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

1. 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

2. Re: Eliminate parameter 't' to find a Cartesian equation

Originally Posted by pikachu26134
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 : $X^2+Y^2=\ldots=1$

3. Re: Eliminate parameter 't' to find a Cartesian equation

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

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

4. Re: Eliminate parameter 't' to find a Cartesian equation

Hello, pikachu26134!

Eliminate the parameter and find a Cartesian equation.

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

Did you get Fernando's hint?

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

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

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

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

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

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

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

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

Where are the y-intercepts?

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

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

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

. . Where is the other one?

Nice puzzle!

6. Re: Eliminate parameter 't' to find a Cartesian equation

Originally Posted by ILikeSerena
Nice puzzle!
In answer #3 I explain it in a more general way. In the particular case of $S^1\equiv x^2+y^2=1$ we have

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

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