# find the coordinates `(x,y)` of the point in the first quadrant at which the tangent line to the curve ` x^3-xy+y^3=0` is parallel to the x-axis.

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### 1 Answer

To find the tangent, differentiate the given curve implicitly:

`x^3-xy+y^3=0`

So, `3x^2-y-xy'+3y^2y'=0`

`rArr 3x^2-y-y'(x-3y^2)=0`

`rArr y'=(y-3x^2)/( 3y^2-x)`

This is the gradient of tangent to the given curve.

By the condition of the problem, the tangent is parallel to the x-axis. So its gradient must be zero.

So, `(y-3x^2)/( 3y^2-x)=0`

`rArr y=3x^2`

Putting this value of y in the curve,

`x^3-x*3x^2+(3x^2)^3=0`

`rArr -2x^3+27x^6=0`

`rArrx^3(27x^3-2)=0`

This equation has two solutions: `x=0` , and

`x=root(3)(2/27)= root(3)(2)/3`

When `x=0, y=3*0^2=0` , and

when `x= root(3)(2)/3, y= 3*(root(3)(2)/3)^2=root(3)(4)/3` .

Thus, there will be two points in the first quadrant where our tangent will be parallel to the curve `x^3-xy+y^3=0` , the coordinates of which are `(0,0)` and `(root(3)(2)/3, root(3)(4)/3)`.

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