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I am currently working on parametric equations. I'm asked to express the parametric equations in Cartesian form and I can't since I can't make $t$ the subject for either equation.

$x=2t + t^2$

$y=2t^2 + t^3$

The question specifically says, by considering $\frac{y}{x}$, find a Cartesian form of the equations.

This lead me to:

$\frac{y}{x}=\frac{2t+t^2}{2+t}$

I'm struggling to see what I need to do from here. Any help is very much appreciated.

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  • $\begingroup$ Can you factor a $t$ out of the numerator? $\endgroup$ – Michael Burr Feb 16 '17 at 18:34
  • $\begingroup$ $$\frac{y}{x}=\frac{2t+t^2}{2+t} \iff \frac{y}x = \frac{t(2+t)}{2+t} = t$$ So $$x= \frac{2y}{x}+\left(\frac yx\right)^2$$ and $y = \;?$ $\endgroup$ – Namaste Feb 16 '17 at 18:55
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...so $\frac yx=t$ and you can substitute for $t$ to get the Cartesian form

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  • $\begingroup$ Thanks, I've been doing these for ages and I knew it'd be something stupid that I couldn't see (embarrassing really). Thanks a lot! $\endgroup$ – Doug Feb 16 '17 at 18:38
  • $\begingroup$ No problem at all $\endgroup$ – David Quinn Feb 16 '17 at 18:39
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HINT: from $$x=2t+t^2$$ we get $$t_{1,2}=-1\pm\sqrt{1+x}$$ and so you can eliminate $t$

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Note that $$y=2t^2+t^3=t(2t+t^2)=tx$$ $$\implies t=\frac{y}{x}$$ So we have $$x=2t+t^2$$ $$x=2\cdot \frac{y}{x}+\left(\frac{y}{x}\right)^2$$ $$x^3-y^2-2xy=0$$

Hope you can complete it now.

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  • $\begingroup$ Thanks, I see it now, I just could not make the connection I needed to. I found $\frac{y}{x}=t$ but couldn't for the life of me see that I had to sub it in to either of the original equations... sorry for wasting your time! $\endgroup$ – Doug Feb 16 '17 at 18:40
  • $\begingroup$ No problem at all ;) $\endgroup$ – SchrodingersCat Feb 16 '17 at 18:41
  • $\begingroup$ @Doug Dont forget to upvote the answers you like and accept the most helpful one. $\endgroup$ – SchrodingersCat Feb 16 '17 at 18:44
  • $\begingroup$ Please, @SchrodingersCat refrain from soliciting an accept vote and/or upvotes. I know you've framed the phrase try and make it look like you have no self-interest invested in it, but please.... It makes you look a bit pathetic to many users on this site. $\endgroup$ – Namaste Feb 16 '17 at 18:49
  • $\begingroup$ @amWhy I understand your concern but I was informing him about the site features. Soliciting an accept vote is not the aim. Doug, if you are reading this, do not misunderstand my previous statement. $\endgroup$ – SchrodingersCat Feb 16 '17 at 18:54

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