JEE MAIN - Mathematics (2025 - 3rd April Evening Shift - No. 5)
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 $P Q R$ be a variable triangle, whose vertex $P$ is on the circle $C$ and the side $Q R$ of length $2 a$ 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 $P Q R$ is :
$8(3+\sqrt{2})$
$8(2+\sqrt{3})$
$6(3+\sqrt{2})$
$6(2+\sqrt{3})$
Explanation
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$$\begin{aligned} &\text { Area }=\frac{1}{2} \times(a+b) \cdot 2 a=a(a+b)\\ &\text { Since, } e=\frac{1}{2} a e=2 \Rightarrow a=4, b=2 \sqrt{3}\\ &\Rightarrow \text { Area }=4(4+2 \sqrt{3})=8(2+\sqrt{3}) \end{aligned}$$
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