Eccentricity of the hyperbola conjugate to th | vedclass

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Question

$e_1^2 = 1 + \frac{{{b^2}}}{{{a^2}}}$ $=$ $1 +$$\frac{{12}}{4}$ $= 4$ $ \Rightarrow$ $ e_1 = 2$ ;
now $\frac{1}{{e_1^2}} + \frac{1}{{e_2^2}}$ $=

Vedclass · Accepted Answer

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

Vedclass · Answer

$e_1^2 = 1 + \frac{{{b^2}}}{{{a^2}}}$ $=$ $1 +$$\frac{{12}}{4}$ $= 4$ $ \Rightarrow$ $ e_1 = 2$ ;
now $\frac{1}{{e_1^2}} + \frac{1}{{e_2^2}}$ $= 1$ 
$\frac{1}{{e_2^2}}$ $=$ $ 1 -$ $\frac{1}{4}$ $ =$ $\frac{3}{4}$
$ \Rightarrow$ $e_2^2$  $=$ $\frac{4}{3}$  $\Rightarrow$  $e_2 $ $= $ $\frac{2}{{\sqrt 3 }}$

Eccentricity of the hyperbola conjugate to the hyperbola $\frac{{{x^2}}}{4} - \frac{{{y^2}}}{{12}} = 1$ is

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