Mathematics · 2026
JEE Main · 24 January 2026, Shift 2 · Q10
Let the length of the latus rectum of an ellipse x^2/a^2+y^2/b^2=1,(a>b), be 30. If its eccentricity is the maximum value of the function f(t)=-3/4+2…
Let the length of the latus rectum of an ellipse $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a>b)$, be 30. If its eccentricity is the maximum value of the function $\displaystyle f(t)=-\frac{3}{4}+2 t-t^2$, then $\displaystyle \left(a^2+b^2\right)$ is equal to
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle 496$
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