Q44 / 75 · JEE Main 2026 (24 January Shift 2) · Mathematics
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
- A276
- B256
- C516
- Correct answerD496
Solution
$f(t) = -t^2 + 2t - \frac{3}{4} = -(t-1)^2 + \frac{1}{4}$.
Maximum value $= \frac{1}{4}$, so $e = \frac{1}{4}$.
$b^2 = a^2(1 - e^2) = a^2\left(1 - \frac{1}{16}\right) = \frac{15a^2}{16}$.
Latus rectum $= \frac{2b^2}{a} = \frac{2 \cdot 15a^2/16}{a} = \frac{15a}{8} = 30 \Rightarrow a = 16$.
$b^2 = \frac{15 \times 256}{16} = 240$.
$a^2 + b^2 = 256 + 240 = 496$.
Maximum value $= \frac{1}{4}$, so $e = \frac{1}{4}$.
$b^2 = a^2(1 - e^2) = a^2\left(1 - \frac{1}{16}\right) = \frac{15a^2}{16}$.
Latus rectum $= \frac{2b^2}{a} = \frac{2 \cdot 15a^2/16}{a} = \frac{15a}{8} = 30 \Rightarrow a = 16$.
$b^2 = \frac{15 \times 256}{16} = 240$.
$a^2 + b^2 = 256 + 240 = 496$.
Hence, the correct answer is option: 'D'.