Let P be the point of intersection of the common tangents to the parabola $$y^2 = 12x$$ and the hyperbola $$8x^2 - y^2 = 8$$. If S and S' denote the foci of the hyperbola where S lies on the positive x-axis then P divides SS' in a ratio:

We first translate the two curves into their standard forms. For the parabola we have $$y^{2}=12x,$$ which can be written as $$y^{2}=4ax$$ with $$4a=12$$ so $$a=3.$$ Thus the focus of the parabola is at $$(3,0).$$

The hyperbola is $$8x^{2}-y^{2}=8.$$ Dividing by 8 gives $$\frac{x^{2}}{1}-\frac{y^{2}}{8}=1,$$ so it is of the standard form $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$ with $$a^{2}=1,\quad b^{2}=8.$$ For a hyperbola the focal distance is obtained from $$c^{2}=a^{2}+b^{2},$$ hence $$c^{2}=1+8=9 \;\Longrightarrow\; c=3.$$ Therefore the two foci of the hyperbola are $$(3,0) \text{ and } (-3,0).$$ We are told that $$S=(3,0)$$ (the focus on the positive $$x$$-axis) and $$S' =(-3,0).$$

We now find the common tangents to both curves.

(i) Tangent to the parabola. For a parabola $$y^{2}=4ax,$$ the slope form of a tangent is $$y=mx+\frac{a}{m}.$$ Substituting $$a=3$$ gives the parabola’s tangent as $$y=mx+\frac{3}{m}\,. \quad -(1)$$

(ii) Tangent to the hyperbola. For a hyperbola $$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1,$$ a straight line $$y=mx+c$$ is tangent when $$c^{2}=a^{2}m^{2}-b^{2}.$$ Here $$a^{2}=1,\; b^{2}=8,$$ so the tangency condition is $$c^{2}=1\cdot m^{2}-8 = m^{2}-8. \quad -(2)$$

Because a common tangent must satisfy both (1) and (2), we must have the same intercept $$c$$ in each. From (1) we see that $$c=\frac{3}{m},$$ and substituting this into (2) gives $$\left(\frac{3}{m}\right)^{2}=m^{2}-8.$$ Carrying out the algebra step by step:

$$\frac{9}{m^{2}} = m^{2}-8,$$ $$9 = m^{4}-8m^{2},$$ $$m^{4}-8m^{2}-9=0.$$

Set $$t=m^{2}$$ to turn it into a quadratic: $$t^{2}-8t-9=0.$$ Using the quadratic formula, $$t=\frac{8\pm\sqrt{64+36}}{2}=\frac{8\pm\sqrt{100}}{2}=\frac{8\pm10}{2}.$$ Thus $$t=9 \quad\text{or}\quad t=-1.$$ Since $$t=m^{2}\ge 0,$$ we accept $$m^{2}=9,$$ giving the two real slopes $$m=3 \quad\text{or}\quad m=-3.$$

For each slope the corresponding $$c$$ is $$c=\frac{3}{m}.$$ Hence for $$m=3,\; c=1,$$ yielding the line $$y=3x+1;$$ for $$m=-3,\; c=-1,$$ yielding the line $$y=-3x-1.$$

These two lines are the common tangents to both curves. Their point of intersection $$P$$ is obtained by solving

$$\begin{cases} y=3x+1,\\[4pt] y=-3x-1. \end{cases}$$

Equating the two expressions for $$y$$ gives $$3x+1 = -3x-1 \;\Longrightarrow\; 6x = -2 \;\Longrightarrow\; x = -\dfrac{1}{3}.$$ Substituting back, $$y = 3\!\left(-\dfrac{1}{3}\right)+1 = -1+1 = 0.$$ Therefore $$P\left(-\dfrac{1}{3},\,0\right).$$

Because $$S(3,0),\;P(-\tfrac{1}{3},0)$$ and $$S'(-3,0)$$ all lie on the $$x$$-axis, we can compute the distances simply from their $$x$$-coordinates.

Distance $$SP = 3 -\!\left(-\dfrac{1}{3}\right) = 3+\dfrac{1}{3} = \dfrac{10}{3},$$ distance $$PS' = -\dfrac{1}{3} -(-3) = -\dfrac{1}{3}+3 = \dfrac{8}{3}.$$

Hence the ratio in which $$P$$ divides $$\overline{SS'}$$ is $$SP : PS' = \dfrac{10}{3} : \dfrac{8}{3} = 10 : 8 = 5 : 4.$$

Hence, the correct answer is Option A.