Mathematics · 2025
JEE Main · 3 April 2025, Shift 2 · Q12
Let C be the circle of minimum area enclosing the ellipse E: x^2/a^2+y^2/b^2=1 with eccentricity 1/2 and foci ( ± 2,0). Let P Q R be a variable…
Let $\displaystyle C$ be the circle of minimum area enclosing the ellipse $\displaystyle E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with eccentricity $\displaystyle \frac{1}{2}$ and foci $\displaystyle ( \pm 2,0)$. Let $\displaystyle P Q R$ be a variable triangle, whose vertex $\displaystyle P$ is on the circle $\displaystyle C$ and the side $\displaystyle Q R$ of length $\displaystyle 2 a$ is parallel to the major axis of $\displaystyle E$ and contains the point of intersection of $\displaystyle E$ with the negative $\displaystyle y$-axis. Then the maximum area of the triangle $\displaystyle P Q R$ is :
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle 8(2+\sqrt{3})$
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JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.