Given, equation of parabola is
$$
\begin{aligned}
y^2 & =8 x \\
a & =2
\end{aligned}
$$
Let $(h, 4)$ be the coordinate of mid-point of chord. Then, equation of chord is
$$
y-4=m(x-h)
$$
If line (ii) passes through the point $P\left(2 t_1^2, 4 t_1\right)$ and $Q\left(2 t_2^2, 4 t_2\right)$ on parabola Eq. (i), then
$$
y\left(t_1+t_2\right)-2 x-4 t_{t_2}=0
$$
having slope
$$
m=\frac{2}{t_1+t_2}
$$
Since, $(h, 4)$ is the mid-point of $P Q$. Therefore,
$$
\begin{aligned}
& 2 \times 4=4\left(t_1+t_2\right) \\
\Rightarrow & t_1+t_2=2
\end{aligned}
$$
Hence, slope of chord $P Q$ is
$$
m=\frac{2}{2}=1 \quad \text { [using Eq. (iv)] }
$$