Question

Find all points on the graph where the tangent line is horizontal?

`x^2y-2x+18y=3`

Enter your answers as a comma separated list of ordered pairs (x,y)

Answer 1

Points at which tangents are horizontal are `(3,1/3)` and `(-6,-1/6)`

Explanation:

Differentiating `x^2y-2x+18y=3`, we get

`2xy+x^2(dy)/(dx)-2+18(dy)/(dx)=0`

or `(x^2+18)(dy)/(dx)=2-2xy`

or `(dy)/(dx)=(2-2xy)/(x^2+18)`

A function has horizontal tangent lines at points where `(dy)/(dx)=0`

i.e. at `(2-2xy)/(x^2+18)=0`

or `2-2xy=0` i.e. `xy=1` or `y=1/x`

Substituting value of `y` in `x^2y-2x+18y=3`, we get

`x^2/x-2x+18/x=3`

or `x^2-2x^2+18=3x`

or `x^2+3x-18=0`

or `(x+6)(x-3)=0`

Hence points are `(3,1/3)` and `(-6,-1/6)`

graph{(x^2y-2x+18y-3)((x-3)^2+(y-1/3)^2-0.01)((x+6)^2+(y+1/6)^2-0.01)=0 [-6.625, 3.375, -1.92, 3.08]}