JEE Main 2026 — Jan 24, Shift 2 — Mathematics solution

JEE Main2026Jan 24, Shift 2Mathematics

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 $f (t) = - \frac{3}{4} + 2 t - t^{2}$ , then $(a^{2} + b^{2})$ is equal to -

A
516
B
256
C
496
D
276

Solution

Maximize $f (t) = - t^{2} + 2 t - \frac{3}{4}$ : $f^{'} (t) = - 2 t + 2 = 0 \Rightarrow t = 1$ . Maximum value $= - 1 + 2 - \frac{3}{4} = \frac{1}{4}$ .

So eccentricity $e = \frac{1}{4}$ and $e^{2} = \frac{1}{16} = \frac{a ^{2} - b ^{2}}{a ^{2}}$ ... (1)

Latus rectum: $\frac{2 b ^{2}}{a} = 30 \Rightarrow b^{2} = 15 a$ ... (2)

From (1): $16 (a^{2} - b^{2}) = a^{2} \Rightarrow 15 a^{2} = 16 b^{2} = 16 (15 a) = 240 a \Rightarrow a = 16$ .

Then $b^{2} = 15 \times 16 = 240$ and $a^{2} = 256$ .

$∴ a^{2} + b^{2} = 256 + 240 = 496$ .

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