MCQMediumJEE Main 2025 · 7 April, Shift 2Conic Sections (Parabola, Ellipse, Hyperbola)

Mathematics Question from JEE Main 2025 · 7 April, Shift 2

Let the length of a latus rectum of an ellipse x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 be 1010. If its eccentricity is the minimum value of the function f(t)=t2+t+1112f(t) = t^2 + t + \frac{11}{12}, tRt \in \mathbb{R}, then a2+b2a^2 + b^2 is equal to:

  • A

    125125

  • B

    126126

  • C

    120120

  • D

    115115

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