JEE Main 2024 Mathematics question
Let the foci of a hyperbola H coincide with the foci of the ellipse E : (x − 1)²/100 + (y − 1)²/75 = 1 and the eccentricity of the hyperbola H be the reciprocal of the eccentricity of the ellipse E. If the length of the transverse axis of H is α and the length of its conjugate axis is β, then 3α² + 2β² is equal to :
Answer: (B)
Solution
For the ellipse e₁ = √(1 − 75/100) = 1/2, and the foci are (1 ± 5, 1) = (6, 1) and (−4, 1). The hyperbola has e₂ = 2 with the same foci. So 2ae₂ =
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