Let the foci and length of the latus rectum of an ellipse x^2/a^2+y^2/b^2=1, a>b be ( ± 5,0) and √ 50, respectively. Then, the square of the…

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