Calculus of a Single Variable, Chapter 10, 10.3, Section 10.3, Problem 25

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