If e and e' are the eccentricities of a hyper | vedclass

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(A) Let the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ $(i)$.Its conjugate hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = -1$ $(ii)$.For

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$(\frac{1}{e})^2 + (\frac{1}{e'})^2 = 1$

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(A) Let the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ $(i)$.Its conjugate hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = -1$ $(ii)$.For the hyperbola $(i)$,the eccentricity $e$ is given by $b^2 = a^2(e^2 - 1)$,which implies $e^2 = 1 + \frac{b^2}{a^2} = \frac{a^2 + b^2}{a^2}$.Thus,$\frac{1}{e^2} = \frac{a^2}{a^2 + b^2}$ $(iii)$.For the conjugate hyperbola $(ii)$,the eccentricity $e'$ is given by $a^2 = b^2(e'^2 - 1)$,which implies $e'^2 = 1 + \frac{a^2}{b^2} = \frac{a^2 + b^2}{b^2}$.Thus,$\frac{1}{e'^2} = \frac{b^2}{a^2 + b^2}$ $(iv)$.Adding $(iii)$ and $(iv)$,we get $\frac{1}{e^2} + \frac{1}{e'^2} = \frac{a^2}{a^2 + b^2} + \frac{b^2}{a^2 + b^2} = \frac{a^2 + b^2}{a^2 + b^2} = 1$.

If $e$ and $e'$ are the eccentricities of a hyperbola and its conjugate hyperbola respectively,then:

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