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In my recent study of Conic Sections, I have come across several proofs (many of those comprise Dandelin spheres) showing that the cross-section of a cone indeed follows the focus-directrix property:

For a section of a cone, the distance from a fixed point (the focus) is proportional to the distance from a fixed line (the directrix), the constant of proportionality being called the eccentricity.

But, in order to truly establish equivalence between the two definitions of the conic sections, I am curious to know whether the converse of this is also true. That is, if the locus of some points follows the focus-directrix property, then is the curve ALWAYS the cross-section of a cone?

Arham Jain
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  • See if this can be of help: https://math.stackexchange.com/questions/3243438/locus-of-right-circular-cone-vertices-resulting-from-constant-conic-intersection/3244321#3244321 and this one too: https://math.stackexchange.com/questions/2368523/find-two-points-and-one-angle-so-that-cone-intersects-the-xy-plane-in-the-para/2369432#2369432 – Intelligenti pauca Jan 10 '24 at 08:24
  • Related: "What is the cone of the conic section?" In particular, see my answer, which covers ellipses but can be adapted to hyperbolas and parabolas. – Blue Jan 10 '24 at 09:04
  • @Blue But, doesn't this already begin with the assumption that the curve is a cross-section of a cone? – Arham Jain Jan 23 '24 at 05:36
  • @Intelligentipauca In this post: https://math.stackexchange.com/questions/3243438/locus-of-right-circular-cone-vertices-resulting-from-constant-conic-intersection/, the questioner says that every conic section is associated uniquely with an intersected right circular cone upto scale, rotation around axis of symmetry and translation. How did we come to this conclusion? – Arham Jain Jan 23 '24 at 05:37
  • @ArhamJain: "doesn't this already begin with the assumption that the curve is a cross-section of a cone?" Not exactly. One can show that the focus-directrix property defines a locus in the form of an ellipse/parabola/hyperbola. (Analytically, the result is immediate, since the property defines a second-degree polynomial $xy$-eqn.) Then, the answer I cited shows —if there's any cone involved— how that cone is situated. If we aren't assuming the cross-sectional nature of the curve, there's a little more work to do to make that connection, but the pieces are all there. (continued) – Blue Jan 23 '24 at 08:31
  • @ArhamJain: (continuing) That said, there's likely a more-direct geometric argument to be made. After all, both the focus and the directrix have geometric meaning in the section-of-a-cone configuration. (See, for instance, this ancient answer of mine.) It should be possible for a configuration (indeed, a whole family of them) to be re-constructed from the focus and directrix (and eccentricity) information. – Blue Jan 23 '24 at 08:37
  • That argument is not so clear to me (you should ask the author of the question), but it is irrelevant for my answer. – Intelligenti pauca Jan 23 '24 at 11:35
  • @Blue I mean, like, if we assume that the conics are "originally" defined as cross-sections of a cone (based on angle made by the cutting plane), then we can easily show using Dandelin spheres that all the points on these curves are distanced from the focus and directrix in a fixed ratio. How do we know that some other weird, random-looking curve (for which it is impossible to draw a cone) doesn't share this ability to have a fixed ratio with a point and line? – Arham Jain Jan 25 '24 at 17:23
  • @ArhamJain: "How do we know that some other weird, random-looking curve [...] doesn't share this ability [...]?" Analytical geometry guarantees this. The focus-directrix property $$\left(;\text{dist to $(f_x,f_y)$};\right) ;=; \text{ecc}\cdot\left(;\text{dist to $px+qy+r$};\right)$$ corresponds (for $e\neq0$) to the eqn $$(x-f_x)^2+(y-f_y)^2;=;e^2;\frac{(px+qy+r)^2}{p^2+q^2}$$ Transforming the eqn using rotations and translations gets it to the recognizable "standard form" of a (possibly-degenerate) ellipse/parabola/hyperbola; no "weird" curves appear. – Blue Jan 25 '24 at 18:26
  • @Blue But, still, that doesn't really establish equivalence...So far, I understand that:
    1. All sections of a cone follow focus-directrix property (i.e. have a fixed ratio of distances from focus and directrix). (Proven using Dandelin Spheres)
    2. All curves that follow focus-directrix property have an equation of the form: Ax^2+Bxy+Cy^2+Dx+Ey+F = 0 (Can be analytically derived)

    However, this doesn't imply that all equations of the form: Ax^2+Bxy+Cy^2+Dx+Ey+F = 0 are sections of a cone.

     – Arham Jain Jan 26 '24 at 03:48
  • @ArhamJain: (Comments aren't for discussion, so this is my last reply.) A degree-2 eqn always represents a familiar (possibly-degenerate) curve we call an ellipse, parabola, or hyperbola. See, eg, Wikipedia's "Matrix representation of conics" or many other resources. The answer of mine that I cited constructs a cone to accompany an ellipse (the para/hyperbola are similar); w/ a little extra work (eg, finding the cone's eqn), one can show that the ellipse is indeed a section of that cone, as desired. ... Good luck! – Blue Jan 26 '24 at 04:30

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