Question

Could someone please help with this:
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 implicitely:\[3x^2-y-xy'+3y^2y'=0\]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.