Background on projective geometry and conic sections; you might want to skip to the actual question
A conic section is analytically described as the zero-locus of points $(x,y)$ in the affine plane of a quadratic polynomial in two variables $x$ and $y$, i.e. the solutions to the equation
$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$
Geometrically, these are the sections obtained by intersecting, in $3$-dimensional affine space, the plane $z=1$ with the cone consisting of the solutions $(x,y,z)$ to the homogeneous quadratic equation
$$Ax^2+Bxy+Cy^2+Dxz+Eyz+Fz^2=0$$
From a different point of view, the cone is composed not of points $(x,y,z)$ in $3$-dimensional space, but of lines through the origin $(0,0,0)$ in $3$-dimensional space. This is reflected by the homogeneity of the equation: if $(x,y,z)$ is a solution, then so is $(\lambda x,\lambda y,\lambda z)$ for any scalar $\lambda$.
Since any line is determined by two distinct points it passes through, lines through the origin $(0,0,0)$ can be represented by homogeneous coordinates $[x,y,z]$ where not all of $x,y,z$ are zero. Each triple $[x,y,z]$ represents the line going through $(0,0,0)$, but two triples represent the same line if they differ by a scalar: i.e. $[x,y,z]$ and $[u,v,w]$ represent the same line if there exists a (non-zero) $\lambda$ such that $x=\lambda u$, $y=\lambda v$, and $z=\lambda w$.
With homogeneous coordinates under our belt, we can define the projective plane to be the space of lines through the origin $(0,0,0)$ in $3$-dimensional affine space. Then we also have a cone in the projective plane given by the solutions $[x,y,z]$ to the homogeneous equation
$$Ax^2+Bxy+Cy^2+Dxz+Eyz+Fz^2=0$$
For the purposes of this question, each point in the projective plane, i.e. line through the origin $(0,0,0)$ is one of two types: the lines $[x,y,z]$ with $z\neq 0$ (which intersect the plane $z=1$ in $3$-dimensional space) and the lines $[x,y,0]$ (which are parallel to the plane $z=1$ in $3$-dimensional space). The former can be identified by homogeneity with the lines $[x,y,1]$, so that each line determines a point in the affine plane and conversely each point in the affine plane determines a line. Thus, the conic section we started with is the intersection of the cone in projective space with the affine plane $z\neq0$.
The latter points, of the form $[x,y,0]$ form the so-called line at infinity around our affine plane. The reason for this lies in how lines in the affine plane interact with points in projective space. A line in the affine plane consists of the solutions $(x,y)$ to an equation
$$ax+by=c$$
Homogenizing this equation gives a homogeneous equation
$$ax+by=cz$$
whose solutions $[x,y,z]$ determine a line in the projective plane. The line in the affine plane is the intersection of the line in the projective plane with the affine plane $z\neq0$, i.e., with the points of the projective plane with homogeneous coordinates $[x,y,1]$.
On the other hand, the intersection of the line in the projective plane with the line at infinity is the intersections with points of the projective plane with homogeneous coordinates $[x,y,0]$. There is only one such intersection point, and that is the point $[-b,a,0]$ on the line at infinity. Since two lines in the affine plane are parallel if and only if their ratios $a:b$ are the same, we see that all lines in the affine plane parallel to the line given by $ax+by=c$ intersect the line at infinity at $[-b,a,0]$. Hence, the points $[x,y,0]$ on the line at infinity are precisely the point at infinity in which parallel lines intersect!
Actual question
It is natural to ask where does the conic section
$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$
intersect the line at infinity? Plugging in $z=0$ in the homogenized equation, we arrive at he solutions $[x,y,0]$ of the homogenized quadratic
$$Ax^2+Bxy+Cy^2=0$$
Presuming we are not working in characteristic $2$, we can compute explicitly the solutions. In the case of real coefficients, the discriminant $B^2-4AC$ being positive, $0$, or negative corresponds to having two intersection points at infinity (hyperbola), one intersection point at infinity (parabola), or no intersection points at infinity (ellipse). Furthermore (still in the case of real coefficients), if the intersection points are actually $[1,\pm i,0]$ then the conic has to be circle (these are known as the circular points at infinity). In particular, this exhibits the fact that all conic sections are uniquely determined by $5$ points as a generalization of the fact that a circle is uniquely determined by $3$ points since a circle is a conic section that has to go through the circular points at infinity.
My question is whether we can give a definition of focus, directrix, and eccentricity using only the algbera-geometric structure of the projective plane, and not the usual metric definition. In particular, since the eccentricy governs (for real coefficients) whether the type of conic, it might be computable from the points at infinity.
Best Answer
In fact, a conic has 4 foci. We can see this if we look at a canonical ellipse,which is wide and short, and start making it smaller in the direction of the x-axis. The two foci get closer, until we reach a circle when they collapse to one point. Then, if we continue they start to have a different trajectory - up and down. This cannot be, and what really happens is that two foci escape to the complex part of the plane, while the other two arrive from it. A purely geometro algebraic definition of foci (well not exactly, because it depends on the projective coordinates you chose!) is the followin: take the points I=[1:i:0] and J=[1:-i:0], and a conic C. By bezout's theorem there are two lines through I that are tangent to C and likewise with J. The two pair of lines intersect at 4 points, which are precisely the foci.
Remark: The fact that there are two tangents from a point to C does not follow immediately from Bezout's, "tangency" is not a linear condition. However, it does follow if first we use projective duality with respect to C.