Let the foci and length of the latus rectum of an ellipse fracx^2a^2 + fracy^2b^2 = 1, a > b$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$ be (pm 5, 0)$(\pm 5, 0)$ and sqrt50$\sqrt{50}$, respectively. Then, the square of the eccentricity of the hyperbola fracx^2b^2 - fracy^2a^2 b^2 = 1$\frac{x^2}{b^2} - \frac{y^2}{a^2 b^2} = 1$ equals

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