JEE Mathematics 2025 Question with Solution

Given: The ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with eccentricity $e = \frac{1}{2}$ and foci $(\pm 2,0)$. The triangle has vertex $P$ on the minimum-area enclosing circle $C$, and side $QR$ is parallel to the major axis and passes through the point where the ellipse meets the negative $y$-axis.

Find: The maximum area of $\triangle PQR$.

For the ellipse, the foci are $(\pm ae,0)$. Since $ae = 2$ and $e = \frac{1}{2}$,

so

Also,

thus

The ellipse intersects the negative $y$-axis at

The minimum-area circle enclosing the ellipse has radius equal to the semi-major axis, so the circle is

If $P = (4\cos\theta,4\sin\theta)$ lies on this circle, then the vertical distance from $P$ to the horizontal line through $(0,-2\sqrt{3})$ is

This is maximum when $\sin\theta = 1$, giving maximum height

Using the solution-page conclusion, the maximum area is taken as

Therefore, the correct option is D.

Note: The solution itself contains a discrepancy because it also mentions $QR = 29$, which would lead to a different numerical area. However, the solution explicitly concludes with option D, so that is the extracted answer.