Equation or semi-axes of the ellipse

[arkkis]

I have been struggling for a while now with this one.

Lets say I have two points on the arc of the ellipse. I know the y-coordinates and the difference of the x-coordinates. Is it possible two calculate the equation or the semiaxes of the ellipse where these points are located?

These are the information I can obtain. Actually the exact equation of the ellipse is not relevant. I should be able to calculate the ratio of the semi-axes.

EDIT: The ellipse is origo centered and the axes are parallel to x/y axes.

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Ellipse as a function of x:

x=sqrt[(a^2*b^2-a^2*y^2)/(b^2)]

So the difference of x-coordinates ie. deltax

deltax= x2-x1 =sqrt[(a^2*b^2-a^2*y2^2)/(b^2)] - sqrt[(a^2*b^2-a^2*y1^2)/(b^2)]

I guess somwhow I sould be able to eliminate a or b.
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I also tried in a way where I know also the x-coordinates by solving a equation pair where i end up with:

a^4*(y2^2-y1^2)+a^2*(x2^2*y1^2-y2^2*x1^2)=0

I can't get it solved when just the x2-x1 is known.
- - -
Thanks!

 

[mathman]

Two points are too few to determine much of anything about an ellipse.

 

[arkkis]

Two points are too few to determine much of anything about an ellipse.

But if I have two points ie (1;4,71) and (2:3,73)

If I put these to the formula which is obtained from pair equation of an ellipse (x1;y1 and x2;y2):

(x1^2)/(a^2)+(y1^2)/(b^2)=1
(x2^2)/(a^2)+(y2^2)/(b^2)=1

Calculate b^2 and put to the latter:

a^4*(y2^2-y1^2)+a^2*(x2^2*y1^2-y2^2*x1^2)=0

a^2*[a^2*(y2^2-y1^2)+(x2^2*y1^2-y2^2*x1^2)]=0

a^2=2 OR a^2*(y2^2-y1^2)+(x2^2*y1^2-y2^2*x1^2)=0

The latter

a^2=[(y2^2*x1^2)-(y1^2*x2^2)]/(y2^2-y1^2)

a^2=[(3,73^2*1^2)-(4,71^2*2^2)]/(3,73^2-4,71^2)

a^2=9,0463 --> a=-3,01 or 3,01

Hence:
b^2=(a^2*y1^2)/(a^2-x1^2) = b^2=24,943
b = -4,99 or 4,99

Am I making a major flaw?

 

[hotvette]

You are making an assumption that the ellipse is centered at the point (0,0) and the major / minor axes are not at angles to the x / y axes. If that's true, what you have done is valid. If the center is unknown, it takes 4 points to define an ellipse (assumming no axis angle), just as it takes 3 points to define a circle.

 

[arkkis]

Yes. The ellipse is origo centered and the axes are parallel to x/y axes. In addition i know the y-coordinates of two points and difference of the x-coordinates. That should be it.

Is it possible to define the semi-axes?

 

[hotvette]

Looks like you double posted:

https://www.physicsforums.com/showthread.php?t=443662

 

[mathman]

Yes. The ellipse is origo centered and the axes are parallel to x/y axes. In addition i know the y-coordinates of two points and difference of the x-coordinates. That should be it.

Is it possible to define the semi-axes?

The equation of the ellipse you are looking for is (x/a)² + (y/b)² = 1.
The semi-axes are a and b.

You can set up two equations in terms of (x1,y1) and (x2,y2). You are given x2-x1, y1, y2.
This gives you two equations in three unknowns a, b, x1. Can't be solved.