Find cartesian equation from parametric equation

[trollcast]

Homework Statement

A curve is defined by the parametric equations:

$$x=tan(t-1)\ \ \ \ \ \ \ y=cot^2(t+1)$$

Homework Equations

The Attempt at a Solution

I think rearranging the first equation for t gives:

$$t=tan^{-1}(x)+1$$

However that doesn't help me as I don't know how to simplify the equation I'd get if I sub t into the y equation?

 

[mfb]

Do you have to simplify the equation? You found an equation y=f(x).

 

[trollcast]

Do you have to simplify the equation? You found an equation y=f(x).

I actually just realized that I've misread the equations so I can see how to solve it now.

The equations should be:

$$x=tan(t)-1\qquad y=cot^2(t)+1$$

So the solution would be:

$$tan(t)=x+1$$
$$y=\frac{1}{tan^2(t)}$$
$$y=\frac{1}{(x+1)^2}$$

 

[Dick]

I actually just realized that I've misread the equations so I can see how to solve it now.

The equations should be:

$$x=tan(t)-1\qquad y=cot^2(t)+1$$

So the solution would be:

$$tan(t)=x+1$$
$$y=\frac{1}{tan^2(t)}$$
$$y=\frac{1}{(x+1)^2}$$

What happened to the '+1' in $y=cot^2(t)+1$?

 

[trollcast]

What happened to the '+1' in $y=cot^2(t)+1$?

Oops, so it should be:

$$x=tan(t)-1\qquad y=cot^2(t)+1\\
tan(t)=x+1\\

y - 1=\frac{1}{tan^2(t)}\\
y=\frac{1}{(x+1)^2} +1\\
y=\frac{(x+1)^2+1}{(x+1)^2}$$

 

[Dick]

Oops, so it should be:

$$x=tan(t)-1\qquad y=cot^2(t)+1\\
tan(t)=x+1\\

y - 1=\frac{1}{tan^2(t)}\\
y=\frac{1}{(x+1)^2} +1\\
y=\frac{(x+1)^2+1}{(x+1)^2}$$

Sure.