Question

What is the equation of the normal line of `f(x)=x/(2-x^2)` at `x = 3`?

Answer 1

`77y=-343x+996`

Explanation:

First we find the y-coordinates;
Substitute x=3
`3/(2-(3)^2)=-3/7`

Then differentiate `x/(2-x^2)`
`dy/dx`=`[x^2+2]/(x^2-2)^2`

Substitute 3 to find the gradient of the tangent
`dy/dx`=`[3^2+2]/[3^2-2]^2`
=`11/49`

General equation of normal:
`(y-y_1)`=-1/(gradient of tangent)`(x-x_1)`

Substituting the values
`(y-(-3/7))=-1/(11/49)(x-3)`
`y+3/7=-49/11x+147/11`
`y=-49/11x+996/77`
`77y=-343x+996`