Explanation

(b)
Let the equation of hyperbola and conjugate hyperbola be
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ and $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = - 1$
Then, the eccentricities are
$e^{2} = \frac{a^{2} + b^{2}}{a^{2}}$ and ${e'}^{2}\$= $\frac{a^{2} + b^{2}}{b^{2}}$
$\therefore\ \frac{1}{e^{2}} + \frac{1}{{e'}^{2}} = \frac{a^{2}}{a^{2} + b^{2}} + \frac{b^{2}}{a^{2} + b^{2}} = 1$