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Thread: Converting from parametric from to implicit

  1. #1
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    Exclamation Converting from parametric from to implicit

    Hi,

    I am stuck on converting from parametric form to implicit.

    For example, if the curve is:

    c = (s^2-s+1, s^2+s+1)

    How should I express as implicit representation?
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  2. #2
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    Re: Converting from parametric from to implicit

    You are saying that x= s^2- s+ 1, y= s^2+ s+ 1. You want to eliminate s from those two equations. You could, for example, solve each equation for s, using the quadratic formula, then set the two equal.
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  3. #3
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    Re: Converting from parametric from to implicit

    Hello, PStudent175!

    I have a solution, but it's quite awkward.


    $\displaystyle \text{Eliminate the parameter: }\:\begin{Bmatrix}x &=& s^2-s+1 & [1] \\ y &=& s^2+s+1 & [2] \end{Bmatrix}$

    Add [1] and [2]: .$\displaystyle x + y \;=\;2s^2 + 2 \quad\Rightarrow\quad 2s^2 \;=\;x+y - 2 $

    . . . . . . . . . . . . . $\displaystyle s^2 \;=\;\frac{x+y-2}{2} \quad\Rightarrow\quad s \;=\;\pm\sqrt{\frac{x+y-2}{2}}$

    Substitute into [2]: .$\displaystyle y \;=\;\frac{x+y-2}{2} \pm\sqrt{\frac{x+y-2}{2}} + 1 $

    . . . . . . . . . . . . . . . $\displaystyle y \;=\;\frac{x+y}{2} \pm\sqrt{\frac{x+y-2}{2}} $

    . . . . . . . . . $\displaystyle y - \frac{x+y}{2} \;=\;\pm\sqrt{\frac{x+y-2}{2}} $


    Square: .$\displaystyle y^2 - y(x+y) + \frac{(x+y)^2}{4} \;=\;\frac{x+y-2}{2}$


    I'll let you clean it up . . .
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  4. #4
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    Re: Converting from parametric from to implicit

    Quote Originally Posted by Soroban View Post
    Hello, PStudent175!

    I have a solution,
    . . .
    Soroban has the right idea, but instead of adding them should have subtracted the equations to give $\displaystyle 2s=y-x$ ...
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  5. #5
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    Re: Converting from parametric from to implicit

    Quote Originally Posted by Soroban View Post
    Hello, PStudent175!

    I have a solution, but it's quite awkward.



    Add [1] and [2]: .$\displaystyle x + y \;=\;2s^2 + 2 \quad\Rightarrow\quad 2s^2 \;=\;x+y - 2 $

    . . . . . . . . . . . . . $\displaystyle s^2 \;=\;\frac{x+y-2}{2} \quad\Rightarrow\quad s \;=\;\pm\sqrt{\frac{x+y-2}{2}}$

    Substitute into [2]: .$\displaystyle y \;=\;\frac{x+y-2}{2} \pm\sqrt{\frac{x+y-2}{2}} + 1 $

    . . . . . . . . . . . . . . . $\displaystyle y \;=\;\frac{x+y}{2} \pm\sqrt{\frac{x+y-2}{2}} $

    . . . . . . . . . $\displaystyle y - \frac{x+y}{2} \;=\;\pm\sqrt{\frac{x+y-2}{2}} $


    Square: .$\displaystyle y^2 - y(x+y) + \frac{(x+y)^2}{4} \;=\;\frac{x+y-2}{2}$


    I'll let you clean it up . . .
    Perfect, thanks a lot that was very useful.

    For the next question it is:

    x = (a-t) / (a+t)
    y = (t) / (a+t)

    And I have tried the same method by adding together to get x+y= a / a+t but am not really sure to go from there?

    Do you know of any resources that has information on this topic as I haven't been taught it yet at university.

    Thanks again for the help!
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  6. #6
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    Re: Converting from parametric from to implicit

    Hello again, PStudent175!

    $\displaystyle \begin{array}{cccc}x &=& \dfrac{a-t}{a+t} & [1] \\ \\[-4mm] y &=& \dfrac{t}{a+t} & [2]\end{array}$

    I could find no "clever" way to eliminate the parameter.

    Solve [2] for $\displaystyle t\!:\;\;y \:=\:\frac{t}{a+t} \quad\Rightarrow\quad ay + ty \:=\:t$

    . . $\displaystyle t - ty \:=\:ay \quad\Rightarrow\quad t(1-y) \:=\:ay \quad\Rightarrow\quad t \:=\:\frac{ay}{1-y}$


    Substitute into [1]: .$\displaystyle x \:=\:\dfrac{a-\frac{ay}{1-y}}{a + \frac{ay}{1-y}}$

    Muliply by $\displaystyle \frac{1-y}{1-y}\!:\;\;x \;=\;\frac{a(1-y) - ay}{a(1-y) + ay} \;=\;\frac{a-ay - ay}{a-ay + ay} \;=\;\frac{a-2ay}{a} $


    Therefore: .$\displaystyle x \;=\;1-2y \quad\Rightarrow\quad \boxed{x + 2y \;=\;1}$


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Interesting observation . . .

    The graph is the straight line $\displaystyle x + 2y \:=\:1$
    . . with a "hole" at (-1, 1).

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  7. #7
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    Re: Converting from parametric from to implicit

    Quote Originally Posted by PStudent175 View Post
    Perfect, thanks a lot that was very useful.

    For the next question it is:

    x = (a-t) / (a+t)
    y = (t) / (a+t)

    And I have tried the same method by adding together to get x+y= a / a+t but am not really sure to go from there?

    Do you know of any resources that has information on this topic as I haven't been taught it yet at university.

    Thanks again for the help!
    Add them to get

    $\displaystyle x+y=\frac{a}{a+t}$

    substitute $\displaystyle 1/(a+t)$ back into the equations:

    $\displaystyle x=(a-t)(x+y)/a$

    $\displaystyle y=t(x+y)/a$

    Solve for $\displaystyle t$ from the second of these and substitute into the first and simplify.
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