The eccentricity of the conjugate hyperbola o | vedclass

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Question

(a) Eccentricity of $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ is $e = \sqrt {\frac{{{a^2} + {b^2}}}{{{a^2}}}} $

Eccentricity of conju

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(a) Eccentricity of $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ is $e = \sqrt {\frac{{{a^2} + {b^2}}}{{{a^2}}}} $

Eccentricity of conjugate hyperbola, $e' = \sqrt {\frac{{{a^2} + {b^2}}}{{{b^2}}}} $

Write the given equation in standard form,

$\frac{{{x^2}}}{1} - \frac{{{y^2}}}{{1/3}} = 1$

==> ${a^2} = 1,\,\,{b^2} = \frac{1}{3}$

$e' = \sqrt {\frac{{1 + 1/3}}{{1/3}}} = \sqrt 4 = 2$.

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

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