Let the tangents at the points $\mathrm{P}$ and $\mathrm{Q}$ on the ellipse $\frac{x^{2}}{2}+\frac{y^{2}}{4}=1$ meet at the point $R(\sqrt{2}, 2 \sqrt{2}-2)$. If $\mathrm{S}$ is the focus of the ellipse on its negative major axis, then $\mathrm{SP}^{2}+\mathrm{SQ}^{2}$ is equal to ___________.

<p>$E \equiv {{{x^2}} \over 2} + {{{y^2}} \over 4} = 1$</p> <p>$$\eqalign{ & T \equiv y = mx\, \pm \,\sqrt {2{m^2} + 4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \left( {\sqrt 2 ,2\sqrt 2 - 2} \right) \cr} $$</p> <p>$$ \Rightarrow \left( {2\sqrt 2 - 2 - m\sqrt 2 } \right) = \pm \,\sqrt {2{m^2} + 4} $$</p> <p>$$ \Rightarrow 2{m^2} - 2m\sqrt 2 \left( {2\sqrt {2 - 2} } \right) + 4(3 - 2\sqrt 2 ) = 2{m^2} + 4$$</p> <p>$\Rightarrow - 2\sqrt 2 m(2\sqrt 2 - 2) = 4 - 12 + 8\sqrt 2$</p> <p>$\Rightarrow - 4\sqrt 2 m(\sqrt 2 - 1) = 8(\sqrt 2 - 1)$</p> <p>$\Rightarrow m = - \sqrt 2$ and $m \to \infty$</p> <p>$\therefore$ Tangents are $x = \sqrt 2$ and $y = - \sqrt 2 x + \sqrt 8$</p> <p>$\therefore$ $P(\sqrt 2 ,0)$ and $Q(1,\sqrt 2 )$</p> <p>and $S = (0, - \sqrt 2 )$</p> <p>$\therefore$ ${(PS)^2} + {(QS)^2} = 4 + 9 = 13$</p>