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Eccentricity of an Ellipse

  1. Dec 4, 2006 #1
    This one question has me totally beaten. And I thought I was pretty good in co-ordinate geometry. Here it is:

    If the equation ax^2 + 2hxy + by^2 =1 represents an ellipse, find the square of the eccentricity of the ellipse.

    I know that the ratio of the distance from the directrix to the focus of a point on the ellipse is the eccentricity. But I cant figure out what the directrix is or where the foci lie. This equation must represent an ellipse with its axes shifted (as the equation with x and y axes as its major axes is (x^2/a*a) + (y*y/b*b) =1). Also, here h*h - ab <0, and abc +2fgh -af*f - bg*g -ch*h is non zero. I just dont know how to go about finding the eccentricity.
     
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  3. Dec 4, 2006 #2

    arildno

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    Do the shift of coordinates first, then worry about the eccentricity!
     
  4. Dec 5, 2006 #3
    How would you do that? By substitituting x+a for x and y+b for y to eliminate the xy term?
     
  5. Dec 5, 2006 #4

    arildno

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    NO!
    First of all, sorry for saying "shifting" the coordinates, I meant "rotating" the coordinates.

    Do you know how to do that?
     
  6. Dec 5, 2006 #5
    Using [tex] x=xcos(t) - ysin(t)[/tex] and [tex] y=xsin(t) + ycos(t) [/tex] then equating the coeff of the xy term to zero from which you would get the value of [tex]tan2(t)[/tex].

    Then, substituting the value of sint and cost, you would get the general equation of the ellipse.... right?
     
  7. Dec 5, 2006 #6

    arildno

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    Right!
    Then use, for example, the relation between the semi-major and semi-minor axes and the eccentricity to determine the latter quantity.
     
  8. Dec 5, 2006 #7
    Thank you, that helps a lot. Its great having such talented people there to look at your problems. Thanks a lot.
     
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