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If e₁ and e₂ are eccentricities of a hyperbola and its conjugate, then
A1/e₁² + 1/e₂² = 1
Be₁² + e₂² = 1
Ce₁ + e₂ = 1
D1/e₁ + 1/e₂ = 1
Answer & Solution
Correct answer: A. 1/e₁² + 1/e₂² = 1
e₁² = 1 + b²/a² and e₂² = 1 + a²/b², and the reciprocals of these squares sum to 1.
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