I am trying to find the cartesian equation of the parametric curve
$$x = 1-t^2, \qquad y = t-2, \qquad -2 \leq t \leq 4$$
I am not sure how to proceed but I think a good direction would be to get everything in terms of $x$, eliminating $t$. I get $t = \sqrt{x-1}$ so I put that in and I get a wrong answer. What is wrong with my solution?
Best Answer
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)