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I need to prove the following theorem:

Theorem Only one conic can be drawn having any two given parallel chords and its centre is at any point on the line bisecting the chords.

I tried to prove it using Analytic Geometry, but I failled. I supposed that the point $P_1(x_1,y_1)$ is in the first chord, that $(a,b)$ is a fixed direction and that the $x$-axis is the axis of the conic. In this way, the center is $C(c,0)$.

I found the point $V_1(x_1-\frac{a}{b}y_1,0)$ as the intersection between the chord and the axis. More than that, $Q_1(x_1-2\frac{a}{b}y_1,-y_1)$ is another intersection between the conic and the chord.

The same procedures can be done for the second chord, getting the points $P_2(x_2,y_2)$, $V_2(x_2-\frac{a}{b}y_2,0)$ (intersection between the chord and the axis) and $Q_2(x_2-2\frac{a}{b}y_2,-y_2)$ (intersection between the conic and the chord).

Thanks for any help.

Cleto

Cleto Pereira
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1 Answers1

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A well-known property of conics states that the midpoints of parallel chords lie on a line passing through the center.

Hence, given two parallel chords $AB$, $CD$ and the line passing through their midpoints, once you choose the center $O$ on that line the conic is fixed: if $A'$ is the reflection of $A$ about $O$, then there is a unique conic through $ABCDA'$ and that conic has $O$ as its center.

EDIT.

As noted by Blue in a comment, the theorem is false in the special case $AB\cong CD$, if center $O$ is chosen at the center of parallelogram $ABDC$: in that case any conic through $ABCD$ satisfies the given conditions. If center $O$ is not the center of parallelogram $ABDC$, the conic is unique but degenerates into a pair of parallel lines.

Intelligenti pauca
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    One quibble: It's possible that reflections of the given chord endpoints about chosen center $O$ don't give any new points. This happens when, and only when, $\square ABCD$ is a parallelogram and $O$ is the intersection of its diagonals. In such a case, the conic is not uniquely determined by $A$, $B$, $C$, $D$, and $O$. So, the theorem, as stated in the question, is false. It can be saved by, say, requiring the given parallel chords to be non-congruent. – Blue Sep 18 '20 at 07:58
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    @blue You are right: if $ABCD$ is a parallelogram and center $O$ is the intersection of its diagonals, then infinitely many conics are possible. But if $O$ is not the intersection of the diagonals, then a unique (degenerate) conic exists: a couple of parallel lines. – Intelligenti pauca Sep 18 '20 at 08:46
  • @Intelligentipauca Where is the point $E$ in your construction? $A, B, C$ and $D$ are given, $A'$ is the reflection, and $E$? Thanks! – Cleto Pereira Sep 22 '20 at 15:29
  • @CletoPereira It was a typo, sorry! I've just corrected it. – Intelligenti pauca Sep 22 '20 at 15:51
  • @Intelligentipauca and Blue. Thanks for your comments. I proved the result that midpoints of parallel chords lie on a line passing through the center. It was something new for me. But I can't use the five points theorem because it is just the main result I need to prove. I'm studying the article https://www.jstor.org/stable/3602341?seq=1 If you have another approach to my problem, please let me know! Thanks a lot! – Cleto Pereira Oct 03 '20 at 21:33
  • @CletoPereira Why don't you follow the proof given there? – Intelligenti pauca Oct 03 '20 at 21:41
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    @CletoPereira Moreover, the five-point theorem can be proved in many different ways (its first proof goes back to Euclid, IIRC). Why do you want to follow that paper? – Intelligenti pauca Oct 03 '20 at 21:45
  • @Intelligentipauca do you have any simple proof to indicate me? I really didn't know anything about the result and it was new for me. As I found this article, i followed it. But unfortunatelly it is very difficult to understanding it and the notation used is complicated. If you have another indication, please let me know! Thanks a lot! – Cleto Pereira Oct 03 '20 at 22:02
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    @CletoPereira If you want a geometric construction for the conic through five points, then you can read this answer of mine. The last part can be changed if needed, to construct the axes instead of computing their lengths. – Intelligenti pauca Oct 03 '20 at 22:56
  • See also this wikipedia page: https://en.wikipedia.org/wiki/Five_points_determine_a_conic – Intelligenti pauca Oct 03 '20 at 22:56