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Math Help - Ellipse problem

  1. #1
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    Ellipse problem

    The tangent at P(a\cos\theta, b\sin\theta) to the ellipse
    b^2x^2+a^2y^2=a^2b^2
    cuts the y-axis at Q. The normal at P is parallel to the line joining Q to one focus S'. if S is the other focus, show that PS is parallel to the y-axis.

    Q(0,b\csc\theta)
    S'(-ae,0) and S(ae,0)
    so gradient of QS' =\frac{b\csc\theta}{ae}=\frac{a\sin\theta}{b\cos\t  heta}
    this means e=\frac{b^2\cos\theta}{a^2\sin^2\theta}
    that would make ae= \frac{b^2\cos\theta}{a\sin^2\theta}
    but in order for PS to be vertical, ae needs to be a\cos\theta
    i.e. e= \cos\theta
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  2. #2
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    Normal PN bisects the angle S'PS.
    Compare the triangles PNS and QPS'.
    They are similar.
    So \angle{PNS} = \angle{QPS'}
    But \angle{QPS} + \angle{S'PN} = 90 degrees
    Hence \angle{PNS} + \angle{NPS} = 90 degrees
    So the PSN is a right angled triangle.
    Last edited by sa-ri-ga-ma; May 7th 2010 at 10:59 PM.
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  3. #3
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    Quote Originally Posted by sa-ri-ga-ma View Post
    Normal PN bisects the angle S'PS.
    Compare the triangles PNS and QPS'.
    They are similar.
    So \angle{PNS} = \angle{QPS'}
    But \angle{QPS} + \angle{S'PN} = 90 degrees
    Hence \angle{PNS} + \angle{NPS} = 90 degrees
    So the PSN is a right angled triangle.
    where is N?
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  4. #4
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    Quote Originally Posted by arze View Post
    where is N?
    Normal at P which cuts the x axis at N.
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