Let the point of intersection of the line $y=a x$ with the chord be $(\alpha, a 0)$, then $\alpha=\frac{4+x_1}{2}$


$$
\Rightarrow x_1=2 \alpha-4 \text { and } a \alpha=\frac{4+y_1}{2} \Rightarrow y_1=2 a \alpha-4
$$
As $\left(x_I, y_1\right)$ lies on the parabola
$$
\begin{array}{rlrl}
& & (2 a \alpha-4)^2=4(2 \alpha-4) \\
\Rightarrow & 4 a^2 \alpha^2+16-16 a \alpha =8 \alpha-16 \\
\Rightarrow & 4 a^2 \alpha^2-16 a \alpha-8 \alpha+32 =0 \\
\Rightarrow & 4 a^2 \alpha^2-(16 a+8) \alpha+32 =0
\end{array}
$$
For two distinct chords $D>0$
$$
\begin{aligned}
& =(16 a+8)^2-4\left(4 a^2\right)(32)>0 \\
& =64(2 a+1)^2-512 a^2>0 \\
& =64\left(4 a^2+1+4 a\right)-512 a^2>0 \\
& =256 a^2+64+256 a-512 a^2>0 \\
& =-256 a^2+256 a+64>0 \\
& =256 a^2-256 a-64 < 0 \\
& =64\left(4 a^2-4 a-1\right) < 0 \\
& =\left(4 a^2-4 a-1\right) < 0 \\
& =4 a^2-4 a+1 < 2 \\
& =(2 a-1)^2 < 2=-\sqrt{2} \leq(2 a-1) \leq \sqrt{2} \\
& =1-\sqrt{2} \leq 2 a \leq 1+\sqrt{2} \\
& =\frac{1-\sqrt{2}}{2} \leq a \leq \frac{1+\sqrt{2}}{2}
\end{aligned}
$$