Let $C$ be the circle of minimum area enclosing the ellipse $E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with eccentricity $\frac{1}{2}$ and foci $( \pm 2,0)$. Let PQR be a variable triangle, whose vertex $P$ is on the circle $C$ and the side $Q R$ of length 29 is parallel to the major axis of $E$ and contains the point of intersection of $E$ with the negative $y$-axis. Then the maximum area of the triangle PQR is :

• A

  $6(3+\sqrt{2})$
• B

  $8(3+\sqrt{2})$
• C

  $62+\sqrt{3}$
• Correct answer
  D

  $82+\sqrt{3}$

Solution

Area of $\triangle P Q R$
$\begin{aligned}<br/>& =\frac{1}{2}(2 a)(a \sin \theta+b) \\ & \therefore \text { maximum area }=a(a+b) \\ & \quad=4(4+2 \sqrt{3})=8(2+\sqrt{3})<br/>\end{aligned}$

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