Equation of the tangent to $y^2=12\left(x+3\right)$ with slope $m$ is 
$y=m\left(x+3\right)+\frac{3}{m},$ which is same as $y=mx+c$
$\Rightarrow c=3m+\frac{3}{m}$
$\Rightarrow 3m^2-mc+3=0$ which is quadratic in $‘m’$

Since, it has both roots equal
$\Rightarrow c^2-36=0\Rightarrow c=\pm 6$
$\Rightarrow c=6 \left\{m>0\Rightarrow c>0\right\}$
$\Rightarrow m=1 \& c=6$
$\Rightarrow \frac{c+m}{c-m}=\frac{7}{5}=1.4$