If e 1 and e 2 are the eccentricities of a hyperbola 3 x 2 - 3 y 2 = 25 and its conjugate, then

Given equation can be written as

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

$\therefore e_1=\sqrt{1+\frac{b^2}{a^2}}$

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

The equation of conjugate hyperbola is

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

$\therefore e_2=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+1}=\sqrt{2}$

$\therefore e_1^2+e_2^2={\left(\sqrt{2}\right)}^2+{\left(\sqrt{2}\right)}^2=4$

Hence, $e_1^2+e_2^2=4$.