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Math Help - 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: . \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: . \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]: . x^2+y^2\:=\:\frac{1 - 2t^2 + t^4}{(1+t^2)^2} + \frac{4t^2}{(1+t^2)^2}

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


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



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

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

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



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

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

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



    Where is the other x-intercept?

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  2. #2
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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 \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 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: . \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: . \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]: . x^2+y^2\:=\:\frac{1 - 2t^2 + t^4}{(1+t^2)^2} + \frac{4t^2}{(1+t^2)^2}

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


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



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

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

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



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

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

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



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