Question

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

Answer 1

Equation of normal is `x-y-1=0`

Explanation:

At `x=1`, `f(x)=(x-1)/(x^2-2)=(1-1)/(1^2-2)=0`

So normal or tangent will be at `x=1` will be at `(1,0)` on the curve.

Slope of tangent is given by `f'(x)` which is given by

`f'(x)=((x^2-2)d/dx(x-1)-(x-1)d/dx(x^2-2))/(x^2-2)^2` or

= `((x^2-2)xx1-(x-1)xx2x)/(x^2-2)^2=(x^2-2-2x^2+2x)/(x^2-2)^2` or

= `-(x^2-2x+2)/(x^2-2)^2`

Hence, at `x=1`, `f'(x)=-(1^2-2*1+2)/(1-2)^2=-1`

As slope of tangent is `-1`, slope of normal would be `-1/-1=1`

Hence using point slope form equation of normal is

`(y-0)=1(x-1)` or `x-y-1=0`