JEE Main 2025 — Ellipse Question with Solution

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 $QR$ of length $2a$ 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 :