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About an hour ago, I discovered a beautiful property of a parabola. enter image description here If a circle intersects a parabola at four points, one of which is the vertex of the parabola, then the center of the triangle, whose vertices are the remaining three points, will belong to the axis of symmetry of the parabola

Is this feature known in advance or not, please mention any references that talk about the topic

I haven't tried to prove it yet, but I don't expect that to be difficult using analytic geometry, but evidence is welcome in the answers.

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  • $\begingroup$ I want to point out that I have been inspired by theorem 11.3.15 in Arseny Akopian's book $\endgroup$
    – زكريا حسناوي
    Commented Jul 9 at 19:16
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    $\begingroup$ This is related to a property of concyclic points on a parabola: if we place the parabola in a coordinate system that renders it as $y=kx^2$, then four points on the parabola are concyclic iff their $x$-coordinates sum to zero. $\endgroup$
    – Oscar Lanzi
    Commented Jul 9 at 22:24
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    $\begingroup$ It seems worth linking to Oscar Lanzi's own question on the subject: math.stackexchange.com/questions/4753014/…. $\endgroup$
    – Semiclassical
    Commented Jul 9 at 22:40

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We can prove a stronger claim, namely that this works for any axis-aligned ellipse which passes through the parabola vertex. First, we choose coordinates such that the parabola's vertex is $(0,0)$ and thus the parabola is of the form $y=ax^2$. Then the equation of a generic ellipse which passes through the origin is $$A(x-h)^2+B(y-k)^2=Ah^2+Bk^2$$ The intersection of this circle with the parabola is now found by solving

$$A(x-h)^2+B(ax^2-k)^2=Ah^2+Bk^2$$ for $x$. Eliminating the trivial solution $x=0$, we get the cubic $$a^2 B x^3+(A-2akB)x-2Ah=0$$ The lack of a quadratic term in this cubic means that the three roots $x_1,x_2,x_3$ sum to zero. Hence the $x$-coordinate of the center of mass of the points $\{(x_k,y_k)\}_{k=1}^3$ is $\frac13(x_1+x_2+x_3)=0$ which indeed falls on the axis of symmetry.

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  • $\begingroup$ Very interesting! Do you recommend any not too advanced textbook for conical surfaces? $\endgroup$
    – Francisco J. Maciel Henning
    Commented Jul 9 at 20:08
  • $\begingroup$ After some experimentation, this is as far as the statement seems to able to go: if we use a generic ellipse (not axis-aligned) then the resulting cubic after imposing $y=ax^2$ can and will have a quadratic term, which contradicts the claim. $\endgroup$
    – Semiclassical
    Commented Jul 9 at 20:09
  • $\begingroup$ @FranciscoJavierMacielHennin Can't say I do: the above is just based on algebra and not any particular textbook. $\endgroup$
    – Semiclassical
    Commented Jul 9 at 20:10
  • $\begingroup$ Oh, thank you anyway. $\endgroup$
    – Francisco J. Maciel Henning
    Commented Jul 9 at 20:14
  • $\begingroup$ As an aside, the case of an axis-aligned ellipse is not actually too much stronger than the original form: we can always rescale the axes to convert such an ellipse into a circle, in which case the original statement suffices. $\endgroup$
    – Semiclassical
    Commented Jul 9 at 20:15

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