# 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**