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Normal to a fixed concentric ellipse

  1. Dec 12, 2012 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    PM and PN are perpendiculars upon the axes from any point 'P' on the ellipse. Prove that MN is always normal to a fixed concentric ellipse

    2. Relevant equations

    3. The attempt at a solution
    I assume point P to be (acosθ, bsinθ)

    The eqn of line MN is then given by
    [itex] bsin\theta x+acos\theta y =absin\theta cos\theta[/itex]
     
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  3. Dec 12, 2012 #2

    haruspex

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    Try writing the generic equation for a normal to an ellipse with same centre and axes.
     
  4. Dec 14, 2012 #3

    utkarshakash

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    [itex]a'sec \phi x - b'cosec \phi y = a^2 - b^2[/itex]
     
  5. Dec 14, 2012 #4

    haruspex

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    I assume you meant [itex]a'sec \phi x - b'cosec \phi y = a'^2 - b'^2[/itex]
    It remains to find expressions for a', b' and phi in terms of a, b and θ that make this the same as the equation for MN. There is a constraint regarding which of a', b' and phi can depend on which of a, b and θ.
     
  6. Dec 15, 2012 #5

    utkarshakash

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    Do you want me to compare the two lines?
     
  7. Dec 15, 2012 #6

    haruspex

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    Yes. You want to make a′x sec(ϕ)−b′y cosec(ϕ)=a′2−b′2 look like bx sin(θ)+ay cos(θ)=ab sinθ cosθ by suitable choices of a', b' and ϕ. But note that this must work keeping a' and b' fixed while ϕ is allowed to vary as a function of θ.
     
  8. Dec 16, 2012 #7

    utkarshakash

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    OK I did exactly what you said and got the following relations after comparison

    [itex] cos\alpha = \dfrac{aa'cos\theta}{a'^2 - b'^2} \\

    sin \alpha = \dfrac{-bb'sin\theta}{a'^2 - b'^2} \\

    tan \alpha = \dfrac{-bb' tan\theta}{aa'}

    [/itex]
     
  9. Dec 16, 2012 #8

    haruspex

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    OK, nearly there. Now you must set a' and b' so that all values of θ and α can occur. (Otherwise there will be a gap in one of the ellipses.)
     
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