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Math Help - More ellipse problems

  1. #1
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    More ellipse problems

    The tangent and normal at P(3\sqrt{2}\cos\theta, 3\sin\theta) to the ellipse \frac{x^2}{18}+\frac{y^2}{9}=1meet the y axis at T and N respectively. If O is the origin, prove the OT.ON is independent of P. Find the coordinates of X, the centre of the circle through P,T and N. Find also the equation of the locus of the point Q on PX produced such that X is the mid-point of PQ.
    I have completed all but the last part of this question and got right answers. X(0,\frac{3}{2}(\frac{1}{\sin\theta}-\sin\theta))
    I let Q(x,y)
    so (0,\frac{3}{2}(\frac{1}{\sin\theta}-\sin\theta))=(\frac{3\sqrt{2}\cos\theta+x}{2},\fra  c{3\sin\theta+y}{2})
    so 0=\frac{3\sqrt{2}\cos\theta+x}{2}---1
    and \frac{3}{2}(\frac{1}{\sin\theta}-\sin\theta)=\frac{3\sin\theta+y}{2}---2
    I worked out 1 and got:
    \cos\theta=\frac{\sqrt{2}x}{6}
    But i cannot express 2 in terms of \sin\theta
    Last edited by arze; May 11th 2010 at 10:31 PM.
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  2. #2
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    Check the coordinates of X, the center of the circle through P,T and N.

    It cannot be
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  3. #3
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    That is the answer given in the answer page.
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  4. #4
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    Quote Originally Posted by arze View Post
    That is the answer given in the answer page.
    TPN is a right angled triangle. So the circumcenter must be the mid point of TN.
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  5. #5
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    yes, but i have found that it is correct, X is the mid-point of TN
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  6. #6
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    Quote Originally Posted by arze View Post
    yes, but i have found that it is correct, X is the mid-point of TN
    Equation of the tangent at P is

    xcosθ/a + ysinθ/b = 1

    It meets the x-axis at T (asecθ, 0).

    Equation of the normal is

    axsec(θ) - bycosec(θ) = a^2 - b^2

    It meets the y-axis at N (0, [b^2 - a^2]/bcsc(θ))

    I am getting the mid point of TN something different. Why?
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  7. #7
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    a=3\sqrt{2}, b=3 so the answers you got is the same are they not?
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  8. #8
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    Quote Originally Posted by arze View Post
    a=3\sqrt{2}, b=3 so the answers you got is the same are they not?
    Co-ordinates of mid point of TN are

    \frac{3\sqrt{2}}{2cosθ}, \frac{-3sinθ}{2}
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  9. #9
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    Quote Originally Posted by arze View Post
    a=3\sqrt{2}, b=3 so the answers you got is the same are they not?
    Co-ordinates of mid point of TN are

    (\frac{3\sqrt{2}}{2\cos\theta}, \frac{-3\sin\theta}{2})
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  10. #10
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    sorry, i forgot to tell you that T and N are both on the y-axis, there was i typo in my original post, i have corrected it already, there is no need to find the point of the x-axis intersect. very sorry
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  11. #11
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    Quote Originally Posted by arze View Post
    sorry, i forgot to tell you that T and N are both on the y-axis, there was i typo in my original post, i have corrected it already, there is no need to find the point of the x-axis intersect. very sorry
    --



    so y = 3/sinθ - 6sinθ and x = - 3sqrt(2) cosθ
    Last edited by sa-ri-ga-ma; May 12th 2010 at 04:21 AM.
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  12. #12
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    i use the \sin^2\theta+\cos^2\theta\equiv 1 formula.
    but i still can't get the answer which is \frac{2x^2}{9}+\frac{9}{16y^2}=1
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