If e _ 1 and e _ 2 are the eccentricities of a hyperbola 3x^ 2 -...

Given equation can be written as

$x^{2} - y^{2} = \dfrac{25}{3}$

$∴ {e}_{1} = \sqrt{1 + \dfrac{b2}{a2}}$

$= \sqrt{1 + 1} = \sqrt{2}$

The equation of conjugate hyperbola is

$- x^{2} + y^{2} = \dfrac{25}{3}$

$∴ {e}_{2} = \sqrt{1 + \dfrac{b2}{a2}} = \sqrt{1 + 1} = \sqrt{2}$

$∴ {e}_{1}^{2} + {e}_{2}^{2} = \left(\sqrt{2}\right)^{2} + \left(\sqrt{2}\right)^{2} = 4$

Hence, ${e}_{1}^{2} + {e}_{2}^{2} = 4$.