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Q)

If $e_1$ and $e_2$ are the eccentricities of the hyperbolas $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$ and $\large\frac{y^2}{b^2}-\frac{x^2}{a^2}$$=1$ Then value of $\large\frac{1}{e_1^2}+\frac{1}{e_2^2}$ is

$\begin{array}{1 1}(A)\; \bigg|\large\frac{a+b}{a-b}\bigg| \\(B)\; \bigg| \large\frac{a^2-b^2}{a^2} \bigg| \\(C)\;1 \\(D)\;\text{none of these} \end{array}$

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A)
$b^2=a^2(e_1^2 -1)$
=> $\large\frac{1}{e_1^2}=\frac{a^2}{a^2+b^2}$
and $a^2=b^2(e_2^2-1)$
=> $\large\frac{1}{e_2^2}=\frac{b^2}{a^2+b^2}$
$\therefore \large\frac{1}{e_2^2}+\frac{1}{e_2^2}$$=1$
Hence C is the correct answer.
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