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:
