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  • The eccentricity of the conjugate hyperbola of the hyperbola  is

The eccentricity of the conjugate hyperbola of the hyperbola \mathrm{x^2-3 y^2=1} is

Option: 1

2


Option: 2

\frac{2}{\sqrt{3}}


Option: 3

4


Option: 4

\frac{4}{3}


Answers (1)

best_answer

The given hyperbola is \mathrm{\frac{x^2}{1}-\frac{y^2}{1 / 3}=1}. Here \mathrm{a^{2}=1} and \mathrm{b^{2}=\frac{1}{3}}
Since \mathrm{b^2=a^2\left(e^2-1\right) \Rightarrow \frac{1}{3}=1\left(e^2-1\right) \Rightarrow e^2=\frac{4}{3} \Rightarrow e=\frac{2}{\sqrt{3}}}
If \mathrm{{e}'} is the eccentricity of the conjugate hyperbola, then 
\mathrm{\frac{1}{e^2}+\frac{1}{e^{\prime 2}}=1 \Rightarrow \frac{1}{e^{\prime 2}}=1-\frac{1}{e^2}=1-\frac{3}{4}=\frac{1}{4} \Rightarrow e^{\prime}=2.}

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jitender.kumar

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