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?
Hello, PStudent175!
I have a solution, but it's quite awkward.
$\displaystyle \text{Eliminate the parameter: }\:\begin{Bmatrix}x &=& s^2-s+1 & [1] \\ y &=& s^2+s+1 & [2] \end{Bmatrix}$
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 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!
Hello again, PStudent175!
$\displaystyle \begin{array}{cccc}x &=& \dfrac{a-t}{a+t} & [1] \\ \\[-4mm] y &=& \dfrac{t}{a+t} & [2]\end{array}$
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).