Let F and $\mathrm{F}^1$ be the foci of the ellipse $\frac{x^2}{4}+\frac{y^2}{b^2}=1(b \lt 2)$ and $B$ is one end of the minor axis. If the area of the triangle $\mathrm{FBF}^1$ is $\sqrt{3}$ sq. units, then the eccentricity of the ellipse is

• Correct answer
  A

  $\frac{\sqrt{3}}{2}$ or $\frac{1}{2}$
• B

  $\frac{1}{\sqrt{3}}$
• C

  $\frac{\sqrt{3}}{4}$ or $\frac{1}{4}$
• D

  $\frac{3}{4}$ or $\frac{1}{4}$

Solution

Given equation of ellipse is $\frac{x^2}{4}+\frac{y^2}{b^2}=1$ and $(b \lt 2)$ so, $F(c, 0), F^{\prime}(-c, 0)$ and $B(0, b)$
Area of $\triangle F B F^{\prime}=\frac{1}{2} \times 2 c \times b$ $\sqrt{3}=b c \Rightarrow b^2 c^2=3 \Rightarrow c^2=\frac{3}{b^2}$ ....(i) We know that $B F+B F^{\prime}=2 a \Rightarrow \sqrt{b^2+c^2}+\sqrt{b^2+c^2}=2 a$ $2 \sqrt{b^2+c^2}=2 a \Rightarrow \sqrt{b^2+c^2}=a$ $b^2+c^2=a^2=4 \Rightarrow b^2+\frac{3}{b^2}=4$ $\Rightarrow b^4-4 b^2+3=0 \Rightarrow\left(b^2-1\right)\left(b^2-3\right)=0$ So, $b=1 \Rightarrow c=\sqrt{3}$ or $b=\sqrt{3} \Rightarrow c=1$ when $c=\sqrt{3} \Rightarrow a e=\sqrt{3} \Rightarrow e=\frac{\sqrt{3}}{2}$ or $c=1 \Rightarrow a e=1 \Rightarrow e=\frac{1}{2}$

Hence, the correct answer is option: 'A'.