Let the length of the latus rectum of an ellipse $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1,(\mathrm{a}>\mathrm{b})$, be 30. If its eccentricity is the maximum value of the function $f(t)=-\frac{3}{4}+2 t-t^{2}$, then $\left(\mathrm{a}^{2}+\mathrm{b}^{2}\right)$ is equal to -

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Ellipse

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 -

Answer: (C)

The maximum of

f (t) = - \frac{3}{4} + 2 t - t^{2}

occurs at

t = 1

, giving

f_{m a x} = \frac{1}{4}

, so

e = \frac{1}{4}

and

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

. From

e^{2} = \frac{a ^{2} - b ^{2}}{a ^{2}}

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

. The latus rectum

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

gives

b^{2} = 15 a

. Substituting,

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

, so

a = 16

a^{2} = 256

and

b^{2} = 240

. Therefore

a^{2} + b^{2} = 496

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Single correct option