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(A) The given ellipse is $\frac{x^{2}}{2} + \frac{y^{2}}{4} = 1$. Here,$a^{2} = 2$ and $b^{2} = 4$. Since $b^{2} > a^{2}$,the major axis is along the

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$13$

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(A) The given ellipse is $\frac{x^{2}}{2} + \frac{y^{2}}{4} = 1$. Here,$a^{2} = 2$ and $b^{2} = 4$. Since $b^{2} > a^{2}$,the major axis is along the $y$-axis.Eccentricity $e = \sqrt{1 - \frac{a^{2}}{b^{2}}} = \sqrt{1 - \frac{2}{4}} = \frac{1}{\sqrt{2}}$.The foci are $(0, \pm be) = (0, \pm 2 	imes \frac{1}{\sqrt{2}}) = (0, \pm \sqrt{2})$.Given $S$ is the focus on the negative major axis,$S = (0, -\sqrt{2})$.The chord of contact of tangents from $R(\sqrt{2}, 2\sqrt{2}-2)$ to the ellipse is given by $\frac{x x_{1}}{a^{2}} + \frac{y y_{1}}{b^{2}} = 1$.Substituting $x_{1} = \sqrt{2}, y_{1} = 2\sqrt{2}-2, a^{2} = 2, b^{2} = 4$,we get $\frac{x \sqrt{2}}{2} + \frac{y(2\sqrt{2}-2)}{4} = 1$,which simplifies to $\frac{x}{\sqrt{2}} + \frac{y(\sqrt{2}-1)}{2} = 1$.Solving this with the ellipse equation,we find the points of contact $P$ and $Q$ as $(1, \sqrt{2})$ and $(\sqrt{2}, 0)$.Now,$SP^{2} = (1-0)^{2} + (\sqrt{2} - (-\sqrt{2}))^{2} = 1^{2} + (2\sqrt{2})^{2} = 1 + 8 = 9$.$SQ^{2} = (\sqrt{2}-0)^{2} + (0 - (-\sqrt{2}))^{2} = 2 + 2 = 4$.Therefore,$SP^{2} + SQ^{2} = 9 + 4 = 13$.

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

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