A parametric puzzler

[soroban]

Hello, everyone!

Here's a classic mind-boggler . . . hope you enjoy it.

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The Missing Intercept

The parametric equations: .\(\displaystyle x \ = \ \frac{1\ -\ t^2}{1\ +\ t^2}\) .and .\(\displaystyle y \ = \ \frac{2t}{1\ +\ t^2}\)

. . represent a unit circle, verified by showing that: \(\displaystyle x^2\ +\ y^2 \:= \:1\)

Find the intercepts.

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Find the y-intercepts. .Let \(\displaystyle x = 0\), solve for \(\displaystyle y.\)

. . We have: .\(\displaystyle x \ = \ \frac{1\ -\ t^2}{1\ +\ t^2} \ = \ 0\qquad\Rightarrow\qquad 1 - t^2 \ = \ 0\quad\Rightarrow\quad t \ = \ \pm1\)

. . Then: .\(\displaystyle y \ = \ \frac{2(\pm1)}{1\ +\ (\pm1)^2} \ = \ \pm 1\)

. . Hence, the y-intercepts are: .\(\displaystyle (0,\pm1)\)


Find the x-intercepts. .Let \(\displaystyle y = 0\), solve for \(\displaystyle x.\)

. . We have: .\(\displaystyle y \ = \ \frac{2t}{1\ +\ t^2} \:= \:0\qquad\Rightarrow\qquad 2t = 0\qquad\Rightarrow\qquad t = 0\)

. . Then: .\(\displaystyle x \ = \ \frac{1\ -\ 0^2}{1\ +\ 0^2} \ = \ 1\)

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


Wait! .We <u>know</u> there is a fourth intercept at \(\displaystyle (-1,0).\)

How did we miss it? .Did we make an error?

 

[zerohour]

Nice one

the equations actually do not representa circle as x^2 + y ^2 do not turn out to be 1

 

[Unco]

\(\displaystyle \left(\frac{1 - t^2}{1 + t^2}\right)^2 + \left(\frac{2t}{1 + t^2}\right)^2\)

\(\displaystyle = \frac{(1 - t^2)^2 + (2t)^2}{(1 + t^2)^2}\)

\(\displaystyle = \frac{1 - 2t^2 + t^4 + 4t^2}{(1 + t^2)^2}\)

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

It's not a unit circle though...

 

[pka]

The problem, (trick?), is in domain of the parameter.
In fact, Unco is correct: it is not the unit circle.
There is no value of t that gives the ‘missing intercept’.
The point (−1,0) corresponds to no value of t.