`x=t^2-t , y=t^3-3t-1` Find the equations of the tangent lines at the point where the curve crosses itself.

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Given parametric equations are:



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))`







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

Using point slope form of the equation,



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




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

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


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