how to solve this question of parabola by only using euclidean geometry?

The original question was to prove that $$A_1A_2$$ and $$B_1B_2$$ intersect at directrix given $$A_1B_1$$ and $$A_2B_2$$ are focal chords of the parabola.

I solved it using the parametric form of parabola and got the answer. But then when I tried using geometry, I'm stuck, in the figure A1B1 is a focal chord, and A2 is a point on parabola, A2A1 intersects directrix at O. Then B2 is the intersection of A1F and B1O,B2' is foot of perp of B2 on directrix. The main target was to prove B2F=B2B'.how am I supposed to prove that the marked angles $$x$$ in the figure are equal? I'm stuck here!

• Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. Sep 25, 2020 at 8:13
• "I solved it using the parametric form of parabola and got the answer." You should include the work you did. It could provide insights that lead to a geometric solution without people duplicating your effort.
– Blue
Sep 25, 2020 at 8:39
• yes I will add it in short period of time. Sep 25, 2020 at 8:46

This theorem is valid not only for a parabola but also for an ellipse or hyperbola:

If $$A_1B_1$$ and $$A_2B_2$$ are focal chords of a conic section, then lines $$A_1A_2$$ and $$B_1B_2$$ intersect on the directrix referred to that focus.

Let $$F$$ be a focus of a generic conic section, $$A_1B_1$$ and $$A_2B_2$$ two chords through $$F$$, $$A_1'$$ and $$A_2'$$ the projections of $$A_1$$ and $$A_2$$ on the directrix, $$O$$ the intersection between line $$A_1A_2$$ and the directrix.

From the similitude of triangles $$OA_1A_1'$$ and $$OA_2A_2'$$ one gets: $$OA_1:OA_2=A_1A_1':A_2A_2'=FA_1:FA_2.$$ By the converse of exterior angle bisector theorem we then obtain that line $$FO$$ is the bisector of external angle $$\angle A_1FB_2$$ of triangle $$FA_1A_2$$.

The same reasoning can be repeated for triangle $$FB_1B_2$$: the line joining focus $$F$$ with the intersection of line $$B_1B_2$$ with the directrix is also the bisector of exterior angle $$\angle B_2FA_1$$. But those exterior angles coincide and have thus the same bisector: it follows that line $$B_1B_2$$ also intersects the directrix at point $$O$$. That completes the proof.

EDIT.

The argument given above can be expanded a bit, to prove some interesting properties of the tangents of a conic section.

If we let $$A_2\to A_1$$ in figure above, then lines $$A_1A_2$$ and $$B_1B_2$$ become the lines tangent to the conic at $$A_1$$ and $$B_1$$ respectively, endpoints of a focal chord (see figure below). The results proven above still hold, hence those tangents meet at a point $$O$$ on the directrix, and line $$FO$$ is the bisector of $$\angle A_1FB_1$$, i.e. $$FO\perp A_1B_1$$:

The line tangent to a conic section at a point $$P$$ meets the directrix relative to focus $$F$$ at a point $$O$$ such that $$\angle PFO=90°$$.

In the particular case of a parabola, it follows that triangles $$A_1FO$$ and $$A_1A_1'O$$ are congruent, implying that the line tangent at $$A_1$$ is the bisector of $$\angle A_1'A_1F$$, a first well known result:

The line tangent to a parabola at a point $$P$$ is the bisector of the angle formed by the line passing through $$P$$ and the focus, with the line through $$P$$ parallel to the axis.

Finally, from $$\angle A_1'A_1F+\angle B_1'B_1F=180°$$ one gets $$\angle OA_1F+\angle OB_1F=90°$$ and consequently $$\angle A_1OB_1=90°$$, which is another famous property:

The lines tangent to a parabola at the endpoints of a focal chord are perpendicular and meet on the directrix.

• Great proof! I wonder if one can also infer somehow that $\angle A_1 O B_1=90^{\circ}$ Sep 25, 2020 at 9:48
• indeed a great proof! Didn't think of it that way....@highgardner won't that be true only when A1O and B1O are tangents?? Sep 25, 2020 at 10:10
• @highgardener That angle becomes $90°$ only in the limit $A_1\to A_2$, when lines $A_1A_2$ and $B_1B_2$ become tangent to the parabola. And this proof is indeed the first step for proving that. Sep 25, 2020 at 10:54
• @KarthikNambiar Yes, my bad. Was misled by the figure. Considering the standard parabola $y^2-4ax=0$ and two points $A_1=(at_1^2, 2at_1)$ and $A_2=(at_2^2, 2at_2)$ on it, where $t_1>t_2>0$ (same side of axis). The slope of this chord is $\frac{2}{t_1+t_2}$. Consider the points $B_1, B_2$, their focal opposites. Their parameters are $\frac{-1}{t_1}$ and $\frac{-1}{t_2}$ resp. Thus the slope as before is: $\frac{-2}{\frac{1}{t_1}+\frac{1}{t_2}}$. Their product will be $-1$ only when $t_1=t_2$ Sep 25, 2020 at 11:04