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Topic:
construction of triangle of given perimeter, given point and angle
Replies:
16
Last Post:
Jun 10, 2011 12:58 PM




Re: parabola through 4 points
Posted:
Nov 12, 1997 6:14 PM


Peter has asked about finding the foci of a hyperbola knowing 5 points that lie on the hyperbola. I can not fully solve this, but I can find the center of the hyperbola. For the ellipse, I can find both the center and the foci. I think some extension about this construction should work for the hyperbola, but it is not yielding itself to me quickly. Ellipse: Construct any chord of the ellipse, construct a parallel chord. The line through the midpoints of two parallel chords passes through the center. (I do not know a proof of this without adding a point at infinity and looking at poles. I can give a reasonable argument using symmetry of the ellipse, but it lacks the proper rigor.) Repeat for a second chord, the interesection of the two lines is the center of the ellipse. This also works for finding the center of teh hyperbola.
Construct any circle centered at this point that intersects the ellipse in 4 points (use the five points, 3 can't be the vertices of the minor axis.) The perpendicular bisectors of the consecutive pairs of points are the major and minor axis. (In the hyperbola case, this will find the axis for the foci) Use compass and straightedge, knowing the relationships between the distance between the foci and the center and the semimajor and minor axis lengths fill find the foci length. This can be constructed easily.
Unfortunately, for the hyperbola where is the length of the minor axis?
Michael Keyton St. Mark's School of Texas
On Tue, 11 Nov 1997, Peter Ash wrote:
> Eileen M. Klimick Schoaff wrote: > > > > A parabola, on the other hand, is determined by any 4 of its points. > > John Conway> > > > > Am I missing something here? Don't 3 points determine a parabola if the axis > > of symmetry is either vertical or horizontal? But if we consider any axes, > > then there are an infinite number of parabolas passing through three points. > > The generic equation is ax^2 + bxy +cy^2 + dx + ey + f = 0. If Jon Roberts is > > considering parabolas of the form y = ax^2 + bx + c, then knowing 3 points > > gives you three equations with three unknowns which can easily be solved  > > unless there is no solution. > > > > In the April 1997 issue of the Mathematics Teacher, a colleague of mine, Dr. > > Ellie Johnson, wrote an article "A Look at Parabolas with a Graphing > > Calculator". In this article she using the calculator to generate many > > solutions to the generic equation. Of course this just shows that given three > > points and restricting yourself to a parabola of the form y = ax^2 + bx + c, > > you can derive the equation. That does not, of course, construct it. > > > > Does the fourth point determine whether the axis of symmetry is vertical, > > horizontal, or rotated? In ax^2 + bxy +cy^2 + dx + ey + f = 0, it looks like > > you need more than 4 points to determine a, b, c, d, e, f. > > > > Then again, I am only a math education person and do not have a PhD in math so > > I am probably far in the dark. > > > > Eileen Schoaff > > Buffalo State College > > Eileen, > > This might be easier to see if you look at the *definition* of the > parabola > and the other conics. A parabola is determined by its focus (a,b) and > its directrix. The directrix can be specified by two parameters as well, > say r and theta, where r is the distance of the line from the origin and > theta is a reference angle. If you know four points on the parabola, you > get four equations in four unknowns. > > Similarly, a circle is also a conic, but it is completely determined by > three parameters: its center (a,b) and radius r, so three points > determine > a circle. > > An ellipse requires five points, as it is determined by two foci (a,b) > and (c,d) and one additional parameter, say the sum of the distances > from > the foci to a point on the ellipse. Similarly, a hyperbola requires five > points as well. > > I have a question on a related topic which I'd like to put to the group. > Suppose I have 5 points on a branch of a hyperbola. Then we know that > there is a unique hyperbola through these five points. I would like to > know > a geometric method of finding the foci of this hyperbola. I seem to > remember reading the solution to this problem somewhere. (When I say > geometric, I > mean coordinate free, not "solve 5 equations in five unknowns". It > doesn't have to be a ruler and compass construction, although I seem to > recall that the solution that I saw was.) I think the solution may be > fairly complex, and I'd be happy just to have a reference. > > Peter Ash >



