If the tangents drawn to the hyperbola $$4y^2 = x^2 + 1$$ intersect the co-ordinate axes at the distinct points A and B, then the locus of the mid point of AB is:

We start from the given hyperbola $$4y^2 = x^2 + 1$$ and write the equation of any line in slope-intercept form as $$y = mx + c$$, where $$m$$ is the slope and $$c$$ is the $$y$$-intercept.

For this line to touch the hyperbola, it must satisfy the condition of tangency. We substitute $$y = mx + c$$ into the hyperbola:

$$4(mx + c)^2 = x^2 + 1.$$

Expanding the left side we obtain

$$4\bigl(m^2x^2 + 2mcx + c^2\bigr) = x^2 + 1,$$ so $$4m^2x^2 + 8mcx + 4c^2 = x^2 + 1.$$ Bringing all terms to one side,

$$(4m^2 - 1)x^2 + 8mcx + (4c^2 - 1) = 0.$$ This is a quadratic in $$x$$. For the line to be a tangent, its discriminant must be zero. The discriminant formula is $$\Delta = B^2 - 4AC$$ for the quadratic $$Ax^2 + Bx + C = 0$$. Here we have

$$A = 4m^2 - 1,\quad B = 8mc,\quad C = 4c^2 - 1.$$

Setting the discriminant to zero gives

$$B^2 - 4AC = 0,$$ so $$(8mc)^2 - 4(4m^2 - 1)(4c^2 - 1) = 0.$$ Dividing by 4 to simplify,

$$16m^2c^2 - (4m^2 - 1)(4c^2 - 1) = 0.$$

Now we expand the product: $$(4m^2 - 1)(4c^2 - 1) = 16m^2c^2 - 4m^2 - 4c^2 + 1,$$ and substitute back:

$$16m^2c^2 - \bigl[16m^2c^2 - 4m^2 - 4c^2 + 1\bigr] = 0.$$ The terms $$16m^2c^2$$ cancel, leaving

$$4m^2 + 4c^2 - 1 = 0,$$ or after dividing by 4,

$$m^2 + c^2 = \dfrac14.$$

Thus any tangent to the hyperbola can be written as $$y = mx + c$$ with $$m^2 + c^2 = \dfrac14.$$

Such a tangent meets the coordinate axes at the points $$A\bigl(x_A, 0\bigr) \quad\text{and}\quad B\bigl(0, c\bigr).$$ Indeed, at $$y = 0$$ we get $$x_A = -\dfrac{c}{m},$$ and at $$x = 0$$ we get $$y = c.$$

Let $$M(h,k)$$ be the midpoint of the segment $$AB$$. Midpoint formulas give

$$h = \dfrac{x_A + 0}{2} = \dfrac{-c/m}{2} = -\dfrac{c}{2m},$$ $$k = \dfrac{0 + c}{2} = \dfrac{c}{2}.$$

From the second relation we have $$c = 2k.$$ Substituting this into the first gives

$$h = -\dfrac{2k}{2m} = -\dfrac{k}{m},$$ so $$m = -\dfrac{k}{h}.$$

We now insert $$m = -\dfrac{k}{h}$$ and $$c = 2k$$ into the tangency condition $$m^2 + c^2 = \dfrac14$$:

$$\left(-\dfrac{k}{h}\right)^2 + (2k)^2 = \dfrac14.$$ Thus $$\dfrac{k^2}{h^2} + 4k^2 = \dfrac14.$$

Multiplying through by $$h^2$$ eliminates the denominator:

$$k^2 + 4k^2h^2 = \dfrac{h^2}{4}.$$

Re-arranging all terms to one side,

$$k^2 + 4k^2h^2 - \dfrac{h^2}{4} = 0.$$

Multiplying by 4 to clear the remaining fraction,

$$4k^2 + 16k^2h^2 - h^2 = 0.$$

Writing the terms in a more standard order gives

$$-h^2 + 4k^2 + 16h^2k^2 = 0,$$ and multiplying by $$-1$$ yields

$$h^2 - 4k^2 - 16h^2k^2 = 0.$$

Since $$(h,k)$$ represents the general midpoint, we now replace $$h$$ by $$x$$ and $$k$$ by $$y$$ to obtain the required locus:

$$x^2 - 4y^2 - 16x^2y^2 = 0.$$

Comparing with the options, this matches Option D.

Hence, the correct answer is Option D.