Let $$L$$ be a tangent line to the parabola $$y^2 = 4x - 20$$ at $$(6, 2)$$. If $$L$$ is also a tangent to the ellipse $$\frac{x^2}{2} + \frac{y^2}{b} = 1$$, then the value of $$b$$ is equal to:

The parabola is $$y^2 = 4x - 20 = 4(x - 5)$$. This is a parabola with vertex at $$(5, 0)$$ and parameter $$a = 1$$, opening rightward.

To find the tangent at $$(6, 2)$$, we differentiate $$y^2 = 4(x-5)$$ implicitly: $$2y \frac{dy}{dx} = 4$$, so $$\frac{dy}{dx} = \frac{2}{y} = \frac{2}{2} = 1$$ at the point $$(6, 2)$$.

The tangent line at $$(6, 2)$$ with slope $$1$$ is: $$y - 2 = 1 \cdot (x - 6)$$, giving $$y = x - 4$$.

For this line to be tangent to the ellipse $$\frac{x^2}{2} + \frac{y^2}{b} = 1$$, we substitute $$y = x - 4$$ into the ellipse equation: $$\frac{x^2}{2} + \frac{(x-4)^2}{b} = 1$$.

Multiplying through by $$2b$$: $$bx^2 + 2(x-4)^2 = 2b$$.

Expanding: $$bx^2 + 2x^2 - 16x + 32 = 2b$$, which gives $$(b+2)x^2 - 16x + (32 - 2b) = 0$$.

For tangency, the discriminant must equal zero: $$(-16)^2 - 4(b+2)(32-2b) = 0$$.

$$256 - 4(32b - 2b^2 + 64 - 4b) = 0$$, so $$256 - 4(-2b^2 + 28b + 64) = 0$$.

$$256 + 8b^2 - 112b - 256 = 0$$, giving $$8b^2 - 112b = 0$$, so $$8b(b - 14) = 0$$.

Since $$b \neq 0$$, we get $$b = 14$$.

The answer is $$14$$, which is Option B.