Let the chord be QR
$$Q(2t_1^2,4t_1)$$ and $$R(2t_2^2,4t_2)$$
When chord sub tends right angle at the origin, $$t_1t_2=-4$$
Also the equation of the chord is $$y-4t_1=\frac{4(t_2-t_1)}{2(t_2^2-t_1^2)}(x-2t_1^2)$$
When finding the common point of intersection, the equation is written in the form of $L_1+\lambda L_2$,
$$t_1^2y-2t_1x-4y+16t_1=0$$
What will the common point be in this case?
The concurrent, or the fixed point $P$, can be identified from the vertical chord whose endpoints are symmetric with respect to the $x$-axis. Knowing that $t_1t_2 = -4$, we have $t_1=-t_2=2$. The fixed point is then $(2\cdot 2^2,0)=(8,0)$.
It remains to show that $(8,0)$ satisfies the chord equation,
$$\frac{y-4t_1}{x-2t_1^2}=\frac{4(t_2-t_1)}{2(t_2^2-t_1^2)}$$
that is, $$\frac{0-4t_1}{8-2t_1^2}=\frac2{t_1+t_2}\implies \frac{-t_1}{4-t_1^2 } = \frac1{t_1+t_2}$$
which is true due to $t_1t_2 = -4$.