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Re: geometry help
Posted:
Apr 25, 2014 1:18 PM


> Thank you, will most of this still work even if the > curve isn't a parabola? It isn't symmetrical, the > yvalue increases slower than it decreases. It can be > split into two halfparabolas though.
Dear kelvin m, Sorry. In my first reply I did not consider the lack of symmetry you mention, taking only the general nature of the curve into account. It could still be a parabola, as the lack of symmetry is consistent with a parabola with the axis inclined to the yaxis. The genaral 2nd degree equation in yox is a.x^2 + b.xy + c.y^2 + d.x + e.y + f = 0 Rotate it thru angle theta giving a new equation in YoX A.X^2 + B.XY + C.Y^2 + D.X + E.Y + F = 0 Choose theta so that B=0 by cot(2.theta)=(ac)/b
There are some invariants in this rotation: 1. F=f 2. A+C=a+c 3. B^24AC=b^24ac
b^24ac is called the discriminant. Its value determines the nature of the conic represented. For a parabola b^24ac=0 so B^2=4AC If B=0 then either A or C = 0
So the equation becomes either C.Y^2 + D.X + E.Y + F = 0 with axis // oX, or A.X^2 + D.X + E.Y + F = 0 with axis // oY
In either case 4 points are necessary to determine the equation. But there are restrictions: 1. No 3 can be in a straight line 2. No one can be within the triangle formed by the other three. You can see this by looking again at the yox plane. Condition 1. implies that 3 points form a triangle. Through those 3 points it is possible to draw 3 different parabolas. Condition 2. implies that the 4th point will determine which of these is the required one, provided the 4th point is not inside the triangle.
Regards, Peter Scales.



