Let $S$ and $S^{\prime}$ be the foci of an ellipse $E$ and $B$ be one end of its minor axis. Let $\left\lfloor\mathrm{S}^{\prime} \mathrm{SB}=\frac{\pi}{6}\right.$ and $(2 \sqrt{3}, 1)$ be a point on $\mathrm{E}$. If $\mathrm{X}$-axis is the major axis and $\mathrm{Y}$-axis is the minor axis of the ellipse $E$, then the sum of the squares of the lengths of major axis and minor axis is

• A

  $20$
• B

  $60$
• Correct answer
  C

  $80$
• D

  $100$

Solution

$(2 \sqrt{3}, 1)$ lies on $E$. $$ \begin{aligned} & E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ & S(a e, 0), S=(-a e, 0) B(0, b) \end{aligned} $$
$\begin{aligned} & \tan \frac{\pi}{6}=\frac{O B}{O S} \Rightarrow \frac{b}{a e}=\frac{1}{\sqrt{3}} \Rightarrow 3 b^2=a^2 e^2 \\ & \because b^2=a^2\left(1-e^2\right) \\ & \Rightarrow 3 b^2=a^2\left(1-\frac{b^2}{a^2}\right) \Rightarrow 3 b^2=a^2-b^2\end{aligned}$ $\Rightarrow a^2=4 b^2$ ...(1) $\begin{aligned} & \text { Also } \frac{12}{a^2}+\frac{1}{b^2}=1 \Rightarrow \frac{12}{4 b^2}+\frac{1}{b^2}=1 \\ & \Rightarrow b^2=4 \Rightarrow a^2=16 \\ & \Rightarrow(2 a)^2+(2 b)^2=4\left(a^2+b^2\right)=4(16+4)=80 .\end{aligned}$

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