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Question
Find the eccentricity of the hyperbola, which is conjugate to the hyperbola x2 – 3y2 = 3
Solution
Given, equation of hyperbola is x2 – 3y2 = 3.
∴ `x^2/3 - y^2/1` = 1
Equation of the hyperbola conjugate to the above hyperbola is `y^2/1 - x^2/3` = 1.
Comparing this equation with `y^2/"b"^2 - x^2/"a"^2` = 1, we get,
b2 = 1 and a2 = 3
Now, a2 = b2(e2 – 1)
∴ 3 = 1(e2 – 1)
∴ 3 = e2 – 1
∴ e2 = 4
∴ e = 2 ...[∵ e > 1]
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