In the projective plane sense, a (projective) hyperbola is just an elliptic cone in 3D with the projection point as its cone peak $O$, and with the "level plane" (onto which to project) tilted so that it intersects with both halves of the cone, where the projection image on the level plane is a hyperbola curve in the usual sense, as the following image illustrates:

enter image description here

Now I want to show that both the asymptotes of the hyperbola, in the projective plane, intersect the hyperbola (the cone) exactly once, ie they are both tangent to the hyperbola (the cone) at infinity.

I have shown that the cone has exactly two infinity points, both of which has exactly a projective tangent line (a tangent plane to the cone). Where I have trouble is I don't know why these two tangent lines (planes) will exactly project onto the asymptotes on the level plane. Sure, one thing to note that is they both intersect the hyperbola (the cone) exactly once at infinity and nowhere on the "ordinary plane" (the level plane), just as both the asymptotes don't intersect the hyperbola curve on the level plane. The problem is that there are unfortunately many other lines other than the asymptotes that don't intersect the hyperbola on the level plane, and it seems they can't be distinguished in this way.

edit just worked out a perturbation technique that can tell apart asymptotes and other lines that don't intersect the hyperbola, observing that asymptotes, if just perturbed a little bit, will intersect the hyperbola curve, whereas other lines are "stable" under perturbation. So this problem is basically solved, but new approaches are welcome and appreciated too.


Here’s a different approach. Working in homogeneous coordinates, the equation of a conic can be written as $\mathbf x^TC\mathbf x=0$, where $C$ is a symmetric matrix. Using pole-polar properties it’s not too difficult to show that all tangent lines $\mathbf l$ to a conic $C$ satisfy the dual equation $\mathbf l^T C^{-1} \mathbf l=0$ and that the point of tangency is the pole of the line, $C^{-1} \mathbf l$.

Since intersection and tangency are affine properties, w.l.o.g. we need only examine the standard rectangular hyperbola $x^2-y^2=1$. Its matrix is $C=\operatorname{diag}(1,-1,-1)$, which is its own inverse. The asymptotes of this hyperbola are the lines $x\pm y=0$, with corresponding homogeneous vectors $\mathbf l=[1:1:0]$ and $\mathbf m=[1:-1:0]$. We then have for the points of tangency $C^{-1}\mathbf l=[1:-1:0]$ and $C^{-1}\mathbf m=[1:1:0]$, both of which are points at infinity, as posited.

Coming at it from a different direction, the tangents through a point $\mathbf p$ to a nondegenerate conic $C$ are given by the degenerate conic $\mathcal M_{\mathbf p}^T C^{-1} \mathcal M_{\mathbf p}$.† Here, $\mathcal M_{\mathbf p}$ is the “cross-product matrix” of $\mathbf p$, i.e., $\mathcal M_{\mathbf p}\mathbf x = \mathbf p\times \mathbf x$. The tangents to our standard hyperbola through its center—in this case the origin $\mathbf p=[0:0:1]$—are then $$\begin{bmatrix} 0&1&0 \\ -1&0&0 \\ 0&0&0 \end{bmatrix} \begin{bmatrix} 1&0&0 \\ 0&-1&0 \\ 0&0&-1 \end{bmatrix} \begin{bmatrix} 0&-1&0 \\ 1&0&0 \\ 0&0&0 \end{bmatrix} = \begin{bmatrix} -1&0&0 \\ 0&1&0 \\ 0&0&0 \end{bmatrix},$$ which corresponds to the Cartesian equation $x^2-y^2=0$. Factoring this into $(x+y)(x-y)=0$, we find that the hyperbola’s asymptotes are the tangents through its center. The points of tangency can then be found as above, or by homogenizing the two equations and solving the resulting system: $$\begin{align} x^2-y^2-w^2 &= 0 \\ x^2-y^2 &= 0 \end{align}$$ with solution $x=\pm y$, $w=0$, i.e., the points with homogeneous coordinates $[1:\pm1:0]$, as above.

Incidentally, all nondegenerate conics are projectively equivalent. Once you’ve chosen the line at infinity, they can then be distinguished by the number of intersections with this line: zero for ellipses, one for parabolas and two for hyperbolas. The center of a conic (if it has one) is the pole of the line at infinity, so from the above result we can find that the points of tangency of a hyperbola’s asymptotes are precisely its intersections with the line at infinity.

†: This is a consequence of the dual conic equation above. A line through two points is given by their cross product, so every point $\mathbf x$ on a tangent through the point $\mathbf p$ satisfies the equation $$(\mathbf p\times \mathbf x)^T C^{-1} (\mathbf p\times\mathbf x) = (\mathcal M_{\mathbf p}\mathbf x)^T C^{-1} (\mathcal M_{\mathbf p}\mathbf x) = \mathbf x(\mathcal M_{\mathbf p}^T C^{-1} \mathcal M_{\mathbf p})\mathbf x = 0.$$

  • $\begingroup$ Thanks. What does it mean when you use a homogeneous coordinate $\mathbf l$ to represent a line? Is it the normal vector of the plane that determines the line? $\endgroup$ – Vim Jan 14 '18 at 1:43
  • $\begingroup$ @Vim Exactly. You pack the coefficients of the equation $ax+by+c=0$ into a vector $\mathbf l=[a:b:c]$ so that the equation of the line becomes $\mathbf l^T\mathbf x=0$. If you model the projective plane as the $z=1$ plane in $\mathbb R^3$, this is the normal of the plane in $\mathbb R^3$ that corresponds to a line in $\mathbb P^2$. Note that these vectors are covariant, so they transform differently from the vectors that represent points. This representation lets you get a lot of mileage out of the point-line duality of the projective plane, too. $\endgroup$ – amd Jan 14 '18 at 2:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.