ellipse equation from 6 points? Watch

Hasufel
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Hey, i was just wondering this - (as i recently tried to answer a question on here about how to sketch a graph, and did my own investigation - as i`ve seen this question twice on here now - and i believe i`ve kicked it`s ass! (in a pdf)).

My question is (haven`t tried it yet):

can you get the cartesion equation of an ellipse if you know 2 points where it intesects both the x and y axes, and the line y=x?

As i`ve never tried this yet (may be about to), i don`t know if the answer is a "simple" yes!

Please enlighten me!

(points being: (BEING EDITED)

\displaystyle (1,0), (-1,0), (0,1), (0,-1), \left

(this is only ONE set of points for one ellipse generated by a version of the equation (implicit) that i was investigating, and together the both ellipses make up the "plot")
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Hasufel
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Bump! new info above.
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TeeEm
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(Original post by Hasufel)
Bump! new info above.
an ellipse is a quadratic whose general form is

ax2+by2+cxy+dx+ey+f=0

or dividing by a

x2+Py2+Qxy+Rx+Sy+T=0

So I will say to "pin" 5 constants you need 5 equations(points)
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tombayes
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(Original post by Hasufel)
Hey, i was just wondering this - (as i recently tried to answer a question on here about how to sketch a graph, and did my own investigation - as i`ve seen this question twice on here now - and i believe i`ve kicked it`s ass! (in a pdf)).

My question is (haven`t tried it yet):

can you get the cartesion equation of an ellipse if you know 2 points where it intesects both the x and y axes, and the line y=x?

As i`ve never tried this yet (may be about to), i don`t know if the answer is a "simple" yes!

Please enlighten me!
Well the ellipse does not necessarily intersect y=x, or the x axis or the y-axis. e.g. circle radius 1 centered (100,-100)

However, if by some chance it does then you can easily prove it is unique given 2 points where it intesects both the x and y axes, and the line y=x? (exercise...). Getting the Cartesian equation is messy though. Why do you want to use Cartesians though, they are super messy for rotated ellipses. Some parametric system should be fine.

If it is not rotated you only need 4 points (can be any 4) to determine what the equation is (easy).

For rotated you need five points (any I believe)

You will need to solve five simultaneous equations in five variables (in theory there is a quicker algorithm for this sort of thing though - i cannot remember its name):

Use the parametric form given in the link:
http://en.wikipedia.org/wiki/Ellipse...arametric_form

Also, don't do it by hand MATLAB is your friend!
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Hasufel
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gheesh! - i never thought of parametrics! - doi!

off to watch the (English) football! - (the Scottish stuff we get up here is like Aldi cola.....(need i say more?) - so, off to watch some GOOD football!

(they torture us up here by showing it late, forcing us to sit through the Scottish rubbish - which no one without a tatoo watches!)
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