What is the slope of the tangent line of # (xy-1/(yx))(xy+1/(xy)) =C #, where C is an arbitrary constant, at #(1,5)#?

Question

1303 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-15T01:58:05

I got #-5#

We are told that the point #(1,5)# lies on # (xy-1/(yx))(xy+1/(xy)) =C #.

So we can find #C# to be #25-1/25#. (We won't actually need this.)

As written, this looks tedious to differentiate implicitly, so let's rewrite it as:

#x^2y^2-1/(x^2y^2) = C#.

Or better, still,

#x^4y^4-1=Cx^2y^2#. So

#4x^3y^4+4x^4y^3 dy/dx = 2Cxy^2+2Cx^2y dy/dx#

#2x^2y^3+2x^3y^2 dy/dx = Cy+Cx dy/dx#

Solving for the derivative gets us

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

Evaluating at #(1,5)#, we get

#dy/dx]_"(1,5)" = -(2(5)^3-5C)/(2(5)^2-C) = -(5(2(5)^2-C))/(2(5)^2-C)=-5#

Bonus

Regardless of the constant, at #x=1#, the slope of the tangent will be #-y#