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Topic: geometry help
Replies: 8   Last Post: Apr 26, 2014 11:56 AM

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Peter Scales

Posts: 192
From: Australia
Registered: 4/3/05
Re: geometry help
Posted: Apr 25, 2014 1:18 PM
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> Thank you, will most of this still work even if the
> curve isn't a parabola? It isn't symmetrical, the
> y-value increases slower than it decreases. It can be
> split into two half-parabolas 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 y-axis.
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)=(a-c)/b

There are some invariants in this rotation:
1. F=f
2. A+C=a+c
3. B^2-4AC=b^2-4ac

b^2-4ac is called the discriminant.
Its value determines the nature of the conic represented.
For a parabola b^2-4ac=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.

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