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Math Help - Ellipse problem

  1. #1
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    Smile Ellipse problem

    Hi,

    So let P(a\cos\theta, b\sin\theta) and Q(a\cos\phi, b\sin\phi) be points on the standard ellipse. I'm trying to show that if PQ subtends a right angle at the point (a, 0) then the chord PQ passes through a fixed point on the x-axis.

    So far I've gotten that x = \frac{a\sin(\theta + \phi)}{\sin\phi - \sin\theta} and I need to show that this is invariant. The condition (that P - (a, 0) - Q forms a right angle) gives me

    b^2 \sin\theta \sin \phi = -a^2 (\cos\theta - 1)(\cos\phi - 1)

    I've messed around with this for well over 2 hours and all I've gotten into is a terrible mess with some very tedious algebra and nothing to show for it!

    Any help would be much appreciated, thanks

    Stonehambey
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  2. #2
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by Stonehambey View Post
    Hi,

    So let P(a\cos\theta, b\sin\theta) and Q(a\cos\phi, b\sin\phi) be points on the standard ellipse. I'm trying to show that if PQ subtends a right angle at the point (a, 0) then the chord PQ passes through a fixed point on the x-axis.

    So far I've gotten that x = \frac{a\sin(\theta + \phi)}{\sin\phi - \sin\theta} and I need to show that this is invariant. The condition (that P - (a, 0) - Q forms a right angle) gives me

    b^2 \sin\theta \sin \phi = -a^2 (\cos\theta - 1)(\cos\phi - 1)

    I've messed around with this for well over 2 hours and all I've gotten into is a terrible mess with some very tedious algebra and nothing to show for it!

    Any help would be much appreciated, thanks

    Stonehambey
    when i calculated for x i got x = \frac{a\sin(\theta - \phi)}{\sin\phi - \sin\theta}. re check yours
    now all it needs to be proved is that this is a constant when b^2 \sin\theta \sin \phi = -a^2 (\cos\theta - 1)(\cos\phi - 1)
    the last equation when manipulated becomes b^2 + a^2(\tan \theta /2)(\tan \phi /2)=0
    this means  \phi is a function of \theta.
    differentiate this both sides and you will get  \frac{d \phi}{d \theta}= \frac{- \sin \phi}{\sin \theta}
    simplyfy x by using difference formulas, that is x= \frac{a\sin(\theta - \phi)}{\sin\phi - \sin\theta}=a\frac{\cos {(\theta - \phi)/2}}{\cos{(\theta + \phi )/2}}
    now x is completely a function of \theta. compute  \frac{dx}{d \theta} buy using the value of \frac{d \phi}{d \theta} calculated above.
    you will get  \frac{dx}{d \theta}=0 which will prove it.
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  3. #3
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    Thanks for the reply

    I managed to get right to the end of that, but when I got to finding \frac{dx}{d\theta} I got

    \frac{dx}{d\theta} = \left(\cos\left( \frac{\theta + \phi}{2} \right)\left[-\sin\left( \frac{\theta - \phi}{2}\right)\left( \frac{1}{2} + \frac{\sin\phi}{2\sin\theta} \right)\right] - \cos\left( \frac{\theta - \phi}{2} \right)\left[-\sin\left( \frac{\theta + \phi}{2}\right)\left( \frac{1}{2} - \frac{\sin\phi}{2\sin\theta} \right)\right]\right) / \cos^2\left( \frac{\theta + \phi}{2} \right)

    which looks horrible! How do I show that this is zero? Or have I made a mistake somewhere?

    Thanks
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  4. #4
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by Stonehambey View Post
    Thanks for the reply

    I managed to get right to the end of that, but when I got to finding \frac{dx}{d\theta} I got

    \frac{dx}{d\theta} = \left(\cos\left( \frac{\theta + \phi}{2} \right)\left[-\sin\left( \frac{\theta - \phi}{2}\right)\left( \frac{1}{2} + \frac{\sin\phi}{2\sin\theta} \right)\right] - \cos\left( \frac{\theta - \phi}{2} \right)\left[-\sin\left( \frac{\theta + \phi}{2}\right)\left( \frac{1}{2} - \frac{\sin\phi}{2\sin\theta} \right)\right]\right) / \cos^2\left( \frac{\theta + \phi}{2} \right)

    which looks horrible! How do I show that this is zero? Or have I made a mistake somewhere?

    Thanks
    I too was getting something like this (although i have not checked this term by term but still .... )
    now to prove that this is zero use \sin \theta + \sin \phi = 2 \sin[ (\theta+\phi)/2] \cos[ (\theta-\phi)/2]
    and
    \sin \theta -\sin \phi=2 \sin [(\theta-\phi)/2] \cos [(\theta +\phi)/2]
    this does it.
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