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Find the eccentricity of this ellipse

  1. Dec 12, 2012 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    The tangent at any point P of a circle meets the tangent at a fixed point A in T, and T is joined to B, the other end of diameter through A. Prove that the locus of point of intersection of AP and BT is an ellipse whose eccentricity is [itex] 1/ \sqrt{2}[/itex]

    2. Relevant equations

    3. The attempt at a solution
    The very first thing I do is assume the equation of a circle. The next thing is to write the equations for tangents and solve them to get T. But it is getting complicated as nothing is known to me. So there are a number of variables which can't be eliminated. Any other ideas?
     
  2. jcsd
  3. Dec 13, 2012 #2

    tiny-tim

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    hi utkarshakash! :smile:
    show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:

    alternatively, since the eccentricity is 1/√2, the minor axis must be 1/√2 time the major axis …

    so have you tried squashing the whole diagram by 1/√2 (along AB), so that the the final result is a circle? :wink:
     
  4. Dec 14, 2012 #3

    utkarshakash

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    For the sake of simplicity, let the equation of the circle be [itex]x^2 + y^2 = c^2 [/itex].

    Let A = (x1,y1) and P = (x2,y2)

    Equation of tangent at P
    [itex] xx_2 + yy_2 - c^2 = 0[/itex]
    Equation of tangent at A
    [itex] xx_1 + yy_1 -c^2 = 0[/itex]

    When I solve these two equations I get

    [itex] x = \dfrac{c^2 (y_2 - y_1)}{x_1y_2-x_2y_1} \\

    y = \dfrac{c^2 (x_1 - x_2)}{x_1y_2-x_2y_1}[/itex]

    OMG It looks so dangerous! I don't want to proceed ahead as the calculation will be complex and rigorous.
     
  5. Dec 14, 2012 #4

    tiny-tim

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    no!

    A is fixed, so you can simplify by letting A = (c,0) :wink:

    (and P = (ccosθ,csinθ) )
     
    Last edited: Dec 14, 2012
  6. Dec 15, 2012 #5

    utkarshakash

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    That did simplify the expression to a large extent. Thanks!
     
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