Question

What is the equation of the line normal to `f(x)=x/(x-2)` at `x=1`?

Answer 1

Equation of normal is `x-2y-3=0`

Explanation:

The line normal to `f(x)` at `x=x_0` is perpendicular to the tangent at the point `(x_0,f(x_0))`.

As slope of the tangent at `x=x_0` is `f'(x_0)`, the slope of normal is `-1/(f'(x_0))` and equation of normal is `y-f(x_0)=-1/(f'(x_0))(x-x_0)`.

Here we are seeking normal at `x=1` and as `f(x)=x/(x-2)`, at this point `f(1)=1/(1-2)=-1`.

For slope let us work out derivative of `f(x)=x/(x-2)=1+2/(x-2)`

and `f'(x)=-2/(x-2)^2` and slope of tangent at `x=1` is `f'(1)=-2/(1-2)^2=-2`

The slope of normal is then `-1/-2=1/2`

and equation of tangent is `y-(-1)=-2(x-1)` or `2x+y-1=0` and that of

normal is `y-(-1)=1/2(x-1)` or `2y+2=x-1` or `x-2y-3=0`
graph{(x-2y-3)(xy-x-2y)(2x+y-1)=0 [-10, 10, -5, 5]}