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