# Parametic y= sqrt(t +1) and y = sqrt(t-1)

1. Nov 5, 2013

### Jbreezy

1. The problem statement, all variables and given/known data

I have to eliminate the parameter and write in Cartesian form.

2. Relevant equations

y= sqrt(t +1) and y = sqrt(t-1)
3. The attempt at a solution

If you were to just go for the gusto and square it out you will end up with 0 = 2. Clearly I'm missing something here. Just stumped. I don't think the right way to go is squaring and setting them equal.

2. Nov 5, 2013

### cepheid

Staff Emeritus
Can you post the *exact* problem statement please?

3. Nov 5, 2013

### Jbreezy

Yep,

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

y = sqrt(t +1) , y = sqrt(t -1)

4. Nov 5, 2013

### cepheid

Staff Emeritus
I think perhaps one of the y's is supposed to be an x.

5. Nov 5, 2013

### Jbreezy

Ha It is possible. There was a misprint on one of the graphs where it was supposed to say seconds and it was like hours. So I suppose. I just didn't know if I was missing something. Seems like you would just do as I said and square it and yatta but you get 0 = 2.

6. Nov 5, 2013

### cepheid

Staff Emeritus
Unless I am missing something obvious, I don't think it has a solution as given.

7. Nov 5, 2013

### Jbreezy

Not sure maybe someone else has ideas too?

8. Nov 6, 2013

### Dick

Yeah, cepheid is right. There is clearly a typo in the problem. It's just expressing y in terms of two contradictory equations in t. There should be an x in there somewhere.

9. Nov 6, 2013

### HallsofIvy

Staff Emeritus
These are NOT "parametric equations" and CANNOT be put in "Cartesian" form without an "x". As others have suggested, it must be $x= \sqrt{t+ 1}$ and $y= \sqrt{t- 1}$. Yes, square both, then solve for t and set the two equations for t equal.

10. Nov 6, 2013

### Jbreezy

OK then there is a typo in my book. Thanks for the help I will bring it up in lecture.