JEE Main 2025 Apr 7 Shift 2, Mathematics Q12: Maxima & Minima

JEE Main2025Apr 7, Shift 2Mathematics

Let the length of a latus rectum of an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ be 10. If its eccentricity is the minimum value of the function $f (t) = t^{2} + t + \frac{11}{12}$ , $t \in R$ , then $a^{2} + b^{2}$ is equal to :

A
125
B
126
C
120
D
115

Solution

$f (t) = t^{2} + t + \frac{11}{12}$ is minimised at $t = - \frac{1}{2}$ , giving $f (- \frac{1}{2}) = \frac{1}{4} - \frac{1}{2} + \frac{11}{12} = \frac{2}{3}$ , so the eccentricity $e = \frac{2}{3}$ .

Latus rectum length $= \frac{2 b ^{2}}{a} = 10 \Rightarrow b^{2} = 5 a$ .

Also $e^{2} = 1 - \frac{b ^{2}}{a ^{2}} \Rightarrow \frac{4}{9} = 1 - \frac{b ^{2}}{a ^{2}} \Rightarrow b^{2} = \frac{5 a ^{2}}{9}$ .

Equating the two expressions for $b^{2}$ : $5 a = \frac{5 a ^{2}}{9} \Rightarrow a = 9$ .

Then $b^{2} = 5 (9) = 45$ .

$∴ a^{2} + b^{2} = 81 + 45 = 126$ .

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JEE Main 2025 Apr 7 Shift 2, Mathematics Q12: Maxima & Minima

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