Normal at $P\left(t_1^2, 2 t_1\right)$ on the parabola $y^2=4 x \ldots(1)$
Meets it again at the point $Q\left(t_2^2, 2 t_2\right)$,
where $t_2=-t_1-\frac{2}{t_1} \ldots(2)$
If $P Q$ subtends a right angle at the vertex $(0,0)$ then $($ Slope of $O P)($ Slope of $O Q)=-1$
$\Rightarrow \frac{2 t_1}{t_1^2} \cdot \frac{2 t_2}{t_2^2}=-1 \Rightarrow t_2=-\frac{4}{t_1} \ldots(3)$
From (ii) and (iii)
$-t_1-\frac{2}{t_1}=-\frac{4}{t_1} \Rightarrow-t_1=-\frac{2}{t_1}$
$\Rightarrow t_1^2=2 \Rightarrow t_1= \pm \sqrt{2} ; \therefore t_2=\pm 2 \sqrt{2}$
$\therefore P$ and $Q$ are $(2 \pm 2 \sqrt{2})$ and $(8,84 \sqrt{2})$
$\therefore P Q=\sqrt{(8-2)^2+(\pm 4 \sqrt{2} \pm 2 \sqrt{2})^2}=\sqrt{36+72}$
$=\sqrt{108}=6 \sqrt{3 .}$