Show that lines created by certain points on the parabola intersect at the directrix?

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Edit: I got the answer by finding points of intersection between the line passing through B and the focus and the parabola, but it didn't seem like the best solution. Any other ideas?

The Segments AA' and BB' are focal chords of the parabola $x^2 = 2y$. The coordinates of A and B are $(4,8)$ and $(-2,2)$ respectively.

The question is then split into parts

a) find the equation of the line through A and B'

b) find the equation of the line through B and A'. But I think that those parts are necessary to arrive at the conclusion that

c) (show that) the two lines you found intersect at the directrix.

I found the focus of the parabola, $(0,1/2)$ and the directrix $y=-1/2$. How do I find the coordinates of B' and A' so that I can find the equations of the two lines in the question? Does it have to do with the fact that the segment from B to the focus has the same slope as the segment from the focus to B'?

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The coordinates of the focal point are correct, as well as the equation of the directrix. Now you can proceed purely analytically. Hints: first show:

The line through $A$ and $A'$ is $y=1\frac{7}{8}x+\frac{1}{2}$

The line through $B$ and $B'$ is $y=-\frac{3}{4}x+\frac{1}{2}$.

From this (by intersecting with the parabola, which gives you a quadratic equation) deduce that:

$A'=(-\frac{1}{4},\frac{1}{32})$ and $B'= (\frac{1}{2},\frac{1}{8})$.

Now show that:

The line through $A$ and $B'$ is $y=2\frac{1}{4}x-1$

The line through $A'$ and $B$ is $y=-1\frac{1}{8}x-\frac{1}{4}$.

These last lines intersect in $(\frac{2}{9},-\frac{1}{2})$, which is a point on the directrix.