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How to find the equations of the axis of the ellipse

  1. Aug 5, 2005 #1
    I was hoping you could give me a hint on how to find the equations of the axis of the ellipse of equation [tex]5x^2-6xy+5y^2-4x-4y-4=0[/tex]. I think this is supposed to be an exercise about lagrange multipliers or something related to the gradient, but i really dont know. I am clueless, i dont know any property about ellipses in general.

    Many Thanks, Paul.
  2. jcsd
  3. Aug 5, 2005 #2


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    Unfortunately, I don't know the English terminology that well regarding analytic geometry.

    There is a thing which we call "middle line", which is a polar (polar line) of a point at infinity with respect to the conic (here the ellipse).

    Axes are a special case of these lines, two of those whose directions are perpendicular, so where [itex]m_1 = - \frac{1}{{m_2 }}[/itex], where m is a direction.

    For a general conic [itex]ax^2 + 2b''xy + a'y + 2by + 2b'x + a'' = 0[/itex] those direction are the solutions of [itex]b''m^2 + \left( {a - a'} \right)m - b'' = 0[/itex].

    In this case, you get [itex]m = 1\,\,\, \vee \,\,m = - 1[/itex].

    Now, the equations of the axes are then:

    F_x ^\prime \left( {x,y} \right) + m_1 \cdot F_y ^\prime \left( {x,y} \right) = 0 \\
    F_x ^\prime \left( {x,y} \right) + m_2 \cdot F_y ^\prime \left( {x,y} \right) = 0 \\

    Here, [itex]F_x ^\prime \left( {x,y} \right)[/itex] mean the partial derivative of the function to the variable x (same for y).

    I tried it and it seems to be working, can you get the equations now?
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