# `x=t^3-6t , y=t^2` Find the equations of the tangent lines at the point where the curve crosses itself. 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`