Let $P\left(t_1^2, 2 t_1\right)$ and $Q\left(t_2^2, 2 t_2\right)$ are extremetre of chord



$$
\therefore \quad h=\frac{3 t_1^2+t_2^2}{4}, k=\frac{6 t_1+2 t_2}{4}
$$
Slope of $P Q=2$
$$
\begin{array}{rlrl}
\therefore & 2 & =\frac{2 t_1-2 t_2}{t_1^2-t_2^2} \\
t_1+t_2 & =1 \Rightarrow t_2=1-t_1
\end{array}
$$
Put the value of $t_2$ in $h$ and $k$, we get
$$
\begin{aligned}
& 4 h=3 t_1^2+\left(1-t_1\right)^2, 4 k=6 t_1+2-2 t_1 \\
& 4 h=4 t_1^2-2 t_1+1,4 k=4 t_1+2
\end{aligned}
$$
Eliminating $t_1$, we get
$$
\begin{aligned}
4 h & =4\left(\frac{2 k-1}{2}\right)^2-2\left(\frac{2 k-1}{2}\right)+1 \\
4 h & =4 k^2-6 k+3 \\
\Rightarrow\left(k-\frac{3}{4}\right)^2 & =4\left(h-\frac{3}{16}\right)
\end{aligned}
$$


$$
\therefore \operatorname{Vertex}\left(\frac{3}{16}, \frac{3}{4}\right)
$$