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KCET2008MathematicsHyperbola

If e₁ and e₂ are the eccentricities of a hyperbola 3 x²-3 y²=25 and its conjugate, then

Options

  1. Ae₁²+e₂²=2
  2. Be ₁²+ e ₂²=4
  3. Ce₁+e₂=4
  4. De₁+e₂= 2

Correct answer

B. e ₁²+ e ₂²=4

Step-by-step solution

Given equation can be rewritten as x²-y²= 25 3 Here, a ²=1, ~b ²=1 e ₁= 1+ b ² a ² = 1+1 = 2 The equation of conjugate hyperbola is gathered - x ²+ y ²= 25 3 e ₂= 1+ a ² ~b ² = 1+1 = 2 e ₁²+ e ₂²=( 2 )²+( 2 )²=4 gathered

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