A tangent to the hyperbola $$\frac{x^2}{4} - \frac{y^2}{2} = 1$$ meets x-axis at P and y-axis at Q. Lines PR and QR are drawn such that OPRQ is a rectangle (where O is the origin). Then R lies on :

We start with the hyperbola equation: $$\frac{x^2}{4} - \frac{y^2}{2} = 1$$. Here, $$a^2 = 4$$ and $$b^2 = 2$$, so $$a = 2$$ and $$b = \sqrt{2}$$. The equation of a tangent to this hyperbola with slope $$m$$ is given by: $$y = mx \pm \sqrt{a^2 m^2 - b^2} = mx \pm \sqrt{4m^2 - 2}$$.

Let the tangent be $$y = mx + c$$, where $$c = \pm \sqrt{4m^2 - 2}$$. This tangent meets the x-axis at point P and the y-axis at point Q.

When the tangent meets the x-axis, $$y = 0$$: $$0 = mx + c \implies x = -\frac{c}{m}$$ So, P has coordinates $$\left(-\frac{c}{m}, 0\right)$$.

When the tangent meets the y-axis, $$x = 0$$: $$y = m \cdot 0 + c = c$$ So, Q has coordinates $$(0, c)$$.

We are given that O is the origin $$(0, 0)$$, and OPRQ is a rectangle. The diagonals of a rectangle bisect each other. The diagonals are OR (from O to R) and PQ (from P to Q). Let R be $$(x, y)$$.

The midpoint of OR is $$\left(\frac{x}{2}, \frac{y}{2}\right)$$.

The midpoint of PQ, where P is $$\left(-\frac{c}{m}, 0\right)$$ and Q is $$(0, c)$$, is $$\left(\frac{-\frac{c}{m} + 0}{2}, \frac{0 + c}{2}\right) = \left(-\frac{c}{2m}, \frac{c}{2}\right)$$.

Setting the midpoints equal: $$\frac{x}{2} = -\frac{c}{2m} \quad \text{and} \quad \frac{y}{2} = \frac{c}{2}$$ From the second equation: $$\frac{y}{2} = \frac{c}{2} \implies y = c$$.

From the first equation: $$\frac{x}{2} = -\frac{c}{2m} \implies x = -\frac{c}{m}$$.

Substituting $$c = y$$ into $$x = -\frac{c}{m}$$ gives $$x = -\frac{y}{m}$$, so $$m = -\frac{y}{x}$$.

We also know $$c = \pm \sqrt{4m^2 - 2}$$. Substituting $$c = y$$ and $$m = -\frac{y}{x}$$: $$y = \pm \sqrt{4\left(-\frac{y}{x}\right)^2 - 2} = \pm \sqrt{4 \cdot \frac{y^2}{x^2} - 2} = \pm \sqrt{\frac{4y^2}{x^2} - 2}$$

Squaring both sides to eliminate the square root: $$y^2 = \left(\pm \sqrt{\frac{4y^2}{x^2} - 2}\right)^2 = \frac{4y^2}{x^2} - 2$$

Rearranging terms: $$y^2 - \frac{4y^2}{x^2} = -2$$ Multiplying both sides by $$x^2$$ to clear the denominator: $$y^2 x^2 - 4y^2 = -2x^2$$ Bringing all terms to one side: $$x^2 y^2 - 4y^2 + 2x^2 = 0$$

Assuming $$x \neq 0$$ and $$y \neq 0$$ (since R is not on the axes), divide both sides by $$x^2 y^2$$: $$\frac{x^2 y^2}{x^2 y^2} - \frac{4y^2}{x^2 y^2} + \frac{2x^2}{x^2 y^2} = 0 \implies 1 - \frac{4}{x^2} + \frac{2}{y^2} = 0$$ Rearranging: $$\frac{4}{x^2} - \frac{2}{y^2} = 1$$

Comparing with the options, this matches option D. Hence, the correct answer is Option D.