Converting from parametric from to implicit

**PStudent175** · December 21st 2012, 07:14 AM

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?

**HallsofIvy** · December 21st 2012, 09:39 AM

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.

**Soroban** · December 21st 2012, 06:29 PM

Hello, PStudent175!

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

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

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

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

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

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

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

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

I'll let you clean it up . . .

**zzephod** · December 21st 2012, 07:49 PM

Originally Posted by Soroban

Hello, PStudent175!

I have a solution, . . .

Soroban has the right idea, but instead of adding them should have subtracted the equations to give $2s=y-x$ ...

**PStudent175** · December 22nd 2012, 07:42 AM

Originally Posted by Soroban

Hello, PStudent175!

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

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

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

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

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

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

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

I'll let you clean it up . . .

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!

**Soroban** · December 22nd 2012, 11:47 AM

Hello again, PStudent175!

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

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

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

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

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

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

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

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

Interesting observation . . .

The graph is the straight line $x + 2y \:=\:1$
. . with a "hole" at (-1, 1).

**zzephod** · December 22nd 2012, 08:04 PM

Originally Posted by PStudent175

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!

Add them to get

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

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

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

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

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

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