Finding the Cartesian equation for a set of parametric equations

I'm trying to solve this problem, but I am having trouble. Can someone give me some pointers?

A curve is defined by the parametric equations:

x = 2t / (1 + t^2)

y = (1 - t^2) / (1 + t^2)

By considering the expression x^2 + y^2, eliminate the parameter *t* to obtain the Cartesian (x-y) equation of the curve.

Thanks

Re: Finding the Cartesian equation for a set of parametric equations

so ... what did you determine for the expression $\displaystyle x^2+y^2$ in terms of t ?

Re: Finding the Cartesian equation for a set of parametric equations

This is the equation I formed: x^2 + y^2 = (4t^2 - t^4 + 1) / (1 + t^4). I don't know how to eliminate *t* from this point.

Re: Finding the Cartesian equation for a set of parametric equations

first off, note that ...

$\displaystyle (1+t^2)^2 \ne 1 + t^4$

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

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

try adding $\displaystyle x^2+y^2$ again ...

Re: Finding the Cartesian equation for a set of parametric equations

Oops, stupid error! Ok, got it now. The ratio cancels out and I get *x*^{2} + y^{2} = 1. Many thanks.

Re: Finding the Cartesian equation for a set of parametric equations

Hello, everyone!

This problem (and solution) involves a paradox.

Paramtric equations: .$\displaystyle \begin{Bmatrix}x &=& \dfrac{2t}{1+t^2} & [1] \\ \\[-4mm] y &=& \dfrac{1-t^2}{1+t^2} & [2] \end{Bmatrix}$

Cartesian equation: .$\displaystyle x^2 + y^2 \:=\:1\;\;[3]$

From [3], we have a circle with center at the Origin and radius 1.

. . It has $\displaystyle x$-intercepts $\displaystyle (\pm1,\,0)$ and $\displaystyle y$-intercepts $\displaystyle (0,\,\pm1)$

Let's determine the intercepts using the parametric equations.

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

Substitute into [1]: .$\displaystyle x \:=\:\frac{2(\pm1)}{1+(\pm1)^2} \;=\;\frac{\pm2}{2} \;=\;\pm1$

The $\displaystyle x$-intercepts are: .$\displaystyle (1,\,0)\text{ and }(-1,\,0)$

For $\displaystyle y$-intercepts, let $\displaystyle x = 0\!:\;\;\frac{2t}{1+t^2} \:=\:0\quad\Rightarrow\quad 2t \:=\:0 \quad\Rightarrow \quad x \:=\:0$

Substitute into [2]: .$\displaystyle y \:=\:\frac{1-0^2}{1+0^2} \quad\Rightarrow\quad y \:=\:1$

The $\displaystyle y$-intercept is $\displaystyle (0,1)$ . . . There is only one!

Where is the *other* $\displaystyle y$-intercept?