Locus of points with parabola

On parabola $$y^2=2px$$ at point $$A$$, a line $$L_1$$ passes that is tangent to the parabola and cuts $$x$$ axis at point $$B$$. From $$A$$, a line $$L_2$$ passes that is perpendicular to $$x$$ axis and cuts the parabola at point $$C$$.

A line $$L_3$$ passes point $$B$$ and is perpendicular to $$x$$ axis. A line $$L_4$$ passes point C and is parallel to $$x$$ axis.

Find the set of points $$F$$ where $$L_3$$ and $$L_4$$ intersect.

Book's answer is $$y^2=-2px$$

Graphic representation as I see it

My attempt to solve it

My intention was to present point $$A$$ as $$(T,K)$$ and eventually express $$L_3$$ and $$L_4$$ that way, find the expression of $$T$$ and $$K$$, and place them in the parabola's equation for the answer. Although I feel like the answer is there, I got lost. Would appreciate help

Let $$A(x_0,y_0)$$. The equation of $$L_2$$ is $$L_2:x=x_0$$. Since the parabola is symmetric about the $$x$$ axis, $$L_2$$ intersects the parabola again at $$A'(x_0,-y_0)$$. Thus, $$L_4:y=-y_0$$.
$$y^2=2px\implies 2yy'=2p\implies y'=p/y$$.
The equation of the tangent at $$A(x_0,y_0)$$ is given by $$L_1:\displaystyle\frac{y-y_0}{x-x_0}=\frac p{y_0}$$.
The $$x$$ intercept of the tangent $$L_1$$ is $$\displaystyle x_0-\frac{y_0^2}p=-x_0\ \because y_0^2=2px_0$$. Thus, $$L_3: x=-x_0$$.
The intersection of $$L_3,L_4$$ is the point $$(-x_0,-y_0)=(x,y)$$. Since $$(x_0,y_0)$$ lies on the parabola, the required locus is:
$$y^2=-2px$$