JEE Main202624 January 2026Morning ShiftMathematicsEllipseActual
Let each of the two ellipses E ₁: x² a² + y² b² =1,(a>b) and E ₂: x² ~A ² + y² ~B ² =1,( ~A < B ) have eccentricity 4 5 . Let the lengths of the latus recta of E₁ and E₂ be l₁ and l₂ , respectively, such that 2 l₁²=9 l₂ . If the distance between the foci of E₁ is 8, then the distance between the foci of E₂ is
Options
- A32 5
- B8 5
- C16 5
- D96 5
Correct answer
A. 32 5
Step-by-step solution
Both ellipses have e = 4/5 . For E₁ ( a > b ): b^2 = a^2(1-e^2) = 9a^2/25 . Latus rectum l₁ = 2b^2/a = 18a/25 . Distance between foci of E₁ : 2ae = 8a/5 = 8 a = 5 , so l₁ = 18/5 . For E₂ ( A A^2 = B^2(1-e^2) = 9B^2/25 . Latus rectum l₂ = 2A^2/B = 18B/25 . Given 2l₁^2 = 9l₂ : 2 324 25 = 9 18B 25 648 = 162B B = 4 . Distance between foci of E₂ = 2Be = 2 4 4 5 = 32 5