Mathematics · 2025
JEE Main · 7 April 2025, Shift 2 · Q12
Let the length of a latus rectum of an ellipse x^2/(a^2)+y^2/(b^2)=1 be 10. If its eccentricity is the minimum value of the function f( t )= t^2+ t…
Let the length of a latus rectum of an ellipse $\displaystyle \frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1$ be 10. If its eccentricity is the minimum value of the function $\displaystyle f(\mathrm{t})=\mathrm{t}^2+\mathrm{t}+\frac{11}{12}, \mathrm{t} \in \mathbf{R}$, then $\displaystyle \mathrm{a}^2+\mathrm{b}^2$ is equal to :
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle 126$
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JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.