JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Ellipse

With eccentricity $\frac{1}{2}$ and foci $(\pm 2,0)$, $ae=2$ gives $a=4$, and $b^2=a^2(1-e^2)=12$, so $b=2\sqrt{3}$. The smallest circle enclosing the ellipse has radius $a=4$, centred at the origin. The base $QR$ has length $2a=8$ and lies on the horizontal line through $(0,-b)=(0,-2\sqrt{3})$, while the apex $P$ lies on the circle. The greatest height from $P$ to this line is attained at the top of the circle, $a+b=4+2\sqrt{3}$. The maximum area is $\frac{1}{2}\cdot 2a\cdot (a+b)=4(4+2\sqrt{3})$, which equals $8(2+\sqrt{3})$.