Let $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1,\ a > b$ be an ellipse, whose eccentricity is $\dfrac{1}{\sqrt{2}}$ and the length of the latus rectum is $\sqrt{14}$. Then the **square of the eccentricity** of $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ is :