> > > I am in a similar quandary with regards to the vertex of a parabola. ................... > John Conway wrote: > > > I'm afraid that 5 is still the number of points required to determine > > an ellipse, because ellipses are distinguished from hyperbolae merely > > by an inequality. (So, for instance, 3 is still the right number > > of points to determine a circle of radius larger than 1.) A parabola, > > on the other hand, is determined by any 4 of its points. John Conway > > > > This information above has lead me to attempt to construct the vertex of > a > parabola from its axes' intercepts. I can do this algebraically by > completing the square on
Let me just think about this more generally. Let's try to find the vertex from any four points. Then projectively, we're asking something about a conic through 4 given points and touching a given line (the line at infinity). We must find a projective definition of the vertex V; namely it's the point whose tangent is orthogonal to the line joining it to the contact-point of the conic with the line at infinity. [Projectively, two lines are orthogonal if they meet the line at infinity in points that are divided harmonically by the circular points at infinity.]
H'm - that seems quite hard. Another thing that worries me is that I thought the parabola through 4 points was unique, which would imply that it could be found by rational operations; but if the 4 points don't form a convex quadrilateral it must be imaginary, which implies to the contrary that some equation-solving is needed. Oh - of course it is - let the 4 points be the intersection of two parabolae!
It seems the equation has degree 2. This is because we're really trying to find which of the pencil of conics through 4 points touch a given line. But a coaxal system of circles is a generic pencil of conics, and for that the answer is that just 2 of them touch a given line.
So at least something is rational - namely the line joining the two vertices. It seems to me that the first task is to construct that, and then one could construct the vertices by intersecting it with some circle.
But obviously I'll have to think with pencil-and-paper or chalk-and-blackboard.