Aspire Faculty ID #18062 · Topic: JEE Main 2026 (24 January Evening Shift) · Just now
JEE Main 2026 (24 January Evening Shift)

Let the length of the latus rectum of an ellipse $ \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 $ f(t) = -\frac{3}{4} + 2t - t^2 $, then $ (a^2 + b^2) $ is equal to

Solution

$ e^2 = \frac{1}{16} $

Then,

$ \frac{b^2}{a^2} = \frac{15}{16} $

$ b^2 = \frac{15}{16} a^2 $

$ \frac{2b^2}{a} = 30 $

$ \frac{2 \cdot 15}{16} a = 30 $

$ \frac{30}{16} a = 30 $

$ a = 16 $

$ b^2 = 240 $

$ a^2 = 256 $

$ a^2 + b^2 = 256 + 240 = 496 $
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