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I am trying to find the cartesian equation of the parametric curve $$x = 1-t^2, y = t-2, -2 \leq t \leq 4$$

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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"So I put that in". Put that where? Also: if $t=\sqrt{x-1}$, then you are restricting $t$ to be positive; what happened to all the values of $t$ between $-2$ and $0$? And what about $y$? –  Arturo Magidin Jun 24 '12 at 22:51
    
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. –  Arturo Magidin Jun 24 '12 at 22:52
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3 Answers

up vote 3 down vote accepted

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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I am a little confused at the flow of this problem, did you know that you had to make $t^2$ equal to something and then put that in or did you realize that later and go back? –  user138246 Jun 24 '12 at 22:59
    
Since I had a $t^2$ in the other equation I just decided to find $t^2$ in terms of $y$, which allows me to eliminate $t$ and obtain an equation in $x,y$ only. –  Pedro Tamaroff Jun 24 '12 at 23:01
    
How do I find the range? –  user138246 Jun 24 '12 at 23:10
    
If $t$ ranges from $-2$ to $4$, what does $x=1-t^2$ range through? Similar for $y$. –  Pedro Tamaroff Jun 24 '12 at 23:13
    
@AndréNicolas True. I'll correct it. –  Pedro Tamaroff Jun 24 '12 at 23:38
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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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