S B^{2} &=S T^{2}
\end{aligned}
$$
$\Rightarrow \quad a^{2} e^{2}+b^{2}=4 a^{2} e^{2}$
$\Rightarrow \quad b^{2}=3 a^{2} e^{2}$
$\Rightarrow \quad a^{2}\left(1-e^{2}\right)=3 a^{2} e^{2}$
$\Rightarrow \quad 1-e^{2}=3 e^{2}$
$\Rightarrow \quad 4 e^{2}=1$
$\Rightarrow \quad e^{2}=\frac{1}{4}$
$\Rightarrow \quad e=\frac{1}{2}$
Eccentricity of the ellipse, $e=\frac{1}{2}$Hence, the correct answer is option: 'C'.