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#1 2006-12-11 10:23:22

Littlebrainer
Guest

Equation of a parabola

Ok so grade Ten Academic math so don't go too heavy on the math terms. I have the points of a parabola. I am modelling the trajectory of a golfball and need to know the equation of the trajectory. Im guessing its a parabola because it evenly moves over but deosn't evenly rise and fall but that could be due to rounding I don't really know. I hope I don't sound too confusing. O and different ways to do it would be helpful. Thanks in advance

#2 2006-12-11 10:31:57

Littlebrainer
Guest

Re: Equation of a parabola

Oh and I can't you a graphing calculator so Quadratic regression is out of the question .... I was given 2 equations h=-4.9t^2+22.8t  and d=29.2t I have used these to find the points on my grid which is height against distance. Could I solve these almost like a liner system. Im just guessing. I know the ball travels about 137 metres and is in the air for about 4.6966 seconds. I just can't remeber way to find the equation of a parabola

#3 2006-12-11 11:03:18

Littlebrainer
Guest

Re: Equation of a parabola

Numbers would be helpful wouldn't they ... srry... anyways ... my points are (0,0) (14.6,10.175)  (29.2,17.1)  (43.8,23.175)  (58.4,26) (68,26.522)    (73,26.375)  (87.6,24.3)  (102.2,19.775)   (116.8,12.8)  (131.4,3.375)  (137.24, -1)..     Got the points by subbing in the time 0.0 , 0.5 , 1.0 , 1.5,   2.0  ,   2.5 ,  3.0 , 3.5 , 4.0 ,  4.5 ,  4.7  into the equations in my above post. Wait I forgot a point the time when ball is at higest point added it.

#4 2006-12-12 06:02:15

Patrick
Real Member
Registered: 2006-02-24
Posts: 1,005

Re: Equation of a parabola

Given three points (x[sub]1[/sub];y[sub]1[/sub]),(x[sub]2[/sub];y[sub]2[/sub]) and (x[sub]3[/sub];y[sub]3[/sub]), you can find the equation of the parabola using the following equation:

that's what my textbook says anyway tongue never used it..

Last edited by Patrick (2006-12-12 06:08:41)


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#5 2006-12-12 21:47:35

gnitsuk
Member
Registered: 2006-02-09
Posts: 121

Re: Equation of a parabola

In general, given n + 1 points there will be a unique nth degree polynomial which passes through those points. So for instance given 2 points (n = 1) where will be a unique 1th degree polynomial (straight line) passing theough those two points.

Specifically for the problem of three points.You know that the parabola must have the form

  y = ax^2 + bx + c

What you need to find out is the values of the "constants" a, b,
and c. Because you don't know them yet, you need to treat a, b, and c
as variables.

What DO you know? You know three points (x,y) on the parabola. Let's
say one of these points is (2,3). Then you can plug these values into
the general equation for the parabola:

  3 = a*2^2 + b*2 + c

  3 = 4a + 2b + c

We have made the "variables" x and y into constants. Now, if you do
the same thing with the other two points on the parabola, you'll have
three linear equations in three unknowns. Solving these will give you the proabola.

Instead of using known points like (2,3) etc. You could use unknown points (x1,y1),(x2,y2) and (x3,y3) and then go through the above process solving for a,b, and c. This would lead you to the big equation given by Patrick.

Last edited by gnitsuk (2006-12-12 21:48:57)

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