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Thread: The Missing Intercept

  1. #1
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    The Missing Intercept

    The Missing Intercept


    I posted this puzzle some time ago,
    but no one provided a satisfactory answer.


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

    We have the parametric equations of a unit circle.



    Verification

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

    Add [1] and [2]: .$\displaystyle x^2+y^2\:=\:\frac{1 - 2t^2 + t^4}{(1+t^2)^2} + \frac{4t^2}{(1+t^2)^2} $

    $\displaystyle = \;\frac{1 + 2t^2+t^4}{(1+t^2)^2} \:=\:\frac{(1+t^2)^2}{(1+t^2)^2} \;=\;1 $


    Hence, a unit circle: .$\displaystyle x^2+y^2\:=\:1$



    To find the $\displaystyle y$-intercepts, let $\displaystyle x = 0.$

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

    Hence, the $\displaystyle y$-intercepts are: .$\displaystyle (0,1),\;(0,\text{-}1)$



    To find the $\displaystyle x$-intercepts, let $\displaystyle y = 0.$

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

    Hence, the $\displaystyle x$-intercept is: .$\displaystyle (1,0)$ ?



    Where is the other $\displaystyle x$-intercept?

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  2. #2
    MHF Contributor
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    Hi

    I don't know which answers have been given previously so I am trying

    Let E the set defined by $\displaystyle \begin{Bmatrix}x \:=\:\dfrac{1-t^2}{1+t^2} \\ \\[-3mm] y \:=\:\dfrac{2t}{1+t^2} \end{Bmatrix}$ for t real

    By showing that $\displaystyle x^2+y^2=1$ you are proving that E is included in the unit circle, but not that it is equal to the unit circle. And it is not equal since the point (-1,0) is not inside E (there is no value of t such that x=-1 and y=0). Only if you are allowed to consider infinite values of t, you can find this point.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Soroban View Post
    The Missing Intercept






    I posted this puzzle some time ago,
    but no one provided a satisfactory answer.


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

    We have the parametric equations of a unit circle.



    Verification

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

    Add [1] and [2]: .$\displaystyle x^2+y^2\:=\:\frac{1 - 2t^2 + t^4}{(1+t^2)^2} + \frac{4t^2}{(1+t^2)^2} $

    $\displaystyle = \;\frac{1 + 2t^2+t^4}{(1+t^2)^2} \:=\:\frac{(1+t^2)^2}{(1+t^2)^2} \;=\;1 $


    Hence, a unit circle: .$\displaystyle x^2+y^2\:=\:1$



    To find the $\displaystyle y$-intercepts, let $\displaystyle x = 0.$

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

    Hence, the $\displaystyle y$-intercepts are: .$\displaystyle (0,1),\;(0,\text{-}1)$



    To find the $\displaystyle x$-intercepts, let $\displaystyle y = 0.$

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

    Hence, the $\displaystyle x$-intercept is: .$\displaystyle (1,0)$ ?



    Where is the other $\displaystyle x$-intercept?
    Not much of a puzzle it's at $\displaystyle t=\pm\infty$ (or rather it's the limit point as $\displaystyle t \to \infty$ ) and there the point is $\displaystyle (-1,0)$. So the curve never reaches the second intercept and the curve is the unit circle with a hole at $\displaystyle (-1,0).$

    CB
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  4. #4
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    Thank you, running-gag and The Cap'n!

    Those are the answers I was hoping for.

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