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BITSAT2018MathematicsHyperbolaActual

If e 1 and e 2 are the eccentricities of a hyperbola 3 x 2 - 3 y 2 = 25 and its conjugate, then

Options

  1. Ae 1 2 + e 2 2 = 2
  2. Be 1 2 + e 2 2 = 4
  3. Ce 1 + e 2 = 4
  4. De 1 + e 2 = 2

Correct answer

B. e 1 2 + e 2 2 = 4

Step-by-step solution

Given equation can be written as x 2 - y 2 = 25 3 ∴   e 1 = 1 + b 2 a 2 = 1 + 1 = 2 The equation of conjugate hyperbola is - x 2 + y 2 = 25 3 ∴   e 2 = 1 + b 2 a 2 = 1 + 1 = 2 ∴   e 1 2 + e 2 2 = 2 2 + 2 2 = 4 Hence, e 1 2 + e 2 2 = 4 .

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