Question

What is the equation of the normal line of `f(x)= 3lnsin2x` at `x = pi/8`?

Answer 1

Equation of normal is `x+6y-36-pi/8=0`

Explanation:

While slope of tangent at curve `f(x)` at `x=a` is given by `f'(a)`, as lope of normal at `x=a` is `(-1)/(f'(a))` as normal is perpendicular to the tangent.

Both pass through the point `(a,f(a))` on the curve.

Here we have `f(x)=3lnsin2x` and hence

`f'(x)=3/(sin2x)xx2cos2x=6cot2x`

and hence we are seeking a normal at `(pi/8,6cot(pi/4))` or `(pi/8,6)`

and as slope of normal is `(-1)/(6cot(pi/4))=-1/6`

and as equation of line passing through `(x_1,y_1)` and having a slope `m` is `y-y_1=m(x-x_1)` and hence equation of normal is

`y-6=(-1)/6(x-pi/8)`

or `6y-36=-x+pi/8`

or `x+6y-36-pi/8=0`