- #1

Gundown64

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EDIT: Figured it out. Stupid me. I should have solved in terms of x, giving me x=1-(y+3)^2 as my answer.

x= 1−[itex]t^{2}[/itex], y= t−3, −2 ≤ t ≤ 2

Eliminate the parameter to find a Cartesian equation of the curve for

−5 ≤ y ≤ −1

N/A

I solved for t and got [itex]\sqrt{1-x}[/itex]. Then I plugged it into y=t-3 and got y=[itex]\sqrt{1-x}[/itex]-3. However, that only gives me half of the parabola when I graph it. I know I need y=-[itex]\sqrt{1-x}[/itex]-3 to get the other half, but how do I make that one equation. I don't think I can use a ± sign in my answer.

Thanks in advance!

## Homework Statement

x= 1−[itex]t^{2}[/itex], y= t−3, −2 ≤ t ≤ 2

Eliminate the parameter to find a Cartesian equation of the curve for

−5 ≤ y ≤ −1

## Homework Equations

N/A

## The Attempt at a Solution

I solved for t and got [itex]\sqrt{1-x}[/itex]. Then I plugged it into y=t-3 and got y=[itex]\sqrt{1-x}[/itex]-3. However, that only gives me half of the parabola when I graph it. I know I need y=-[itex]\sqrt{1-x}[/itex]-3 to get the other half, but how do I make that one equation. I don't think I can use a ± sign in my answer.

Thanks in advance!

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