Question

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)`?

Answer 1

`-5`

Explanation:

`f(x,y) = (x y - 1/(y x)) (x y + 1/(x y))-C=0` so

`(dy)/(dx) = - f_x/(f_y) = (2/(x^3 y^2) + 2 x y^2)/(2/(x^2 y^3) + 2 x^2 y) = -y/x`

In the point `{x,y} = {1,5}` we have

`(dy)/(dx)=-5`