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Math Help - Geometry question

  1. #1
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    Geometry question

    Show that an equation of the chord joining the points [latex] P(acos\omega, bsin\omega) and Q(acos\theta, b sin\theta) [/latex] on the ellipse with equation [latex] b^2x^2+a^2y^2=a^2b^2 is
    Bxcos\frac{1}{2}(\theta+\omega)+aysin\frac{1}{2}(\ theta +\omega)=abcos\frac{1}{2}(\theta-\omega) [/latex]

    I just can't seem to make it work. Help please!!!
    Last edited by speedfreak; December 6th 2009 at 07:24 AM.
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  2. #2
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    Quote Originally Posted by speedfreak View Post
    Show that an equation of the chord joining the points [latex] P(acos\omega, bsin\omega) and Q(acos\theta, b sin\theta) [/latex] on the ellipse with equation [latex] b^2x^2+a^2y^2=a^2b^2 is
    Bxcos\frac{1}{2}(\theta+\omega)+aysin\frac{1}{2}(\ theta +\omega)=abcos\frac{1}{2}(\theta-\omega) [/latex]

    I just can't seem to make it work. Help please!!!
    HI

    First of all , find the gradient of PQ

    m=\frac{b\sin \omega-b\sin \theta}{a\cos \omega-a\cos \theta}

    Then make use of this formula : y-y_1=m(x-x_1)

    y-b\sin \omega=\frac{b\sin \omega-b\sin \theta}{a\cos \omega-a\cos \theta}(x-a\cos \omega)

    (y-b\sin \omega)(a\cos \omega-a\cos \theta)=(b\sin \omega-b\sin \theta)(x-a\cos \omega)

    ya\cos \omega-ya\cos \theta-bx\sin \omega+bx\sin \theta=ba\sin \theta\cos \omega-ba\sin \omega\cos \theta

    ya(\cos \omega-\cos \theta)+bx(\sin \theta-\sin \omega)=ba(\sin \theta\cos \omega-\sin \omega\cos \theta)

    Then use the sin and cos formulas .

     <br />
ya[-2\sin \frac{\omega+\theta}{2}\sin \frac{\omega-\theta}{2}]+bx[2\cos \frac{\theta+\omega}{2}\sin \frac{\theta-\omega}{2}]=ba\sin (\theta-\omega)<br />

    ay\sin \frac{\omega+\theta}{2}+bx\cos \frac{\omega+\theta}{2}=\frac{ba\sin (\theta-\omega)}{-2\sin \frac{1}{2}(\theta-\omega)}

     <br />
ay\sin \frac{\omega+\theta}{2}+bx\cos \frac{\omega+\theta}{2}=\frac{ba\cdot 2\sin \frac{\omega-\theta}{2}\cos \frac{\omega-\theta}{2}}{2\sin \frac{\omega-\theta}{2}}<br />

    Then cancel the like terms to get the equation . Phew ..
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  3. #3
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    Quote Originally Posted by mathaddict View Post
    HI

    First of all , find the gradient of PQ

    m=\frac{b\sin \omega-b\sin \theta}{a\cos \omega-a\cos \theta}

    Then make use of this formula : y-y_1=m(x-x_1)

    y-b\sin \omega=\frac{b\sin \omega-b\sin \theta}{a\cos \omega-a\cos \theta}(x-a\cos \omega)

    (y-b\sin \omega)(a\cos \omega-a\cos \theta)=(b\sin \omega-b\sin \theta)(x-a\cos \omega)

    ya\cos \omega-ya\cos \theta-bx\sin \omega+bx\sin \theta=ba\sin \theta\cos \omega-ba\sin \omega\cos \theta

    ya(\cos \omega-\cos \theta)+bx(\sin \theta-\sin \omega)=ba(\sin \theta\cos \omega-\sin \omega\cos \theta)

    Then use the sin and cos formulas .

     <br />
ya[-2\sin \frac{\omega+\theta}{2}\sin \frac{\omega-\theta}{2}]+bx[2\cos \frac{\theta+\omega}{2}\sin \frac{\theta-\omega}{2}]=ba\sin (\theta-\omega)<br />

    ay\sin \frac{\omega+\theta}{2}+bx\cos \frac{\omega+\theta}{2}=\frac{ba\sin (\theta-\omega)}{-2\sin \frac{1}{2}(\theta-\omega)}

     <br />
ay\sin \frac{\omega+\theta}{2}+bx\cos \frac{\omega+\theta}{2}=\frac{ba\cdot 2\sin \frac{\omega-\theta}{2}\cos \frac{\omega-\theta}{2}}{2\sin \frac{\omega-\theta}{2}}<br />

    Then cancel the like terms to get the equation . Phew ..
    Thanks most helpful!!
    I am also stuck on the second part of the question:
    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.
    So far I have work out the gradients of both AP and AQ and deduced that because they are perpendicular:
     b^2\sin\theta\sin\omega = a^2(\cos\theta + \cos\omega - 1 - \cos\theta\cos\omega)
    and also that  xcos\frac{\theta+\omega}{2} = acos\frac{\theta-\omega}{2}
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