JEE Main 2024 Mathematics question
Let the foci and length of the latus rectum of an ellipse x²/a² + y²/b² = 1, a > b be (± 5, 0) and √50, respectively. Then, the square of the eccentricity of the hyperbola x²/b² − y²/(a²b²) = 1 equals .
Answer: 51
Solution
ae = 5 and 2b²/a = √50, so b² = (5√2/2)a. With b² = a²(1 − e²) and a = 5/e: (5√2/2) = a(1 − e²) = (5/e)(1 − e²), so √2e = 2(1 − e²), i.e. 2e² + √2e − 2
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