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Saturday, 31 October

JEE Main 2025 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 7 April 2025, Shift 2, Maths Q12

Question 12 of 75 in this shift, Maths question 12 of 25, Section A.

Let the length of a latus rectum of an ellipse x2a2+y2b2=1\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)=t2+t+1112f(t)=t^2+t+\frac{11}{12}, t∈Rt\in\mathbf{R}, then a2+b2a^2+b^2 is equal to :
  1. (1)115
  2. (2)120
  3. (3)125
  4. (4)126Official answer

Official answer

Option 4

NTA final key.

Same idea in other shifts

Asked 12× in all
  1. 4 Apr 2026, Shift 1 · Q22Consider the parabola P:y2=4kxP: y^2=4kx and the ellipse E:x2a2+y2b2=1E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1. Let the line segment joining the points of…MediumNumerical value
  2. 5 Apr 2024, Shift 1 · Q15Let the line 2x+3y−k=02x+3y-k=0, k>0k>0, intersect the xx-axis and yy-axis at the points A and B, respectively. If the equation of the circle…MediumSingle correct
  3. 9 Apr 2024, Shift 1 · Q12Let f(x)=x2+9f(x)=x^2+9, g(x)=xx−9g(x)=\frac{x}{x-9} and a=f∘g(10)a=f\circ g(10), b=g∘f(3)b=g\circ f(3). If ee and ll denote the eccentricity and the length of the…EasySingle correct
  4. 1 Feb 2024, Shift 1 · Q13Let x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>ba>b be an ellipse, whose eccentricity is 12\frac{1}{\sqrt{2}} and the length of the latusrectum…MediumSingle correct
  5. 31 Jan 2024, Shift 1 · Q28Let the foci and 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 (±5,0)(\pm5,0) and 50\sqrt{50},…MediumNumerical value
  6. 31 Jan 2024, Shift 2 · Q14Let PP be a parabola with vertex (2,3)(2,3) and directrix 2x+y=62x+y=6. Let an ellipse E:x2a2+y2b2=1, a>bE:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\ a>b, of…HardSingle correct
  7. 6 Apr 2023, Shift 2 · Q27Let the eccentricity of an ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 is reciprocal to that of the hyperbola 2x2−2y2=12x^2-2y^2=1. If the…HardNumerical value
  8. 27 Jul 2022, Shift 1 · Q24An ellipse E:x2a2+y2b2=1E : \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 passes through the vertices of the hyperbola…HardNumerical value

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 108 of the 174 JEE Main shifts held from 2017 to 2026.