curvesparametric
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.
If $x=1-t^2$ and $y=t-2$ then
$$y+2=t$$
$$(y+2)^2=t^2$$
We replace this in the other equation
$$x=1-t^2$$
$$x=1-(y+2)^2$$
$$x-1=-(y+2)^2$$
$$1-x=(y+2)^2$$
NOTE: Although one might solve for $y$, it is not necessary to do so, since the expression $\sqrt{1-x}$ will have to be taken as $\pm \sqrt{1-x}$. Thus, it is better to stick to the above parabola in "$(y,x)$" rather than to a squareroot function in "$(x,y)$".
Now you need to find what are then ranges of $x$ and $y$ for the respective values of $t$. Note you'll have a curve which will not be a function (It will be an horizontal cropped parabola)
Your answer for (a) is correct. Putting that into standard form, we get
$$\frac {x^2}2 - \frac {y^2}2 = 1$$
We therefore see that the graph is a hyperbola centered at the origin with branches to the left and the right. (You do know enough analytical geometry to see that, I hope!) At least, your graph is part of that hyperbola. You need to find which part.
Since both $x$ and $y$ result from square roots, we see that $x \ge 0$ and $y \ge 0$. This means the graph is limited to the first quadrant, cutting out the other three-fourths of the hyperbola. We next want to find the domain for the parameter $t$. From the square roots in the formulas for $x$ and $y$ we see that
$$t+1 \ge 0 \text { and } t-1 \ge 0$$
The intersection for that is
$$t \ge 1$$
Setting $t=1$ we get $x = \sqrt 2$, $y=0$. As $t$ increases both $x$ and $y$ increase, up to infinity. We therefore get confirmation that our graph is the portion of the hyperbola in the first quadrant.
You can plot a few more values of $x$ and $y$ for $t$-values greater than 1, and combine that with your knowledge of analytical geometry that the asymptote is $y=x$ to sketch the graph. Plotting these points in order will confirm the direction of our arrow, which will point up and to the right, toward the corner of your graph area.
Of course, if you have a graphing calculator or program you can easily check all this, but we saw that was not actually necessary. Here is the graph on my calculator, which agrees with our analysis.
Best Answer
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?