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Math Help - Parametric Equations for Conic Section

  1. #1
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    Parametric Equations for Conic Section

    How would I solve this:

    Find a parametric equation for the conic section:
    9x^2 - 18x + 6y^2 - 96y + 9 = 0

    Thanks!
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  2. #2
    Super Member bigwave's Avatar
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    this is an ellepse

    the equation is in the form of

    \frac{\left(x-h\right)^2}{b^2} + \frac{\left(y-k\right)^2}{a^2} = 1

    you will need to complete the square to get it to this form
    I was a not exactly sure what you meant by parametric equation
    this is just a modified equation from standard form to make is easy to recognize the charactoristics.
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  3. #3
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    Hello iheartyou119

    Welcome to Math Help Forum!
    Quote Originally Posted by iheartyou119 View Post
    How would I solve this:

    Find a parametric equation for the conic section:
    9x^2 - 18x + 6y^2 - 96y + 9 = 0

    Thanks!
    As bigwave has said, you'll need to complete the square to write the equation in a form suitable for parametric equations. This form is usually written:
    \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1
    which is indeed an ellipse, centre (h,k). The parametric form of this is:
    x = h+a\cos\theta, y = k+b\sin\theta
    which works, of course, because \cos^2\theta + \sin^2\theta = 1.

    Here's the first couple of lines showing you how to complete the square:
    9x^2 - 18x + 6y^2 - 96y + 9 = 0

    \Rightarrow 9(x^2-2x) +6(y^2-16y)+9=0

    \Rightarrow 9\Big((x-1)^2-1\Big)+6\Big((y-8)^2-64\Big)+9=0
    Can you take it from here and find the values of a, b, h, k?

    Grandad
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  4. #4
    Super Member bigwave's Avatar
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    Talking ellipse where the foci are paralel to the y axis

    I think this an ellipse where the foci are paralel to the y axis
    so the b^2 would be under the x^2

    just being cranky
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  5. #5
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    still wrong

    I have tried this, and I still can't seem to get the right answer. The answer is supposed to be:

    x = 1 + 4 cos t, y = 3 + 3 sin t
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  6. #6
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    Hello iheartyou119
    Quote Originally Posted by iheartyou119 View Post
    I have tried this, and I still can't seem to get the right answer. The answer is supposed to be:

    x = 1 + 4 cos t, y = 3 + 3 sin t
    The easiest way to check this is to work backwards. These two equations give:

    \cos^2t+\sin^2t =\frac{(x-1)^2}{4^2}+\frac{(y-3)^2}{3^2}=1

    \Rightarrow 9(x^2-2x+1)+16(y^2-6y+9)=144

    \Rightarrow 9x^2 -18x+\color{red}16\color{black}y^2-96y+9=0

    So there's the problem: either you copied the question down wrong, or there's a typo in the book.

    Grandad
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  7. #7
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    wrong equation

    Yes, you are correct. The equation is supposed to be:

    9x^2 - 18x + 16y^2 - 96y + 9 = 0
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