Given parametric equations are:

`x=t^2-t`

`y=t^3-3t-1`

We have to find the point where the curves cross.

Let's draw a table for different values of t, and find different values of t which give the same value of x and y ,this will be the point where the curves cross. (Refer the attached image).

So the curves cross at the point (2,1) for t= -1 and 2

Derivative `dy/dx` is the slope of the line tangent to the parametric graph `(x(t),y(t))`

`x=t^2-t`

`dx/dt=2t-1`

`y=t^3-3t-1`

`dy/dt=3t^2-3`

`dy/dx=(dy/dt)/(dx/dt)`

`dy/dx=(3t^2-3)/(2t-1)`

At t=-1, `dy/dx=(3(-1)^2-3)/(2(-1)-1)=0`

Using point slope form of the equation,

`y-1=0(x-2)`

`=>y=1`

At t=2, `dy/dx=(3(2)^2-3)/(2(2)-1)=(12-3)/(4-1)=9/3=3`

`y-1=3(x-2)`

`y-1=3x-6`

`=>y=3x-5`

Equations of the tangent lines where the given curve crosses itself are:

`y=1 , y=3x-5`

## See eNotes Ad-Free

Start your **48-hour free trial** to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Already a member? Log in here.