Equation or semi-axes of the ellipse

[arkkis]

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

Homework Statement

Let's 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?

Homework Equations

I should be able to calculate the ratio of the semi-axes.

The Attempt at a Solution

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.
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Thanks!

 

[Delphi51]

I don't think it can be done! Consider the simplest case where you have two points on a circle. There are an infinite number of different circles that fit the two points. Algebraically, there are 3 unknown parameters in the equation of a circle (x-a)² + (y-b)² = r², so you need three equations to find them, and therefore you need 3 points to substitute in. For the ellipse, you have 4 unknown parameters, so you need 4 points.

 

[hotvette]

What Delphi51 said is true in the case where the origin of the ellipse / circle is unknown. If it is (0,0), then 2 points are all that's needed for an ellipse and 1 point for a circle. You need to have the same number of points as unknowns.

 

[arkkis]

Yes, let's say the ellipse centered in origo. In that way I have only two unknonwns.

As I said only that matters are the length of the semi-axes.

If know the two points exactly it easy to calculate a and b. But I know y-coordinates and the difference of x-coordinates.

My mind says that a and b are able to be calculated since the I know the y-coordinates and their "projection" to x-axis but I know only the difference. Isn´t there only two number ie. x-coordinates where their difference matches the one which is known.

 

[Delphi51]

If you plug (x1,y1) and (x1+d,y2) into the equation for an ellipse centered on the origin and not rotated, you will still have unknown x1 in addition to a and b. Looks like 2 equations and 3 unknowns. But conceptually it looks like there is only one ellipse that will fit. Maybe you will see some simplification when you are looking at the two equations.

 

[hotvette]

I didn't realize that it was the difference in x values that was known. In that case, not enough information. Reason is the center/origin of the ellipse is fixed (at 0,0). Sometimes it helps to visualize. Imagine an ellipse centered about the origin and two points on the ellipse. Move the points to the left or right. The y's don't change and the distance between x's doesn't change. However, the points no longer lie on the original ellipse as they move, which means the values for a and b need to change as the points move.

Let's say the origin can move right/left along with the points. As long as you know the difference between the origin and each x-value, the problem can be solved. But, just knowing the difference in x-values still isn't enough for the same reason as above.

 

[Delphi51]

Nicely reasoned, hotvette! I see it clearly now.

 

[arkkis]

Yes, hotvettes example was clear. I figured that the known y-coordinates has their counter parts aas below.

y1=-y4 and y2=-y3

Isn´t it only one ellipse that fits these four points?

 

[D H]

Those new points do not add any information. You still do not have enough information.