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

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Answer 1 · 2016-03-13T07:19:46

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

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#