Let $m$ is slope of common tangent and an equation of tangent to hyperbola $\frac{x^2}{1}-\frac{y^2}{8}=1$ is

$y=mx\pm \sqrt{1\cdot m^2-8} ...\left(1\right)$

equation of tangent to parabola $y^2=12x$ is

$y=mx+\frac{3}{m} ...\left(2\right)$

$\because \left(1\right) \& \left(2\right)$ is same

$\therefore \pm \sqrt{m^2-8}=\frac{3}{m}$

$\Rightarrow m^4-8m^2-9=0\Rightarrow m^2=-1$ (Reject)

$\Rightarrow m^4-8m^2-9=0\Rightarrow m^2=9\Rightarrow m=\pm 3$

$\because$ equation of common tangents is $y=3x+1\&y=-3x-1$

$\therefore$ point of intersection of common tangents is $R\left(-\frac{1}{3},0\right)$ & Foci of hyperbolas are $\left(\mathrm{3,0}\right) \& \left(-\mathrm{3,0}\right)$

$\Rightarrow \frac{-3\lambda +3}{\lambda +1}=-\frac{1}{3}\Rightarrow \lambda =\frac{5}{4}$

Hence, the required ratio is $5 : 4$.