I don't know much about this, but that never stopped me before.
From things I've heard, I get the feeling that the best way to think of conic sections in the plane is as if they belonged to the projective plane P^2.
P^2 is a surface that is not easy to visualize, but one way to describe it is: P^2 = the ordinary plane R^2 plus the "circle at infinity". The circle at infinity (aka P^1) consists of one point for each family of parallel lines in R^2. This is equivalent to the set of antipodal pairs {exp(it, -exp(it)} for 0 <= t <= π, except we have to identify the pairs for t = 0 and t = π; hence a "circle". Then ellipses are hyperbolas and hyperbolas are ellipses (when you include the points at infinity that belong in the locus). For example, the hyperbola (x/4)^2 - (y/3)^2 = 1 includes which points at infinity? Plug in the equation of a line through 0: y = m x to get ((1/4)^2 - (m/3)^2) = 1 / x^2 Now let x —> oo to determine the values m = ±3/4, which tell us just which points on the circle at infinity we should add to our hyperbola to "complete" it. Of course, these are the same points in the circle at infinity determined by the asymptotes. Now consider conic sections, what you get in 3-space by cutting a cone with a plane. The projective wisdom seems to apply here, too. By adding new points to R^3 we get projective 3-space P^3. The new points this time will be one point for each complete family of mutually parallel planes in R^3, or in other words one point for each antipodal pair of points on the unit sphere S^2 in R^3. Now instead of a *circle* at infinity, we have to add a copy of P^2 at infinity. Let's see what we get when we try to complete a quadratic surface Q in R^3. For any x in Q, let T(x) denote the tangent plane to Q at the point x. Since T(x) is a plane in R^3 it belongs to just one point of the P^2 at infinity. What happens to T(x), as a point of P^2, as |x| —> oo ? If Q is a hyperboloid of one sheet, it's easy to see from a picture that the directions of the tangent planes T(x) approach two simple closed curves in the sphere S^2, each curve the antipodal image of the other. In P^2 itself this is just one simple closed curve at infinity to add to Q. Thus Q gets "joined to itself at infinity". But wait - things are seldom what they seem (cream is milk and milk is cream). In order to add this simple closed curve at infinity, you have to *twist* the hyperboloid, because look at which points have parallel tangent planes! Wait, that's no ellipsoid at all, that's topologically ... ... a torus! And while we're at it, what is a cone once it's been "completed" this way? It's a ... a ... I have no idea what it's called. Its a torus with a narrow waist (circumference zero). And going back to the hyperbola in the plane, we should have "twisted" that, too, when completing it. The strangeness we are detecting all (OK, mostly) goes away in the setting that algebraic geometers have decided is most appropriate to look at this kind of thing: Complex projective space and homogeneous polynomial equations. Say P(X,Y,Z) is a complex quadric polynomial in complex variables X, Y, Z such that every monomial is the same degree. E.g. P(X,Y,Z) = X^2 + YZ. Then whenever X, Y, Z make the equation P(X,Y,Z) = 0 true, then so do cX, cY, cZ for any nonzero complex number c, since P(cX,cY,cZ) = c^2 P(X,Y,Z) = 0. Ignoring the triple X = Y = Z = 0, we may as well take equivalence classes in the space C^3 - {(0,0,0)} via (X,Y,Z) ~ c * (X,Y,Z). When we do this, the set of equivalence classes is called "the complex projective plane and denoted CP^2. And now I have to eat dinner. —Dan