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Thread: Eliminate parameter 't' to find a Cartesian equation

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

    $\displaystyle E-\{A\}\equiv \begin{Bmatrix}x=a\dfrac{1-t^2}{1+t^2}\\{}\\y=b\dfrac{2t}{1+t^2}\end{matrix}$
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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.

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

    Did you get Fernando's hint?

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

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

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

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

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

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

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

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


    Where are the y-intercepts?

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

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

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

    . . 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 $\displaystyle S^1\equiv x^2+y^2=1$ we have

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

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