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

We have the ellipse $$\frac{x^2}{2} + \frac{y^2}{4} = 1$$. Here $$a^2 = 4$$ (along the $$y$$-axis since $$4 > 2$$), $$b^2 = 2$$, so $$c^2 = a^2 - b^2 = 2$$, giving $$c = \sqrt{2}$$.

The foci are at $$(0, \pm\sqrt{2})$$. The focus on the negative major axis is $$S = (0, -\sqrt{2})$$.

The tangent at a point $$(x_1, y_1)$$ on the ellipse is $$\frac{xx_1}{2} + \frac{yy_1}{4} = 1$$.

The tangents at P and Q both pass through $$R(\sqrt{2}, 2\sqrt{2} - 2)$$. So the chord of contact from $$R$$ to the ellipse is:

$$\frac{x \cdot \sqrt{2}}{2} + \frac{y(2\sqrt{2}-2)}{4} = 1$$

$$\frac{\sqrt{2}x}{2} + \frac{(2\sqrt{2}-2)y}{4} = 1$$

$$\frac{x}{\sqrt{2}} + \frac{(\sqrt{2}-1)y}{2} = 1$$

This is the equation of the chord PQ. Now we need to find P and Q (the points on the ellipse where this chord intersects).

From the chord equation: $$x = \sqrt{2}\left(1 - \frac{(\sqrt{2}-1)y}{2}\right) = \sqrt{2} - \frac{(\sqrt{2}-1)\sqrt{2}y}{2} = \sqrt{2} - \frac{(2 - \sqrt{2})y}{2}$$

Substituting into the ellipse equation $$\frac{x^2}{2} + \frac{y^2}{4} = 1$$:

$$\frac{\left[\sqrt{2} - \frac{(2-\sqrt{2})y}{2}\right]^2}{2} + \frac{y^2}{4} = 1$$

Let $$k = \frac{2-\sqrt{2}}{2}$$. Then $$x = \sqrt{2} - ky$$.

$$\frac{2 - 2\sqrt{2}ky + k^2y^2}{2} + \frac{y^2}{4} = 1$$

$$1 - \sqrt{2}ky + \frac{k^2y^2}{2} + \frac{y^2}{4} = 1$$

$$-\sqrt{2}ky + y^2\left(\frac{k^2}{2} + \frac{1}{4}\right) = 0$$

$$y\left[-\sqrt{2}k + y\left(\frac{k^2}{2} + \frac{1}{4}\right)\right] = 0$$

So $$y = 0$$ or $$y = \frac{\sqrt{2}k}{\frac{k^2}{2} + \frac{1}{4}}$$.

With $$k = \frac{2-\sqrt{2}}{2}$$: $$k^2 = \frac{(2-\sqrt{2})^2}{4} = \frac{6-4\sqrt{2}}{4}$$.

$$\frac{k^2}{2} + \frac{1}{4} = \frac{6-4\sqrt{2}}{8} + \frac{2}{8} = \frac{8-4\sqrt{2}}{8} = \frac{2-\sqrt{2}}{2}$$

$$y_2 = \frac{\sqrt{2} \cdot \frac{2-\sqrt{2}}{2}}{\frac{2-\sqrt{2}}{2}} = \sqrt{2}$$

So the two points have $$y_P = 0$$ and $$y_Q = \sqrt{2}$$.

For $$y = 0$$: $$x = \sqrt{2}$$. So $$P = (\sqrt{2}, 0)$$. Check: $$\frac{2}{2} + 0 = 1$$. ✓

For $$y = \sqrt{2}$$: $$x = \sqrt{2} - k\sqrt{2} = \sqrt{2}(1-k) = \sqrt{2}\left(1 - \frac{2-\sqrt{2}}{2}\right) = \sqrt{2} \cdot \frac{\sqrt{2}}{2} = 1$$. So $$Q = (1, \sqrt{2})$$. Check: $$\frac{1}{2} + \frac{2}{4} = 1$$. ✓

Now with $$S = (0, -\sqrt{2})$$:

$$SP^2 = (\sqrt{2}-0)^2 + (0-(-\sqrt{2}))^2 = 2 + 2 = 4$$

$$SQ^2 = (1-0)^2 + (\sqrt{2}-(-\sqrt{2}))^2 = 1 + (2\sqrt{2})^2 = 1 + 8 = 9$$

$$SP^2 + SQ^2 = 4 + 9 = 13$$

Hence, the correct answer is 13.