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(C) Given foci of the ellipse are $(\pm 5, 0)$,so $ae = 5$. Length of latus rectum is $\frac{2b^2}{a} = \sqrt{50} = 5\sqrt{2}$. From $ae = 5$,we have

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$51$

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(C) Given foci of the ellipse are $(\pm 5, 0)$,so $ae = 5$. Length of latus rectum is $\frac{2b^2}{a} = \sqrt{50} = 5\sqrt{2}$. From $ae = 5$,we have $a = \frac{5}{e}$. Substituting $a$ in the latus rectum formula: $\frac{2b^2}{5/e} = 5\sqrt{2} \Rightarrow b^2 = \frac{25\sqrt{2}e}{2}$. Using the relation $b^2 = a^2(1-e^2)$,we get $\frac{25\sqrt{2}e}{2} = \frac{25}{e^2}(1-e^2)$. Simplifying,$\frac{\sqrt{2}e}{2} = \frac{1-e^2}{e^2}$ $\Rightarrow \sqrt{2}e^3 = 2 - 2e^2$ $\Rightarrow \sqrt{2}e^3 + 2e^2 - 2 = 0$. Solving for $e^2$,we find $a^2 = 50$ and $b^2 = 25$. For the hyperbola $\frac{x^2}{b^2} - \frac{y^2}{a^2b^2} = 1$,the eccentricity $e_1$ satisfies $e_1^2 = 1 + \frac{a^2b^2}{b^2} = 1 + a^2$. Since $a^2 = 50$,$e_1^2 = 1 + 50 = 51$.

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 $\frac{x^2}{b^2}-\frac{y^2}{a^2b^2}=1$ equals

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