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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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    If you are trying to eliminate the parameter, it would be much simpler to solve for $t$ form the equation $y=t-2$, and then substitute *that* into the equation for $x$; that will give you an expression that involves only $x$ and $y$, and no $t$s.2012-06-24

3 Answers 3

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

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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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    @AndréNicolas True. I'll correct it.2012-06-24