Hi,

I am stuck on converting from parametric form to implicit.

For example, if the curve is:

c = (s^2-s+1, s^2+s+1)

How should I express as implicit representation?

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- Dec 21st 2012, 07:14 AMPStudent175Converting from parametric from to implicit
Hi,

I am stuck on converting from parametric form to implicit.

For example, if the curve is:

c = (s^2-s+1, s^2+s+1)

How should I express as implicit representation? - Dec 21st 2012, 09:39 AMHallsofIvyRe: Converting from parametric from to implicit
You are saying that x= s^2- s+ 1, y= s^2+ s+ 1. You want to eliminate s from those two equations. You could, for example, solve each equation for s, using the quadratic formula, then set the two equal.

- Dec 21st 2012, 06:29 PMSorobanRe: Converting from parametric from to implicit
Hello, PStudent175!

I have a solution, but it's quite awkward.

Quote:

$\displaystyle \text{Eliminate the parameter: }\:\begin{Bmatrix}x &=& s^2-s+1 & [1] \\ y &=& s^2+s+1 & [2] \end{Bmatrix}$

Add [1] and [2]: .$\displaystyle x + y \;=\;2s^2 + 2 \quad\Rightarrow\quad 2s^2 \;=\;x+y - 2 $

. . . . . . . . . . . . . $\displaystyle s^2 \;=\;\frac{x+y-2}{2} \quad\Rightarrow\quad s \;=\;\pm\sqrt{\frac{x+y-2}{2}}$

Substitute into [2]: .$\displaystyle y \;=\;\frac{x+y-2}{2} \pm\sqrt{\frac{x+y-2}{2}} + 1 $

. . . . . . . . . . . . . . . $\displaystyle y \;=\;\frac{x+y}{2} \pm\sqrt{\frac{x+y-2}{2}} $

. . . . . . . . . $\displaystyle y - \frac{x+y}{2} \;=\;\pm\sqrt{\frac{x+y-2}{2}} $

Square: .$\displaystyle y^2 - y(x+y) + \frac{(x+y)^2}{4} \;=\;\frac{x+y-2}{2}$

I'll letclean it up . . .*you*

- Dec 21st 2012, 07:49 PMzzephodRe: Converting from parametric from to implicit
- Dec 22nd 2012, 07:42 AMPStudent175Re: Converting from parametric from to implicit
Perfect, thanks a lot that was very useful.

For the next question it is:

x = (a-t) / (a+t)

y = (t) / (a+t)

And I have tried the same method by adding together to get x+y= a / a+t but am not really sure to go from there?

Do you know of any resources that has information on this topic as I haven't been taught it yet at university.

Thanks again for the help! - Dec 22nd 2012, 11:47 AMSorobanRe: Converting from parametric from to implicit
Hello again, PStudent175!

Quote:

$\displaystyle \begin{array}{cccc}x &=& \dfrac{a-t}{a+t} & [1] \\ \\[-4mm] y &=& \dfrac{t}{a+t} & [2]\end{array}$

I could find no "clever" way to eliminate the parameter.

Solve [2] for $\displaystyle t\!:\;\;y \:=\:\frac{t}{a+t} \quad\Rightarrow\quad ay + ty \:=\:t$

. . $\displaystyle t - ty \:=\:ay \quad\Rightarrow\quad t(1-y) \:=\:ay \quad\Rightarrow\quad t \:=\:\frac{ay}{1-y}$

Substitute into [1]: .$\displaystyle x \:=\:\dfrac{a-\frac{ay}{1-y}}{a + \frac{ay}{1-y}}$

Muliply by $\displaystyle \frac{1-y}{1-y}\!:\;\;x \;=\;\frac{a(1-y) - ay}{a(1-y) + ay} \;=\;\frac{a-ay - ay}{a-ay + ay} \;=\;\frac{a-2ay}{a} $

Therefore: .$\displaystyle x \;=\;1-2y \quad\Rightarrow\quad \boxed{x + 2y \;=\;1}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Interesting observation . . .

The graph is the straight line $\displaystyle x + 2y \:=\:1$

. . with a "hole" at (-1, 1).

- Dec 22nd 2012, 08:04 PMzzephodRe: Converting from parametric from to implicit
Add them to get

$\displaystyle x+y=\frac{a}{a+t}$

substitute $\displaystyle 1/(a+t)$ back into the equations:

$\displaystyle x=(a-t)(x+y)/a$

$\displaystyle y=t(x+y)/a$

Solve for $\displaystyle t$ from the second of these and substitute into the first and simplify.