Here are my ideas. Let $P$ be the point of incidence. $P$ is the intersection of the line $y=b$ and the parabola $y^2=4ax$.
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The point $P$ will be $\left( \frac{b^2}{4a},b\right)$.
The equation of tangent $PT$ at $P$ is
$y \cdot b=2a\left( x+\frac{b^2}{4a}\right)$
The slope of this line is $\displaystyle \tan(\theta)=\frac{2a}{b}$(see the diagram)
Let the slope of the reflected ray be $m$.
$$ \begin{aligned} \therefore \ \tan \theta &= \Bigg| \frac{m-\frac{2a}{b}}{1+m\frac{2a}{b}}\Bigg| \\ \frac{2a}{b} &=\Bigg| \frac{m-\frac{2a}{b}}{1+m\frac{2a}{b}}\Bigg|\end{aligned} $$
From here, $\displaystyle m= \frac{4ab}{b^2-4a^2}$
Therefore, the equation of the reflected ray is
$\displaystyle y-b=\frac{4ab}{b^2-4a^2} \left( x-\frac{b^2}{4a}\right)$
This line satisfies the point $(a,0)$, therefore the reflected ray passes through the focus.