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Simple ellipse question

  1. Dec 4, 2009 #1
    given an ellipse in vector form
    r(phi)=a*cos(phi)i +b*sin(phi)j
    where i and j are the unit vectors for x and y,
    then y= b*sin(phi), and x = a*cos(phi).
    tan(phi) = y / x ,
    but y/x=(b/a)*tan(phi)
    which implies 1 = b/a or b=a
    which is false.

    What is the deal?
    doing the same for a circle leads to the true statement, 1=1.
  2. jcsd
  3. Dec 4, 2009 #2


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    What is wrong is that in your parameterization:

    r(phi)=a*cos(phi)i +b*sin(phi)j

    the parameter phi is not equal to the polar coordinate angle theta unless a = b.
  4. Dec 6, 2009 #3
    I never mentioned anything about polar coordinates, or its angle theta.
    Since r is in cartesian components,
    y must be equal to b*sin([tex]\phi[/tex]),
    and x must be equal to a*cos([tex]\phi[/tex]), right?
    and since they are perpendicular, they form a right triangle with r as the hypotenuse... then the tangent of that angle, tan([tex]\phi[/tex]) must equal y/x, no?
  5. Dec 6, 2009 #4


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    phi is a parameter, not an actual angle. That is, if you choose a certain value for phi, you'll get a point (x,y) that's on the ellipse. However, if you actually plot this point and connect it to (0,0), the resulting line is NOT at an angle phi from the x-axis. There's no reason it should be, since phi is an arbitrarily picked value.
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