If the eccentricities of the hyperbolas frac{ | vedclass

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(A) For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{\frac{a^2 + b^2

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(A) For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{\frac{a^2 + b^2}{a^2}}$.Therefore,$\frac{1}{e^2} = \frac{a^2}{a^2 + b^2}$.For the hyperbola $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$,the eccentricity $e_1$ is given by $e_1 = \sqrt{1 + \frac{a^2}{b^2}} = \sqrt{\frac{b^2 + a^2}{b^2}}$.Therefore,$\frac{1}{e_1^2} = \frac{b^2}{a^2 + b^2}$.Adding these two expressions,we get $\frac{1}{e^2} + \frac{1}{e_1^2} = \frac{a^2}{a^2 + b^2} + \frac{b^2}{a^2 + b^2} = \frac{a^2 + b^2}{a^2 + b^2} = 1$.

If the eccentricities of the hyperbolas $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$ are $e$ and $e_1$ respectively,then $\frac{1}{e^2} + \frac{1}{e_1^2} = $

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