Let x^2/(a^2)+y^2/(b^2)=1, a > b be an ellipse, whose eccentricity is 1/(√ 2) and the length of the latusrectum is √ 14. Then the square of the…

Let $\displaystyle \frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1, \mathrm{a}>\mathrm{b}$ be an ellipse, whose eccentricity is $\displaystyle \frac{1}{\sqrt{2}}$ and the length of the latusrectum is $\displaystyle \sqrt{14}$. Then the square of the eccentricity of $\displaystyle \frac{x^2}{\mathrm{a}^2}-\frac{y^2}{\mathrm{~b}^2}=1$ is: