Cartesian equation of a parametric curve

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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?

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3 Answers

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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)

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Hint: Let $t = y + 2$. Then plug into the equation for $x$.

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Solving for $t$ using the $x$ equation involves square roots, which involve sign issues. Properly, you have $$ t = \pm \sqrt{1-x} $$ To avoid this multiple value issue, use the $y$ equation instead to solve for $t$ and plug that into the $x$ equation. I think you're going to get a parabola opening horizontally.

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