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Find the eccentricity of the hyperbola that is conjugate to x²/4 - y²/12 = 1.

1. 2/sqrt(3)
2. 2
3. sqrt(3)
4. 4/3

Correct answer: 2/sqrt(3)

Solution

Conjugate: y²/12 - x²/4 = 1, so B² = 12 (transverse), A² = 4 (conjugate). e² = 1 + 4/12 = 4/3, giving e = 2/sqrt(3).

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