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Sketch the curve by using the parametric equation to plot points. Indicate with arrow the direction in which the curve is traced as $t$ increases.

$$x=t^2+t$$ $$y=t^2-t$$

I tried to convert this parametric equation into a Cartesian equation, but I didn't know how. Please help.

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    $\begingroup$ How did you try? What manipulations of it did you use? Did any of them get you a weird result? What have you found? $\endgroup$ – JMCF125 Dec 27 '13 at 12:14
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    $\begingroup$ I don't think converting to Cartesian coordinates is what you were intended to do. Rather, select values of $t$ and then plot the points $(x,y)$ given by the parameterization. Then, informally deduce what the curve is. Note if you just look at the Cartesian equation, you won't be able to determine the direction in which the curve is traced out. $\endgroup$ – David Mitra Dec 27 '13 at 12:16
  • $\begingroup$ I have tried this x=t(t+1) and y=t(t-1), then t=y/(t-1)=x/(t+1) but I didn't know how to eliminate t $\endgroup$ – user32104 Dec 27 '13 at 12:18
  • $\begingroup$ @DavidMitra thank you but how to know the shape without use points (because our teacher says we will not be able to draw table for points in the exam) $\endgroup$ – user32104 Dec 27 '13 at 12:21
  • $\begingroup$ Well, the problem is oddly phrased then, it clearly indicates (to me at least) to use a "table method". $\endgroup$ – David Mitra Dec 27 '13 at 12:25
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HINT:

On addition we have $\displaystyle 2t^2=x+y\iff t^2=\frac{x+y}2$

On subtraction, $\displaystyle 2t=x-y\iff t=\frac{x-y}2$

Can you eliminate $t$ from here?

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    $\begingroup$ Thank you, do you mean $((x-y)/2)^2=((x+y)2)$? $\endgroup$ – user32104 Dec 27 '13 at 12:12
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    $\begingroup$ @user32104, my pleasure. Nice to hear that you could complete $\endgroup$ – lab bhattacharjee Dec 27 '13 at 14:44
  • $\begingroup$ how about x = 4 + 2t and y = 5 + 2t? $\endgroup$ – Hasan Iqbal Apr 4 '16 at 11:01
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HINT : $t^2=x-t=y+t$. So, you can represent $t$ by $x,y$.

Since we have $t=\frac{x-y}{2}$, we have $$x=\left(\frac{x-y}{2}\right)^2+\frac{x-y}{2}.$$

We can get the following equation with $x$ and $y$ :

$$x^2-2xy+y^2-2x-2y=0.$$

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