Let the foci and length of the latus rectum of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b b e( \pm 5,0)$ 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

<p>$$\begin{aligned} & \text { focii } \equiv( \pm 5,0) ; \frac{2 b^2}{a}=\sqrt{50} \\ & a=5 \quad b^2=\frac{5 \sqrt{2} a}{2} \\ & b^2=a^2\left(1-e^2\right)=\frac{5 \sqrt{2} a}{2} \end{aligned}$$</p> <p>$$\begin{aligned} & \Rightarrow \mathrm{a}\left(1-\mathrm{e}^2\right)=\frac{5 \sqrt{2}}{2} \\ & \Rightarrow \frac{5}{\mathrm{e}}\left(1-\mathrm{e}^2\right)=\frac{5 \sqrt{2}}{2} \\ & \Rightarrow \sqrt{2}-\sqrt{2} \mathrm{e}^2=\mathrm{e} \\ & \Rightarrow \sqrt{2} \mathrm{e}^2+\mathrm{e}-\sqrt{2}=0 \\ & \Rightarrow \sqrt{2} \mathrm{e}^2+2 \mathrm{e}-\mathrm{e}-\sqrt{2}=0 \\ & \Rightarrow \sqrt{2} \mathrm{e}(\mathrm{e}+\sqrt{2})-1(1+\sqrt{2})=0 \\ & \Rightarrow(\mathrm{e}+\sqrt{2})(\sqrt{2} \mathrm{e}-1)=0 \\ & \therefore \mathrm{e} \neq-\sqrt{2} ; \mathrm{e}=\frac{1}{\sqrt{2}} \end{aligned}$$</p> <p>$$\begin{aligned} & \frac{\mathrm{x}^2}{\mathrm{~b}^2}-\frac{\mathrm{y}^2}{\mathrm{a}^2 \mathrm{~b}^2}=1 \quad \mathrm{a}=5 \sqrt{2} \\ & b=5 \\ & a^2 b^2=b^2\left(e_1^2-1\right) \Rightarrow e_1^2=51 \\ & \end{aligned}$$</p>