Question

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

Answer 1

Equation of normal at `x=pi/3` is `(3sqrt3-8)y-sqrt3x-2sqrt3pi+8pi/3+3sqrt3=0`

Explanation:

If `f(x)=tan^2x-3x`,

`f(pi/3)=tan^2(pi/3)-3xxpi/3=(sqrt3)^2-pi=3-pi`.

Hence normal should pass through `(3-pi,pi/3)`

Slope of the curve at `x` is given by differential of `f(x)` i.e.

`f'(x)=2tanx xxsec^2x-3`.

Hence slope of tangent at `x=pi/3` is

`f'(pi/3)=2tan(pi/3)xxsec^2(pi/3)-3=2sqrt3xx(2/sqrt3)^2-3=8/sqrt3-3=(8-3sqrt3)/sqrt3` and as product of slopes of tangent and normal should be `-1`

Hence, slope of normal is `sqrt3/(3sqrt3-8)`

Hence, equation of normal using point slope form is

`(y-pi/3)=sqrt3/(3sqrt3-8)(x-3+pi)` or

`(3sqrt3-8)y-sqrt3x-sqrt3pi+8pi/3+3sqrt3-sqrt3pi=0`

or `(3sqrt3-8)y-sqrt3x-2sqrt3pi+8pi/3+3sqrt3=0`