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Math Help - Eliminate parameter 't' to find a Cartesian equation

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    Eliminate parameter 't' to find a Cartesian equation

    I have made the attempt at solving it, in the form of being left with even more ts.
    X= (2t)/(t^2+1) Y= (1-t^2)/(t^2+1)
    Eliminate the parameters and find a Cartesian equation
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Eliminate parameter 't' to find a Cartesian equation

    Quote Originally Posted by pikachu26134 View Post
    I have made the attempt at solving it, in the form of being left with even more ts. X= (2t)/(t^2+1) Y= (1-t^2)/(t^2+1) Eliminate the parameters and find a Cartesian equation
    Hint : X^2+Y^2=\ldots=1
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Re: Eliminate parameter 't' to find a Cartesian equation

    More general, consider the ellipse E\equiv x^2/a^2+y^2/b^2=1 . One parametric representation of E is x=a\cos \varphi,\;y=b\sin \varphi . Let M(\varphi) be the point (a\cos \varphi,b\sin \varphi ) , the map \varphi\to M(\varphi) from [\pi,-\pi) to E is bijective ; on the other hand, the map t\to 2\arctan t from \mathbb{R} to (-\pi,\pi) is also bijective. Then, the map t\to \left(a\frac{1-t^2}{1+t^2},b\frac{2t}{1+t^2}\right) from \mathbb{R} to \mathbb{R}^2 is bijective from \mathbb{R} to E except for the point M(-\pi) ; that is, except for the point A(-a,0) . As a consequence, a parametric representation of E-\{A\} is:

    E-\{A\}\equiv \begin{Bmatrix}x=a\dfrac{1-t^2}{1+t^2}\\{}\\y=b\dfrac{2t}{1+t^2}\end{matrix}
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  4. #4
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    Re: Eliminate parameter 't' to find a Cartesian equation

    Hello, pikachu26134!

    Eliminate the parameter and find a Cartesian equation.

    . . \begin{Bmatrix}x &=& \dfrac{2t}{t^2+1} & [1] \\ \\[-3mm] y &=& \dfrac{1-t^2}{t^2+1} & [2] \end{Bmatrix}

    Did you get Fernando's hint?

    . . \text{Square [1]: }\:x^2 \:=\:\left(\dfrac{2t}{t^2+1}\right)^2 \:=\:\frac{4t^2}{(t^4+2t^2+1)^2}

    . . \text{Square [2]: }\:y^2 \:=\:\left(\frac{1-t^2}{t^2+1}\right)^2 \:=\:\frac{t^4 - 2t^2 + 1}{(t^4+2t^2+1)^2}

    Add: . x^2 + y^2 \;=\;\frac{4t^2}{t^4+2t^2+1} + \frac{t^4-2t^2+1}{t^4+2t^2+1}

    m . . . x^2+y^2 \;=\;\frac{t^4+2t^2+1}{t^4+2t^2+1}

    m . . . x^2+y^2 \;=\;1

    We have a circle with radius 1 and its center at the origin.


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


    I've posed this puzzle before, but it's worth repeating.


    Where are the x-intercepts?

    When y = 0\!:\;\frac{1-t^2}{t^2+1} \:=\:0 \quad\Rightarrow\quad t \:=\:\pm1

    Then: . x \:=\:\frac{2(\pm1)}{(\pm1)^2+1} \:=\:\frac{\pm2}{2} \:=\:\pm1

    The x-intercepts are: . (-1,0),\:(1,0)


    Where are the y-intercepts?

    When x = 0\!:\;\frac{2t}{t^2+1} \:=\:0 \quad\Rightarrow\quad t \;=\;0

    Then: . y \:=\:\frac{1-0^2}{0^2+1} \:=\:1

    There is one y-intercept: . (0,1)

    . . Where is the other one?

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  5. #5
    Super Member ILikeSerena's Avatar
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    Re: Eliminate parameter 't' to find a Cartesian equation

    Nice puzzle!
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Re: Eliminate parameter 't' to find a Cartesian equation

    Quote Originally Posted by ILikeSerena View Post
    Nice puzzle!
    In answer #3 I explain it in a more general way. In the particular case of S^1\equiv x^2+y^2=1 we have

    S^1-\{(0,-1)\}\equiv\begin{Bmatrix}x=\dfrac{2t}{t^2+1}\\{}\\ y=\dfrac{1-t^2}{t^2+1}\end{matrix}\quad(t\in\mathbb{R})

    S^1\equiv\begin{Bmatrix}x=\dfrac{2t}{t^2+1}\\{}\\ y=\dfrac{1-t^2}{t^2+1}\end{matrix}\quad (t\in\mathbb{R}\cup\{\infty\})
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