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I have a doubt on the classification of non-degenerate conics (parabola, ellipse, hyperbola) in projective geometry (my textbook is "Multiple View Geometry in Computer Vision", which, as the title implies, is not specifically targeted at projective geometry).

As far as I understand, we can classify conics from a projective and from an affine point of view:

From the projective point of view, all non-degenerate conics can be transformed to a circle, so there's no distinction between the three "classic" conic types.

From the affine point of view, the classification depends on the number of intersections (0: ellipse, 1: parabola, 2: hyperbola) between the conic and the line at infinity, which is fixed for affine transformations.

Summarizing, it seems that a generic projective transformation alters the number of intersections between a conic and the line at infinity, whereas an affinity does not change it.

My question is: why does the number of intersection between a conic and the line at infinity change for a projectivity? Shouldn't intersection be an invariant?

Thanks in advance!

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Summarizing, it seems that a generic projective transformation alters the number of intersections between a conic and the line at infinity, whereas an affinity does not change it.

I'd not put it like this, even though you are essentially right. Instead I'd say that a generic projective transformation alters the line at infinity.

So let's say you have a hyperbola, which intersects the line at infinity in two points. Then you apply a projective transformation to map that to the unit circle. The image of the line at infinity under that transformation will be a finite line which intersects the unit circle in two points, namely the images of the original points of intersection. The new line at infinity, after the transformation was applied, has no real points in common with the unit circle. (Now the points of intersection would be the complex ideal circle points $[1:\pm i:0]$, but you don't need this for your question I guess.)

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    $\begingroup$ When you say "the new line at infinity", do you mean the [0 0 1] vector? I (as an ignorant in the field) would say that the new line at infinity, in your example, is the image of the original line at infinity, i.e. the line intersecting the circle. $\endgroup$
    – rand
    Aug 5, 2014 at 13:13
  • $\begingroup$ In other words, does the classification depend on the intersection between the conic and [0 0 1], or on the current line at infinity, whatever it is? $\endgroup$
    – rand
    Aug 5, 2014 at 13:15
  • $\begingroup$ @SimonePalazzo: You're right, by “new line at infinity” I mean the line at infinity in the new image, as opposed to the “image of the line at infinity” which I used in the sentence before that. I'd say the “current line at infinity” is always the vector $[0:0:1]$, at least in the standard embedding. Other embeddings are possible, in which case you'd classify with respect to their line at infinity. $\endgroup$
    – MvG
    Aug 5, 2014 at 13:22
  • $\begingroup$ @SimonePalazzo: I tend to think as a projective transformation very much like a change of basis: it doesn't really change the relationship between things. In this sense, the image of the old line at infinity is closer related to the original line at infinity than the “new” line at infinity, i.e. the line which has the same coordinates after the transformation that the old line had before that transformation. But this is probably a very subjective view. $\endgroup$
    – MvG
    Aug 5, 2014 at 13:23
  • $\begingroup$ @SimonePalazzo: You classify with respect to the $[0:0:1]$ line if you want to know “what is this conic right now?” You classify with the image of that line under some transformation if you want to know “what was this conic before I applied the transformation?” The former will see a change in kind caused by a projective transformation, while the latter maintains the result. The former is like a transformation of your objects themselves, the latter more like a change in coordinate system. $\endgroup$
    – MvG
    Aug 5, 2014 at 13:27

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