Let P(2, 4), Q (18,-12) be the points on the parabola ${\mathrm{y}}^2=8\mathrm{x}.$ The equation of straight line having slope $\frac{1}{2}$ and passing through the point of intersection of the tangents to the parabola drawn at the points P and Q is 

$$\begin{gathered} \mathrm{Given} \mathrm{parabola} {\mathrm{y}}^2=8\mathrm{x} \\ \Rightarrow \mathrm{a}=2 \\ \mathrm{P}=\left(2,4\right)=\left({a{t}_1}^2,2{\mathrm{at}}_1\right) \\ \Rightarrow 2{\mathrm{at}}_1=4\Rightarrow {\mathrm{t}}_1=1 \\ \mathrm{Q}=\left(18,-12\right)=\left({{\mathrm{at}}_2}^2,2{\mathrm{at}}_2\right) \\ \Rightarrow 2{\mathrm{at}}_2=-12\Rightarrow {\mathrm{t}}_2=-3 \\ Now point of intersection of \tan gents at A and B is \\ \mathrm{A}=\left({\mathrm{at}}_1{\mathrm{t}}_2, \mathrm{a}\left({\mathrm{t}}_1+{\mathrm{t}}_2\right)\right)=\left(-6,-4\right) \end{gathered}$$

Equation of required line is

  $$\begin{gathered} \mathrm{y}+4=\frac{1}{2}\left(x+6\right) \\ \Rightarrow \mathrm{x}-2 \mathrm{y}=2 \\ \Rightarrow x-2y-2=0 \end{gathered}$$