Parametric equations with t^2=/- t

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orsi22

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In dealing with parametric equations, I believe the normal route is to solve for t, then incorporate that into the x or y equations to eliminate the t parameter. However, I have x=t^2+t, y=t^2-t. I can't figure out how to eliminate the t. It's probably simple algebra, but I'm stuck. Any advice on how to get these equations into rectangular form?
 
\(\displaystyle x=t^{2}+t\)

Just use the quadratic formula.

\(\displaystyle t=\frac{\sqrt{4x+1}-1}{2}, \;\ t=\frac{-(\sqrt{4x+1}+1)}{2}\)
 
Sorry, but I don't really understand your answer. I don't know how having x=an equation with x in it is going to work. Also, I'm not quite sure how you used the quadratic and ended up with the answer that you did. When using the quadratic, you're solving an equation set = 0 right? In which case, to solve when x=0, you could simply factor to t(t+1), so x=0,-1. Right? Ultimately I need to have an equation as y=mx+b.
Thanks,
Nick
 
I don't know how having x=an equation with x in it is going to work.
Click to expand...

Orsi22,

It's just a minor typo. Galactus said he was using the quadratic formula. You should have figured that out. It should be "t =", not "x = ".

Now apply the same thing for the equation with y.
 
I feel like an idiot, but I still don't get it. Out of the equation t^2+t, I'm assuming that we're using 1 as a, 1 as b, and 0 as c right? So the quadratic would look like -1+/-(square root 1-4(0))/2, which would simplify as to 0, -1. I think I'm missing the key step on how to throw the x into the equation, since I've only used the formula when an equation is set to 0.
Does this make any sense?
 
See the trick. You have to set \(\displaystyle t^{2}+t-x=0\) and use c = -x as the constant.
 
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