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:



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