What are the coordinates of the turning points of #y^3+3xy^2-x^3=3#?

Question

I got one solution of #(1,1)#, but the other solution is at #(root(3)(3),-root(3)(3))# and i have no idea how to get it.

5723 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-13T01:42:32

#(1,1)# and #(1,-1)# are the turning points.

#y^3+3xy^2-x^3=3#

Using implicit differentiation,
#3y^2times(dy)/(dx)+3xtimes2y(dy)/(dx)+3y^2-3x^2=0#

#(dy)/(dx)(3y^2+6xy)=3x^2-3y^2#

#(dy)/(dx)=(3(x^2-y^2))/(3y(y+2x))#

#(dy)/(dx)=(x^2-y^2)/(y(y+2x)#

For turning points, #(dy)/(dx)=0#

#(x^2-y^2)/(y(y+2x)=0#

#x^2-y^2=0#

#(x-y)(x+y)=0#

#y=x# or #y=-x#

Sub #y=x# back into the original equation

#x^3+3x*x^2-x^3=3#

#3x^3=3#

#x^3=1#

#x=1#
Therefore #(1,1)# is one of the 2 turning points

Sub #y=-x# back into the original equation

#x^3+3x*(-x)^2-x^3=3#

#3x^3=3#

#x^3=1#

#x=1#
Therefore, #(1,-1)# is the other turning point

#root(3)3=1#
#-root(3)3=-1#
So you were missing the turning point #(1,-1)#