The eccentricity of the conjugate hyperbola to the hyperbola \(x^2-3y^2\) = 1 is

Solution :

Equation of the conjugate hyperbola to the hyperbola \(x^2-3y^2\) = 1 is

\(-x^2-3y^2\) = 1 \(\implies\) \(-x^2\over 1\) + \(y^2\over {1/3}\) = 1

Here \(a^2\) = 1, \(b^2\) = \(1\over 3\)

\(\therefore\)  eccentricity e = \(\sqrt{1 + a^2/b^2}\) = \(\sqrt{1+3}\) = 2


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