Question

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

Answer 1

`x+(sqrt3+(4pi)/9)y-(4pi)/3(1+pi/(3sqrt3))=0`

Explanation:

Let us find the slope of the tangent, which should give us slope of normal (as it is perpendicular to tangent and hence product of their slopes would be `-1`).

Slope of tangent is given by first derivative and this is

`f'(x)=1xxtanx+x xxsec^2x` and as we need slope of tangent at `x=pi/3`,

`f'([i/3)=tan(pi/3)+pi/3xxsec^2(pi/3)=sqrt3+pi/3xx(2/sqrt3)^2`

= `sqrt3+(4pi)/9`

and slope of normal is `-1/(sqrt3+(4pi)/9)`

As normal is needed at `(x,y)` with `x=pi/3`, `(x,y)` is `(pi/3,pi/3xxtan(pi/3))` or `(pi/3,pi/sqrt3)`

Hence, using point slope form of equation `(y-y_1)=m(x-x_1)`

Equation of normal is `(y-pi/sqrt3)=-1/(sqrt3+(4pi)/9)(x-pi/3)` or

`(sqrt3+(4pi)/9)(y-pi/sqrt3)=pi/3-x`

`sqrt3y-pi+(4piy)/9-(4pi^2)/(9sqrt3)-pi/3+x=0` or

`x+(sqrt3+(4pi)/9)y-(4pi)/3(1+pi/(3sqrt3))=0`