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Math Help - parametric form of equation

  1. #1
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    parametric form of equation

    how do i express x^2+y^2-28x+4y+120=0 in parametric form
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  2. #2
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    Its also (x-14)^2+(y+2)^2-80 = 0...
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  3. #3
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    Hello, sigma1!

    How do i express x^2+y^2-28x+4y+120\:=\:0 in parametric form?

    We have: . x^2 - 28x + y^2 + 4y \;=\;-120

    Complete the square: . x^2 - 28x {\color{blue}\:+\: 196} + y^2 + 4y {\color{red}\:+\: 4} \;=\;-120 {\color{blue}\:+\: 196} {\color{red}\:+\: 4}

    And we have: . (x-14)^2 + (y+2)^2 \;=\;80

    This is a circle with center (14,-2) and radius 4\sqrt{5}


    The parametric equations are: . \begin{Bmatrix}x &=& 14 + 4\sqrt{5}\,\cos\theta \\ \\[-3mm] y &=& \text{-}2 + 4\sqrt{5}\,\sin\theta \end{Bmatrix}

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  4. #4
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    i inderstand uto the point where you complete the square but can you please explain the principle you used to arrive at your answer. thanks.
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  5. #5
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    In general,

    x^2+y^2 +ax+by +c=0


    x^2+ 2\frac a2x + (\frac a2)^2+y^2+2\frac b2y +(\frac b2)^2+c-(\frac a2)^2-(\frac b2)^2=0


    (x-\frac a2)^2+(y-\frac b2)^2=(\frac a2)^2+(\frac b2)^2-c

    which you can check by expanding the squares. In your case, a=-28, b=4, c=120.

    http://www.algebralab.org/lessons/le...gthesquare.xml
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  6. #6
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    i do understand that point but i dont understand how you use that to get to the parametric equations. with the trig variables.
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  7. #7
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    Oh. A circle of radius r centred at the origin is given by the parametric equations

    \begin{Bmatrix}x &=& r\,\cos\theta \\ \\[-3mm] y &=& r\,\sin\theta \end{Bmatrix}.

    This is pretty much just a definition. Can you translate these so that the centre is at the correct point?
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  8. #8
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    Quote Originally Posted by sigma1 View Post
    i do understand that point but i dont understand how you use that to get to the parametric equations. with the trig variables.
    a circle centered at the origin, x^2 + y^2 = r^2 , can be written parametrically as

    x = r\cos{t}

    y = r\sin{t}

    shifting the center from the origin to the point (h,k) yields the equations

    x = h + r\cos{t}

    y = k + r\sin{t}

    ... that's all.
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  9. #9
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    Quote Originally Posted by skeeter View Post
    a circle centered at the origin, x^2 + y^2 = r^2 , can be written parametrically as

    x = r\cos{t}

    y = r\sin{t}

    shifting the center from the origin to the point (h,k) yields the equations

    x = h + r\cos{t}

    y = k + r\sin{t}

    ... that's all.

    thanks alot i inderstand it clearly now.
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