Family of parabolas through three points?
I saw this question about a week ago. We were asked to derive the equation for the parabola through these three points in the Cartesian plane:
(1, 11), (0, 6), (2, 18)
My problem with the question is that there is more than one parabola fitting those three points. In fact, there are infinitely many. Everyone else was assuming a vertical axis, which is probably what the asker intended, but that condition was not stated. So what about the others?
Using a single variable parameter, derive an equation representing the family of parabolas passing through the three given points.
Please read it carefully. The objective is not a single parabola, but rather a family of parabolas. I asked this same question two days ago, but was compelled to delete it because nobody was addressing the question.
- RaffaeleLv 79 years ago
You need 5 points, in general, to determinate one and only one parabola
These 5 points cannot be anywhere in the plane
In the following drawing
A,B and C are the given points (1, 11), (0, 6), (2, 18)
A generic conic section in a xy-plane has equation
ax^2 + bxy + cy^2 + dx + fy + g = 0 (*)
(dividing by one of the parameters which is not zero, actually they reduce to five)
1st condition is that (*) is not the equation of a degenerate conic section
like x^2 - y^2 = 0 which represents 2 straight lines
it happens when the plane passes through cone simmetry axis
(*) is not degenerate iff a determinant formed with its coefficients is not zero
look here for details
2nd condition is that discriminant of (*) is zero
b^2 - 4ac = 0
as you can see it's very hard, with 3 points only, found the double parameters family of parabolas
If we knew that one of them is the vertex, it should be easier and the parabolas will deend by only one parameter
thank you for the interesting question
many teachers 'forget' to specify what is the symmetry axis
leaving it as an 'implicit' conditionSource(s): a great free software for any platform http://www.geogebra.org/cms/