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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

> 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.

Date Subject Author
4/23/14 kelvin_m
4/23/14 Narasimham
4/24/14 Peter Scales
4/24/14 kelvin_m
4/24/14 kelvin_m
4/24/14 Narasimham
4/25/14 Peter Scales
4/26/14 Narasimham
4/26/14 Peter Scales