alleryRelation between eccentricities of conjugate hyperbolas:
Ae + e' = 2
Be × e' = 1
C1/e² + 1/e'² = 1
De = e'
Answer & Solution
Correct answer: C. 1/e² + 1/e'² = 1
For hyperbola x²/a² - y²/b² = 1
(e) and conjugate y²/b² - x²/a² = 1 (e'): 1/e² + 1/e'² = 1. Proof: e² = 1 + b²/a², e'² = 1 + a²/b².
1/e² = a²/(a²+b²), 1/e'² = b²/(a²+b²). Sum = 1. ✓
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