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If $e$ and $e’$ are eccentricities of hyperbola and its conjugate respectively, then
${\left( {\frac{1}{e}} \right)^2} + {\left( {\frac{1}{{e'}}} \right)^2} = 1$
$\frac{1}{e} + \frac{1}{{e'}} = 1$
${\left( {\frac{1}{e}} \right)^2} + {\left( {\frac{1}{{e'}}} \right)^2} = 0$
$\frac{1}{e} + \frac{1}{{e'}} = 2$
Solution
(a) Let hyperbola is $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1$…..$(i)$
Then its conjugate will be, $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = – 1$…..$(ii)$
If $e$ is eccentricity of hyperbola $(i),$ then ${b^2} = {a^2}({e^2} – 1)$
or $\frac{1}{{{e^2}}} = \frac{{{a^2}}}{{({a^2} + {b^2})}}$ …..$(iii)$
Similarly if e' is eccentricity of conjugate $(ii),$ then ${a^2} = {b^2}(e{'^2} – 1)$ or
$\frac{1}{{e{'^2}}} = \frac{{{b^2}}}{{({a^2} + {b^2})}}$…..$(iv)$
Adding $(iii)$ and $(iv),$
$\frac{1}{{{{(e')}^2}}} + \frac{1}{{{e^2}}}$
$= \frac{{{a^2}}}{{{a^2} + {b^2}}} + \frac{{{b^2}}}{{{a^2} + {b^2}}} = 1.$
