# Math Help - Parameter Elimination Problem - # 3

1. ## Parameter Elimination Problem - # 3

$x = t^{2} - 2t$

$y = t$

Solving for t

$x = t^{2} - 2t$

$2t = t^{2}$

$2 = \dfrac{t^{2}}{2t}$

$2 = t$

$y = \sqrt{2}$ - This must be converted into an "x =" equation

Ultimately this last equation needs to be converted to an "x =" thing, because it's part of a bigger area problem in terms of the y axis not x axis. However, something went wrong on this thing.

2. ## Re: Parameter Elimination Problem - # 3

Originally Posted by Jason76
$x = t^{2} - 2t$

$y = t$
$x=t^2-2t=(t-1)^2-1=(y-1)^2-1$

that do it?

3. ## Re: Parameter Elimination Problem - # 3

Originally Posted by romsek
$x=t^2-2t=(t-1)^2-1=(y-1)^2-1$

that do it?
How did you get that? Was that completing the square?

4. ## Re: Parameter Elimination Problem - # 3

Originally Posted by Jason76
How did you get that? Was that completing the square?
yes

5. ## Re: Parameter Elimination Problem - # 3

Originally Posted by Jason76
$x = t^{2} - 2t$

$y = t$

Solving for t

$x = t^{2} - 2t$

$2t = t^{2}$

$2 = \dfrac{t^{2}}{2t}$

$2 = t$

$y = \sqrt{2}$ - This must be converted into an "x =" equation

Ultimately this last equation needs to be converted to an "x =" thing, because it's part of a bigger area problem in terms of the y axis not x axis. However, something went wrong on this thing.
Surely you can see that you can just replace the "t" with "y"...

6. ## Re: Parameter Elimination Problem - # 3

Originally Posted by Jason76
$x = t^{2} - 2t$

$y = t$

Solving for t

$x = t^{2} - 2t$

$2t = t^{2}$

$2 = \dfrac{t^{2}}{2t}$

$2 = t$

$y = \sqrt{2}$ - This must be converted into an "x =" equation

Ultimately this last equation needs to be converted to an "x =" thing, because it's part of a bigger area problem in terms of the y axis not x axis. However, something went wrong on this thing.
What you've done here is: at the lines:

$x = t^{2} - 2t$

$2t = t^{2}$

you are effectively trying to solve for t where x = 0. So at the end of this piece, you have established that $y = \sqrt 2$ when $x = 0$, or that the curve moves through the point $(0, \sqrt 2)$.