Let e _ 1 and e _ 2 be the eccentricities of the ellipse x ^ 2 25 +...

${e}_{1} = \sqrt{1 - \dfrac{{b}^{2}}{25}}$ and ${e}_{2} = \sqrt{1 + \dfrac{{b}^{2}}{16}}$

Given ${e}_{1}{e}_{2} = 1$

$⇒ {\left({e}_{1}{e}_{2}\right)}^{2} = 1 ⇒ \left(1 - \dfrac{{b}^{2}}{25}\right)\left(1 + \dfrac{{b}^{2}}{16}\right) = 1$

$⇒ \dfrac{{b}^{2}}{16} - \dfrac{{b}^{2}}{25} - \dfrac{{b}^{4}}{25 \times 16} = 0$

$⇒ \dfrac{9}{16 \times 25}{b}^{2} - \dfrac{{b}^{4}}{25 \times 16} = 0 ⇒ {b}^{2} = 9$

Thus, ${e}_{1} = \sqrt{1 - \dfrac{9}{25}} = \dfrac{4}{5}$ and ${e}_{2} = \sqrt{1 + \dfrac{9}{16}} = \dfrac{5}{4}$

Therefore, $\alpha = 2\left(5\right)\left({e}_{1}\right) = 8$ and $\beta = 2\left(4\right)\left({e}_{2}\right) = 10$

$∴ \left(\alpha, \beta\right) = \left(8, 10\right)$