Given parabola is $y^2 = 4x$ ....(1)
Let $P \equiv\left(t^{2}_{1} , 2t_{1}\right)$ and $ Q \equiv \left(t^{2}_{2}, 2t_{2}\right) $
Slope of $ OP = \frac{2t_{1}}{t_{1}^{2}} = \frac{2}{t_{1}} $
and slope of $ OQ = \frac{2}{t_{2}} $
Since $ OP \bot OQ$,
$ \therefore \frac{4}{t_{1}t_{2}} = - 1 t_{1 }t_{2} = - 4 $ ....(2)
Let $R (h, k)$ be the middle point of $PQ$, then
$h = \frac{t_{1}^{2} + t^{2}_{2}}{ 2}$ ....(3)
and $ k = t_{1} +t_{2} $ ....(4)
From (4) , $ k^{2} = t^{2}_{1} + t^{2}_{2} + 2t_{1}t_{2} = 2h-8$
[ From (2) and (3)]
Hence locus of $R (h, k) $ is $y^2 = 2x - 8$.