Calculus of a Single Variable, Chapter 10, 10.3, Section 10.3, Problem 26

The parametric equations are:
x=t^3-6t ------------------(1)
y=t^2 -----------------(2)
From equation 2,
t=+-sqrt(y)
Substitute t=sqrt(y) in equation (1),
x=(sqrt(y))^3-6sqrt(y)
=>x=ysqrt(y)-6sqrt(y) ----------------(3)
Now substitute t=-sqrt(y) in equation (1),
x=-ysqrt(y)+6sqrt(y) ----------------(4)
The curve will cross itself at the point, where x and y values are same for different values of t.
So setting the equations 3 and 4 equal will give the point,
ysqrt(y)-6sqrt(y)=-ysqrt(y)+6sqrt(y)
=>ysqrt(y)+ysqrt(y)=6sqrt(y)+6sqrt(y)
=>2ysqrt(y)=12sqrt(y)
=>2y=12
=>y=6
Plug in the value of y in equation 4,
x=-6sqrt(6)+6sqrt(6)
=>x=0
So the curve crosses itself at the point (0,6). Note that,we can find this point by plotting the graph also.
Now let's find t for this point,
t=+-sqrt(y)=+-sqrt(6)
The derivative dy/dx is the slope of the line tangent to the parametric graph (x(t),y(t))
dy/dx=(dy/dt)/(dx/dt)
y=t^2
=>dy/dt=2t
x=t^3-6t
=>dx/dt=3t^2-6
dy/dx=(2t)/(3t^2-6)
For t=sqrt(6) , dy/dx=(2sqrt(6))/(3(sqrt(6))^2-6)=(2sqrt(6))/(18-6)=sqrt(6)/6
Equation of the tangent line can be found by using point slope form of the line,
y-6=sqrt(6)/6(x-0)
=>y=sqrt(6)/6x+6
For t=-sqrt(6) , dy/dx=(2(-sqrt(6)))/(3(-sqrt(6))^2-6)=(-2sqrt(6))/(18-6)=(-sqrt(6))/6
Equation of the tangent line will be:
y-6=(-sqrt(6))/6(x-0)
=>y=(-sqrt(6))/6x+6
Equations of the tangent line where the curve crosses itself are:
y=sqrt(6)/6x+6 and y=-sqrt(6)/6x+6