Conics and the Absolute Conic

Motivation

Point a camera at a round plate, a coin, or a wheel and the image is almost never a circle — it is an ellipse. Tilt the view and the ellipse opens up toward a parabola; tilt further and it becomes a hyperbola. All three are the same object seen through the same lens: a conic. Circles, ellipses, parabolas, hyperbolas, and even pairs of straight lines form one projective family, and in homogeneous coordinates they all share a single, compact description.

Conics matter here for three reasons. First, they are the natural quadratic object of projective geometry: where a point is a vector and a line is a vector, a conic is a symmetric matrix, and the incidence rule is a quadratic form. Second, a projective map sends a conic to a conic — the property survives every homography — so conics are a safe currency to carry between coordinate frames. Third, and most important: one particular conic, the absolute conic, lives out at infinity and is the only geometric object that stays exactly the same no matter where the camera is pointed or how it is moved. That makes it the cleanest possible stand-in for the intrinsic matrix K — the object a later part of this series actually solves for.

The core object

A point x̃ = (x, y, w) lies on a conic exactly when its coordinates satisfy a homogeneous quadratic equation. There are six degree-2 monomials in three variables — x², xy, y², xw, yw, w² — so the general conic is

which is precisely a single quadratic form in x̃:

C is symmetric — the off-diagonal entries are half the cross-term coefficients, exactly what makes x̃^TCx̃ reproduce each middle term twice. Because the equation is homogeneous, doubling every entry of C describes the same conic, so C is defined only up to scale: C ≅ λC. A symmetric 3×3 matrix has 6 independent entries, minus 1 for the overall scale, leaving 5 degrees of freedom — a count we will meet again, unmistakably, as the degrees of freedom of the calibration matrix.

Tangent lines. Differentiating the quadratic form at a point shows that the tangent line at a point x̃ on C is the matrix times the point:

This is the conic analogue of the point/line duality from the previous page: as x̃ ranges over the conic, Cx̃ sweeps out all of its tangent lines. It is worth checking that x̃ itself satisfies incidence: x̃^T(Cx̃) = x̃^TCx̃ = 0, so the point lies on its own tangent.

The dual conic. Dually, the set of tangent lines whose point of contact lies on the conic satisfies a quadratic condition of its own. Writing x̃ = C⁻¹l and substituting back gives lTC⁻¹l = 0. So a conic and its dual are related by inverting the matrix — the fact that will let us pass freely between "the points on ω" and "the lines tangent to ω".

Rank and degeneracy. The rank of C classifies which object it is:

The congruence rule

Suppose a projective map sends points as x̃′ = H x̃. We want the matrix C′ of the image conic — the set of image points x̃′ lying on the image of the original conic. The derivation is a single substitution. Invert the map, x̃ = H⁻¹x̃′, and substitute it into the original equation:

Reading off the matrix sandwiched in the middle gives the whole rule:

and, inverting both sides (using (H⁻^T)⁻¹ = H^T), the same rule for the dual conic:

One practical warning hides in the notation: the rule is H⁻T on the left and H⁻¹ on the right, not H C HT. Getting the inverses or their order wrong is the standard first mistake, and it announces itself immediately as a conic that does not line up with the transformed points.

What is preserved. Multiplying C on the left by H⁻T and on the right by H⁻¹ — both invertible — cannot change rank. So rank is a projective invariant: a nondegenerate conic stays nondegenerate, a pair of lines stays a pair of lines. What is not preserved is the metric "type": an ellipse can become a parabola or a hyperbola under a general homography, because those labels are really statements about where the conic meets the (itself transformed) line at infinity. Being nondegenerate is projective; being specifically elliptical is not.

The one conic that matters

Everything so far holds for any conic. Now pick one specific conic, chosen because it encodes the very notion of right angles and equal lengths in the world. Go up to P³ and look at the plane at infinity:

Inside that plane, define the absolute conic Ω∞ as the set of directions satisfying

Over the reals this equation has only the trivial solution at the origin, which is not a point of P³ — so Ω∞ has no real points at all. It is purely imaginary, and that is fine: projective geometry never required real coordinates, and Ω∞ is a perfectly well-defined conic that merely happens to be invisible. It is the imaginary circle at infinity, and every metric statement you could want — angles, ratios of lengths, orthogonality — is hidden inside it.

How a camera sees it. A camera is x̃ = K[R|t]X̃. For a point at infinity, W = 0, so the translation column multiplies zero and drops out entirely:

Work in the camera's own frame, so R = I and therefore X̃ = K⁻¹x̃. Substituting that direction relation into X² + Y² + Z² = 0 gives the equation of the image conic in the image plane:

so the image of the absolute conic is

with dual conic

The crucial observation. Look at what ω depends on: K, and nothing else. Not R, not t. Rotate the camera, walk across the room, turn it toward the ceiling — ω is identical, because it is the image of an object sitting at infinity, where the camera's position and orientation leave no fingerprint. It has 5 degrees of freedom, exactly the 5 that K has, which is the strongest possible hint that it is K in disguise.

It is a disguise you can undo. The relation ω = K⁻TK⁻¹ says that K⁻¹ is (up to rotation) a Cholesky factor of ω: factor the symmetric positive-definite ω as ω = LTL with L upper triangular, and K⁻¹ = L hands you K. So measuring ω and measuring K are the same task, and a later part of this series — calibration — does precisely that, by finding the conic to which images of special planar patterns must be tangent.

Interactive

On the left, moving a handle changes H and therefore C′ = H⁻TCH⁻¹; notice that however badly you stretch the quad, the image conic stays nondegenerate unless you deliberately flatten the quad into a line, and the source circle's rank-3 nature is always preserved. On the right, note that f_x ≠ f_y alone is enough to make ω non-circular — the ellipse's aspect ratio is the camera's pixel aspect, read straight off the conic — while a nonzero skew rotates the ellipse's axes off the image axes, and c_x, c_y shift its center to the principal point.

Recap