Find the equations of the tangent lines at the point where the curve crosses itself.

The parametric equations are:

  ------------------(1)

          -----------------(2)

From equation 2,

Substitute in equation (1),

  ----------------(3)

Now substitute in equation (1),

   ----------------(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,

Plug in the value of y...

The parametric equations are:

  ------------------(1)

          -----------------(2)

From equation 2,

Substitute in equation (1),

  ----------------(3)

Now substitute in equation (1),

   ----------------(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,

Plug in the value of y in equation 4,

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,

The derivative is the slope of the line tangent to the parametric graph 

For , 

Equation of the tangent line can be found by using point slope form of the line,

For ,  

Equation of the tangent line will be:

Equations of the tangent line where the curve crosses itself are:

  and