For a function #y=f(x)#, the slope of tangent at #x=x_0# i.e. at #(x_0.f(x_0))# is given by #f'(x_0)#, where #f'(x)=(dy)/(dx)#
and hence slope of normal is #-1/(f'(x_0))# as normal is perpendicular to tangent.
and while equation of tangent is #y-f(x_0)=f'(x_0)(x-x_0)#
and equation of normal is #y-f(x_0)=-1/(f'(x_0))(x-x_0)#
Here we have #f(x)=x^3/sqrt(16x^2-x)# and at #x_0=2#. As #f(x_0)=8/sqrt62#, hence we are seeking normal at point #(2,8/sqrt62)# or #(2,1.016)#
Further using quotient rule #f'(x)=(3x^2sqrt(16x^2-x)-x^3((32x-1)/(2sqrt(16x^2-x))))/(16x^2-x)#
and #f'(2)=(12sqrt62-8(63/(2sqrt62)))/62#
and slope of normal is #-62/(12sqrt62-8(63/(2sqrt62)))#
= #-sqrt62/(12-504/124)=-sqrt62/(12-126/31)#
= #-(31sqrt62)/(372-126)=-(31sqrt62)/246=-0.992253#
and equation of normal is #y-1.016=-0.992253(x-2)#
graph{(y-1.016+0.992253(x-2))(x^3/sqrt(16x^2-x)-y)=0 [-10, 10, -5, 5]}
