Question

The eccentricity of the conjugate hyperbola of the hyperbola $x^2-3y^2=1$ is

• A. 2
• B.
• C.
• D.

Answer 1

The correct option is C

the given hyperbola is,

$\frac{\frac{x^2-y^2}{1}}{\frac{1}{3}}=1$

If $e_1$ is the eccentricity of given hyperbola and $e_2$ is the eccentricity of the conjugate hyperbola then,

$e_1^{-2}+e_2^{-2}=1$ ...(1)

Here,

$e_1=1+\frac{b^2}{a^2}=1+\frac{1}{3}$

$=\frac{4}{3}$

$\frac{1}{{\left(\frac{4}{3}\right)}^2}+\frac{1}{e_2^2}=1$

$\frac{1}{16}+\frac{1}{e_2^2}=1$

$\frac{1}{e_2^2}=1-\frac{9}{16}$

$=\frac{7}{16}$

$e_2=\frac{4}{\sqrt{7}}$