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Friday, 9 October

JEE Main 2026 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 8 April 2026, Shift 2, Maths Q20

Question 20 of 75 in this shift, Maths question 20 of 25, Section A.

Let x2f(a2+7a+3)+y2f(3a+15)=1\frac{x^2}{f(a^2 + 7a + 3)} + \frac{y^2}{f(3a + 15)} = 1 represent an ellipse with major axis along yy-axis, where ff is a strictly decreasing positive function on R\mathbf{R}. If the set of all possible values of aa is R−[α,β]\mathbf{R} - [\alpha, \beta], then α2+β2\alpha^2 + \beta^2 is equal to :
  1. (1)28
  2. (2)40Official answer
  3. (3)61
  4. (4)24

Official answer

Option 2

NTA final key.

Chapter
Conic Sections
Topic
No close match in our taxonomy
Idea tested
No close match in our taxonomy

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.