Trammel method for constructing an ellipse
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Hi all, I read about several methods for constructing an ellipse with straightedge and compass, and one seemed very peculiar to me. It was called the trammel method...and it's the only one that I can't prove. It seemed so simple, but I just can't prove it. ;P If anyone could find a proof for me or point me in the right direction? And also if this is in the wrong section please kindly inform me which section it should be in. Thanks. :) :)
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Hi all, I read about several methods for constructing an ellipse with straightedge and compass, and one seemed very peculiar to me. It was called the trammel method...and it's the only one that I can't prove. It seemed so simple, but I just can't prove it. ;P If anyone could find a proof for me or point me in the right direction? And also if this is in the wrong section please kindly inform me which section it should be in. Thanks. :) :)
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Hi all, I read about several methods for constructing an ellipse with straightedge and compass, and one seemed very peculiar to me. It was called the trammel method...and it's the only one that I can't prove. It seemed so simple, but I just can't prove it. ;P If anyone could find a proof for me or point me in the right direction? And also if this is in the wrong section please kindly inform me which section it should be in. Thanks. :) :)
Hi, the proof is rather simple: let the coordinates of A,B and O be called (x,y), (xB,0) and (0,yO) respectively. Then (based on AB=b and BO=AO-AB=a-b): xB= x*(a-b)/a yO= -y*(a-b)/b (congruent triangles, or rule of three) Now triangle "B,O,origin" is rectangular, hence xB*xB + yO*yO = (a-b)*(a-b) (Pythagoras) substituting the earlier formulae for xB and yO, and simplifying yields (x/a)^2 + (y/b)^2 = 1 which is exactly the equation of an ellipse with radii a and b. QED.
Luc Pattyn [Forum Guidelines] [My Articles]
Fixturized forever. :confused:
modified on Sunday, October 26, 2008 3:08 PM
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Hi, the proof is rather simple: let the coordinates of A,B and O be called (x,y), (xB,0) and (0,yO) respectively. Then (based on AB=b and BO=AO-AB=a-b): xB= x*(a-b)/a yO= -y*(a-b)/b (congruent triangles, or rule of three) Now triangle "B,O,origin" is rectangular, hence xB*xB + yO*yO = (a-b)*(a-b) (Pythagoras) substituting the earlier formulae for xB and yO, and simplifying yields (x/a)^2 + (y/b)^2 = 1 which is exactly the equation of an ellipse with radii a and b. QED.
Luc Pattyn [Forum Guidelines] [My Articles]
Fixturized forever. :confused:
modified on Sunday, October 26, 2008 3:08 PM