Question

Find the coordinates 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.

Answer 1

Differentiate implicitly:

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

Solve for y

`y'(3y^2-x)=y-3x^2 \\ \\y'={y-3x^2\over3y^2-x}`

When the tangent is parallel to the x-axis we have y'=0, so we must solve

`y'={y-3x^2\over3y^2-x}=0\implies y=3x^2`

To find the actual value of x we plug this expression for y into the original equation

`x^3-3x^3+27x^6=0 \\ \\x^3(27x^3-2)=0\implies x=\{0,{\sqrt[3]2\over3}\}`

Plugging this into the formula for y above gives the points

`(0,0)\text{ and }({\sqrt[3]2\over3},{\sqrt[3]4\over3})`

which is where our tangent will be parallel to the x-axis.