What is the slope of the tangent line of # (xy-y/x)(xy-x/y) =C #, where C is an arbitrary constant, at #(-2,1)#?
1 Answer
The slope of the tangent is
Explanation:
We have:
# (xy-y/x)(xy-x/y) =C #
At
# (-2+1/2)(-2+2) =C => C=0#
Multiplying out gives:
# (xy)(xy) - (xy)(x/y) - (xy)(y/x) + (y/x)(x/y) = C#
# x^2y^2 - x^2 - y^2 + 1 = C#
Differentiating wrt
# (x^2)(2ydy/dx) + (2x)(y^2) -2x - 2ydy/dx = 0 #
# :. x^2ydy/dx + xy^2 -x - ydy/dx = 0 #
# :. (x^2y-y)dy/dx + xy^2 -x = 0 #
# :. (x^2y-y)dy/dx = x - xy^2 #
# :. dy/dx = (x - xy^2)/(x^2y-y) #
So at
# :. dy/dx = (-2 - (-2))/(4-1) = 0#
So the slope of the tangent is
The curve of this family are quote interesting:
C=3
C=1
C=0.5
C=0.25
C=0 - The specific curve that (-2,1) lies on
C=-1
C=-2