Question

What is the equation of the line that is normal to `f(x)= sqry(x-2)` at `x=5`?

Answer 1

Equation of tangent is `x-2sqrt3y+1=0`

Explanation:

We are seeking equation at `x=5` on the curve `f(x)=sqrt(x-2)` i.e. at `(5,sqrt3)` as `f(5)=sqrt(5-2)=sqrt3`

As slope of tangent at `f(5)` is `f'(5)` let us first find `f'(x)`

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

`1/(2sqrt(5-2))=1/(2sqrt3)`

and equation of tangent is `y-sqrt3=1/(2sqrt3)(x-5)`

or `x-2sqrt3y+1=0`

graph{(x-2sqrt3y+1)(y-sqrt(x-2))=0 [0.81, 10.81, -0.56, 4.44]}