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JEE Main202628 January 2026Morning ShiftMathematicsHyperbolaActual

For some (0, 2 ) , let the eccentricity and the length of the latus rectum of the hyperbola x²-y² ² =8 be e₁ and l₁ , respectively, and let the eccentricity and the length of the latus rectum of the ellipse x² ² +y²=6 be e₂ and l₂ , respectively. If e₁²=e₂² ( ² +1 ) , then ( l₁ l₂ e₁ e₂ ) ² is equal to _ _ _ _

Correct answer

0

Step-by-step solution

x^2 8 - y^2 8 ^2 = 1 , e₁ = 1 + 8 ^2 8 ₁ = 2b^2 a = 2 (8 ^2 ) 2 2 x^2 6 + y^2 6 ^2 = 1 ; e₂ = 1 - 6 ^2 6 = ₂ = 2b^2 a = 2 6 ^2 6 e₁^2 = e₂^2(1 + ^2 ) 1 + ^2 = ^2 (1 + 1 ^2 ) 1 + ^2 = ^2 + ^2 Solving we get = 4 ₁ = 2 2 e₁ = 3 2 ₂ = 6 e₂ = 1 2 ( ₁ ₂ e₁ e₂ ) ^2 = 8 (By putting values)

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