alleryHyperbola x²/a² - y²/b² = 1 has eccentricity:
Ae = sqrt(1 + b²/a²) > 1
B< 1
C0
DCannot determine
Answer & Solution
Correct answer: A. e = sqrt(1 + b²/a²) > 1
**For hyperbola:** e = sqrt(1 + b²/a²), always > 1. Foci at (±ae, 0). Asymptotes y = ±(b/a)x.
**Rectangular hyperbola:** a = b, e = sqrt(2).
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