How does geogebra quickly determine the analytical formula of a parabola passing through three point

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How does geogebra quickly determine the analytical formula of a parabola passing through three points?For example, a parabola passes through three points: A (-2,0), B (4,0), and C (0,2).

Best Answer
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or

Form polynomial with the 3 points (A,B,C). Then place 2 additional points (D,E) on the function. Then form the conic section with the 5 points (A,B,C,D,E).

Addition:

create the directrix and focus to the Conic . Use these two to create the parabola (yields shortened coefficients)

Finally, a user tool can be created (ggt)

Comments (9)

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You need to do exactly what you would do on paper, e.g. create a system of 3 equations in 3 variables using the general equation of the parabola (first of all figure out which one it is (x=... or y=...), depending on the position of the three points, then telling to the parabola to pass through A, B and C respectively.

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Thank you, this is general practice.

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2

or

Form polynomial with the 3 points (A,B,C). Then place 2 additional points (D,E) on the function. Then form the conic section with the 5 points (A,B,C,D,E).

Addition:

create the directrix and focus to the Conic . Use these two to create the parabola (yields shortened coefficients)

Finally, a user tool can be created (ggt)

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1

Thank you。

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Thank you

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2

What about

FitPoly({A,B,C}, 2)

?

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Thank you。

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Probably Lagrange method! The idea is to express the sought parabola as a sum of three different parabolas, each having zeros at two of the x-coordinates of the points and passing through the third point (different for each parabola).


For example, if the points are P1=(x1,y1), P2=(x2,y2) and P3=(x3,y3), then a parabola with zeros at x1 and x2 is simply given by:


y=(x-x1)(x-x2)

That parabola doesn't necessarily go through P3, but it is easy to force it through P3 by vertical scaling, and that will obviously preserve the zeros at x1 and x2. So we need to divide the parabola by "wrong", i.e. its value at x3, and multiply it by "right", i.e. y3:


y = y3 (x-x1)(x-x2)/((x3-x1)(x3-x2))

The other two parabolas are obtained by cycling the indices in this formula, then you add the three quadratic polynomials together and that's the parabola through three points. Each of those three parabolas "contributes" one of the three points through which the sum passes, and it doesn't affect the other two points because we constructed it so that it is equal to 0 at those other points' x coordinates.


Of course this method works for polynomials of degree n-1 through n points, for any n.

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