If \(e\) and \(e^{\prime}\) be the eccentricities of a hyperbola and its conjugate, then \(\frac{1}{e^{2}}+\frac{1}{e^{\prime 2}}=\)
\(e^{2}=\left(1+\frac{b^{2}}{a^{2}}\right)\) and \(e^{2}=\left(1+\frac{a^{2}}{b^{2}}\right)\) Hence \(\frac{1}{e^{2}}+\frac{1}{e^{\prime 2}}=1\) \[\] Choice (A)\[\]
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