Find the slope of the tangent line to the graph at the given point.

Question

x^3 + y^3 – 6xy = 0, (4/3, 8/3)

Answer 1

x^3 + y^3 - 6xy = 0
3x^2 + 3y^2 y' - 6y - 6xy' = 0
(3y^2-6x)y' = 6y-3x^2

y' = (2y-x^2)/(y^2-2x)
So, at (4/3,8/3) the slope is (2(8/3)-(4/3)^2)/((8/3)^2-2(4/3)) = 4/5

Using the point-slope form of the line, its equation is

y - 8/3 = 4/5 (x - 4/3)

see the graphs at

www.wolframalpha.com/input/?i=plot+x%5E3+%2B+y%5E3+-+6xy+%3D+0,+y+%3D+4%2F5+(x-4%2F3)%2B8%2F3