Let the foci and length of the latus rectum of an ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$$ be $$(\pm 5, 0)$$ and $$\sqrt{50}$$, respectively. Then, the square of the eccentricity of the hyperbola $$\frac{x^2}{b^2} - \frac{y^2}{a^2 b^2} = 1$$ equals