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JEE Main 202624 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Ellipse

JEE Main 2026 · 24 January · Evening Shift · Mathematics · Ellipse

Let the length of the latus rectum of an ellipse x^ 2 a^ 2 + y^ 2...

Let the length of the latus rectum of an ellipse x2a2+y2b2=1,(a>b)\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)=−34+2t−t2f(t)=-\frac{3}{4}+2 t-t^{2}, then (a2+b2a^{2}+b^{2}) is equal to

Options

  1. A276
  2. B256
  3. C516
  4. D496

Correct answer

D. 496

Step-by-step solution

f(t)=−t2+2t−34=−(t−1)2+14f(t) = -t^2 + 2t - \frac{3}{4} = -(t-1)^2 + \frac{1}{4}.
Maximum value =14= \frac{1}{4}, so e=14e = \frac{1}{4}.
b2=a2(1−e2)=a2(1−116)=15a216b^2 = a^2(1 - e^2) = a^2\left(1 - \frac{1}{16}\right) = \frac{15a^2}{16}.
Latus rectum =2b2a=2⋅15a2/16a=15a8=30⇒a=16= \frac{2b^2}{a} = \frac{2 \cdot 15a^2/16}{a} = \frac{15a}{8} = 30 \Rightarrow a = 16.
b2=15×25616=240b^2 = \frac{15 \times 256}{16} = 240.
a2+b2=256+240=496a^2 + b^2 = 256 + 240 = 496.
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