If the tangent to the conic, $$y - 6 = x^2$$ at $$(2, 10)$$ touches the circle, $$x^2 + y^2 + 8x - 2y = k$$ (for some fixed $$k$$) at a point $$(\alpha, \beta)$$; then $$(\alpha, \beta)$$ is

We have the parabola $$y-6 = x^2,$$ that is $$y = x^2 + 6.$$

For a tangent at any point, we first need the slope. The derivative of $$y = x^2 + 6$$ with respect to $$x$$ is $$\dfrac{dy}{dx} = 2x.$$ At the given point $$(2,\,10)$$, the slope therefore equals $$2\cdot 2 = 4.$$

Now we write the equation of the tangent line by using the point-slope form $$y - y_1 = m(x - x_1).$$ Substituting the point $$(2,\,10)$$ and the slope $$m = 4,$$ we obtain

$$y - 10 = 4(x - 2).$$

Simplifying,

$$y - 10 = 4x - 8 \quad\Longrightarrow\quad y = 4x + 2.$$

So the required tangent line is $$y = 4x + 2.$$

This straight line is also tangent to the circle $$x^2 + y^2 + 8x - 2y = k,$$ where $$k$$ is some constant yet to be fixed. To impose the condition of tangency, we substitute the linear relation $$y = 4x + 2$$ into the circle and demand that the resulting quadratic in $$x$$ have exactly one real root (i.e., discriminant zero).

Substituting $$y = 4x + 2$$ gives

$$x^2 + (4x + 2)^2 + 8x - 2(4x + 2) = k.$$

We expand the squared term and the product:

$$(4x + 2)^2 = 16x^2 + 16x + 4,$$

$$-\,2(4x + 2) = -8x - 4.$$

Putting everything back into the circle equation, we have

$$x^2 + \bigl(16x^2 + 16x + 4\bigr) + 8x + \bigl(-8x - 4\bigr) = k.$$

Observe that $$+8x$$ and $$-8x$$ cancel, and likewise $$+4$$ and $$-4$$ cancel. We are left with

$$x^2 + 16x^2 + 16x = k.$$

Combining like terms,

$$17x^2 + 16x = k.$$

We now regard this as a quadratic equation in $$x$$:

$$17x^2 + 16x - k = 0.$$

For tangency, the discriminant $$D$$ of this quadratic must be zero. Using the discriminant formula $$D = b^2 - 4ac$$ for a quadratic $$ax^2 + bx + c = 0,$$ here $$a = 17,\; b = 16,\; c = -k.$$ Therefore

$$D = 16^2 - 4\cdot 17 \cdot (-k) = 256 + 68k.$$

Setting $$D = 0$$ for a repeated root, we get

$$256 + 68k = 0 \quad\Longrightarrow\quad k = -\dfrac{256}{68} = -\dfrac{64}{17}.$$

With the discriminant zero, the unique root $$x = \alpha$$ is obtained via the formula $$x = -\dfrac{b}{2a}:$$

$$\alpha = -\dfrac{16}{2\cdot 17} = -\dfrac{16}{34} = -\dfrac{8}{17}.$$

Now we find $$\beta$$ by feeding $$\alpha$$ into the tangent line $$y = 4x + 2$$:

$$\beta = 4\left(-\dfrac{8}{17}\right) + 2 = -\dfrac{32}{17} + 2 = -\dfrac{32}{17} + \dfrac{34}{17} = \dfrac{2}{17}.$$

Thus $$(\alpha,\beta) = \left(-\dfrac{8}{17},\,\dfrac{2}{17}\right).$$

Comparing with the given options, this matches Option B.

Hence, the correct answer is Option B.