Mathematics · 2022
JEE Main · 24 June 2022, Shift 2 · Q9
Let the maximum area of the triangle that can be inscribed in the ellipse x^2/a^2+y^2/4=1, a>2, having one of its vertices at one end of the major…
Let the maximum area of the triangle that can be inscribed in the ellipse $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{4}=1, a>2$, having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the $\displaystyle y$-axis, be $\displaystyle 6 \sqrt{3}$. Then the eccentricity of the ellipse is:
Official answer
(1)
$\displaystyle \frac{\sqrt{3}}{2}$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.