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

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The parametric equations are:

`x=t^3-6t`  ------------------(1)

`y=t^2`          -----------------(2)

From equation 2,


Substitute `t=sqrt(y)` in equation (1),


`=>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,






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 `dy/dx` is the slope of the line tangent to the parametric graph `(x(t),y(t))`







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,



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:



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

`y=sqrt(6)/6x+6`  and `y=-sqrt(6)/6x+6`

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