Question

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

Answer 1

`y=1/3x-4/3`

Explanation:

In order to find the normal to the tangent we first need to find the gradient of the tangent line at `x=1`

Differentiate `f(x)`

`d/dx(x(x^2-2)^(-1))=(x^2-2)^(-1)-(x^2-2)^(-2)*2x*x`

`->=1/((x^2-2))- (2x^2)/(x^2-2)^2=((x^2-2)-2x^2)/(x^2-2)^2=(-x^2-2)/(x^2-2)^2=- (x^2+2)/(x^2-2)^2`

Gradient of tangent at `x=1`

`- (1^2+2)/(1^2-2)^2=-3`

gradient of normal:

`-3*m_2=-1=>m_2=1/3`

Coordinates at `x=1`

`( 1 , -1 )`

Equation of normal:

`y+1=1/3(x-1)=>y=1/3x-4/3`

Graph: