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# If $e$ and $e'$ be the eccentricities of a hyperbola and its conjugate then $\large\frac {1}{e^2}+\frac{1}{e^{12}}$ is equal to

$\begin{array}{1 1}(A)\;0 \\(B)\;1 \\(C)\;2 \\(D)\;none\;of\;these \end{array}$

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A)
$\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1 b^2=a^2(e^2-1) (or) e^2=1+\large\frac{a^2+b^2}{a^2} \qquad= \large\frac{a^2+b^2}{a^2} Conjugate hyperbola \large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=-1$
(or) $\large\frac{y^2}{b^2}-\frac{x^2}{a^2}$$=1$
$\therefore e^{'2}=1+\large\frac{a^2}{b^2}$
$\qquad= \large\frac{a^2+b^2}{b^2}$
$\therefore \large\frac{1}{e^2} +\frac{1}{e^{'2}}=\frac{a^2+b^2}{a^2+b^2}$
$\qquad=1$
Hence B is the correct answer.