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Consider a parabola $y^2 = 4ax$ from which we draw two focal chords at $t_1, t_2$ respectively . lets say $P1P2$ and $C1C2$ where $C1,C2$ are the other end points intersecting the parabola again . Now my question is it is easy to show that the lines $P1C1$ and $P2C2$ meets at directrix of the parabola? I think this property holds even for ellipse/hyperbola too , but can the proof be such that it may give insight in case of those too ? If this property is valid . Though i have tried using coordinate finding the equations of those lines like one would be $y - 2at_1$ = $\frac{2}{t_1 + t_2} (x-at_1^2)$ but its getting lengthy when solving for intersection point . I hope there is a better geometric approach to it . My coordinate method was this : as above the line joining it is given , the other equation would be $y + \frac{2a}{t_1}$ = $\frac{-2t_1 t_2}{t_1 + t_2} (x-\frac{a}{t_1^2})$ , just using the propety of $t_1 t'_1 = -1$ , where t'_1 is the other point other than t_1 intersecting parabola. Equating the y value we get $x\frac{2(t_1 t_2 + 1)}{t_1 + t_2} = 2a \frac {t_2 - t_1 - t_2 - t^3_1 - t^2_1 t_2 + t^3_1}{(t_1+t_2 )t_1}$ gives x = -a

Orion_Pax
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    All parabolas are similar, in the sense that one can make one coincide with the other after a shift, rotation, and homothety. So it seems if you could prove your statement for the parabola $y=x^2$ it would suffice. That might be easier, at least less notation. – coffeemath May 02 '22 at 20:11
  • Yeah i understand that , may you check the above and tell what mistake i am making @coffeemath ? although i am looking for a geometric method in the problem – Orion_Pax May 02 '22 at 20:28
  • Nevermind i rectified it , may you give a geometric proof ? – Orion_Pax May 02 '22 at 20:48
  • What is a "focal chord" of a parabola? I know a chord is a line segment joining two points on the parabola, but I don't happen to know what the modifier "focal" is in the phrase "focal chord". – coffeemath May 02 '22 at 21:44
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    Focal chord means a chord which passes through focus of parabola @coffeemath – Orion_Pax May 02 '22 at 21:48
  • Here's a geometric solution (see the Edit): https://math.stackexchange.com/questions/3839729/how-to-solve-this-question-of-parabola-by-only-using-euclidean-geometry/3839766#3839766 Note that the question is about a parabola, but the answer is valid for any conic section. – Intelligenti pauca May 03 '22 at 08:02
  • Thats what i was looking for , thanks a lot @Intelligentipauca – Orion_Pax May 03 '22 at 14:32

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