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Ellipse chord subtending a right angle

  1. Aug 13, 2016 #1
    1. The problem statement, all variables and given/known data
    Show that the equation of the chord joining the points [itex]P(a\cos(\phi), b\sin(\phi)) [/itex]and [itex]Q(a\cos(\theta), b\sin(\theta))[/itex] on the ellipse [itex]b^2x^2+a^2y^2=a^2b^2 [/itex] is [itex] bx cos\frac{1}{2}(\theta+\phi)+ay\sin\frac{1}{2}(\theta+\phi)=ab\cos\frac{1}{2}(\theta-\phi)[/itex].
    Prove that , if the chord PQ subtends a right angle at the point (a,0), then PQ passes through a fixed point on the x axis.

    2. Relevant equations


    3. The attempt at a solution
    The second part of the question is where I am having difficulty.

    The question seems to suggest a parametric as opposed to cartesian approach.

    PQ subtends a right angle ⇒

    [itex]
    (\frac{b^2}{a^2})(\frac{\sin(\phi)}{1-\cos(\phi)})(\frac{\sin(\theta)}{1-\cos(\theta)})=-1
    [/itex]

    The coordinates of the intersection of PQ with the x axis are

    [itex]
    (a\frac{\cos(\frac{\theta-\phi}{2})}{\cos(\frac{\theta+\phi}{2})}, 0)
    [/itex]

    Presumably I should be able to manipulate the first expression to yield a constant value for [itex]
    \frac{\cos(\frac{\theta-\phi}{2})}{\cos(\frac{\theta+\phi}{2})}
    [/itex] however, I don't seem to be making much progress in this respect.
     
  2. jcsd
  3. Aug 13, 2016 #2

    haruspex

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    Since one of the expressions involves half angles, try substituting theta=2A, phi=2B everywhere.
     
  4. Aug 14, 2016 #3
    That seems to have done the trick. So the point of intersection is the constant value

    [itex]
    a(\frac{a^2+b^2}{a^2-b^2},0)
    [/itex]

    Thanks for your help.
     
  5. Aug 14, 2016 #4

    haruspex

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    Well done.
     
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