We have, equation of parabola
$$
y^2=8 x
$$

Differentiate w.r.t. ' $x$ '
$$
2 y \frac{d y}{d x}=8 \quad \Rightarrow \quad \frac{d y}{d x}=\frac{4}{y}
$$

At point $P(2,4)$ slope of tangent
$$
m_1=\frac{4}{4}=1
$$

So, equation of lst tangent
$$
\begin{array}{rlrl}
& & y-4 & =1(x-2) \\
\Rightarrow & y & =x+2
\end{array}
$$

At point $Q(18,-12)$ slope of tangent
$$
m_2=\frac{4}{-12}=-\frac{1}{3}
$$

So, equation of 2 nd tangent
$$
\begin{aligned}
y+12 & =-\frac{1}{3}(x-18) \\
\Rightarrow \quad 3 y+36 & =-x+18
\end{aligned}
$$


$$
\Rightarrow \quad 3 y+18=-x
$$

By substituting from Eqs. (i) and (ii),
$$
\begin{aligned}
& & 3 y+18 & =-(y-2) \\
\Rightarrow & & 4 y & =-16 \\
\Rightarrow & & y & =-4
\end{aligned}
$$

Put in Eq. (i),
$$
x=-4-2=-6
$$
$\therefore$ Tangents intersects at $(-6,-4)$
Required equation of tangent passes through $(-6,-4)$ and slope $\frac{1}{2}$.
$$
\begin{gathered}
y-(-4)=\frac{1}{2}[x-(-6)] \\
\Rightarrow \quad 2 y+8=x+6 \Rightarrow \quad x-2 y=2
\end{gathered}
$$