Recall that, #f'(2)# gives the slope of the Tangent Line to the
Curve C : #y=f(x)=2x^2-xsqrt(x^2-x)#
at the pt. #P(2,f(2))=P(2,8-2sqrt2)#.
As Normal at pt. #P# is #bot# to the tgt., its slope is given by,
#1/(f'(2)), if f'(2)!=0#.
Now, #f(x)=2x^2-xsqrt(x^2-x) = 2x^2-sqrt(x^4-x^3).#
#:. f'(x)=2(2x)-1/(2sqrt(x^4-x^3))*(x^4-x^3)'#
#=4x-1/(2sqrt(x^4-x^3))*(4x^3-3x^2)#
#=4x-(x^2(4x-3))/(2xsqrt(x^2-x))#
#:. f'(x)=4x-(x(4x-3))/(2sqrt(x^2-x))#
#rArr f'(2)=8-(2(8-3))/(2sqrt(4-2))=8-5/sqrt2=(8sqrt2-5)/sqrt2 !=0.#
Hence, the Reqd. Slope #=sqrt2/(8sqrt2-5)=(sqrt2(8sqrt2+5))/{(8sqrt2)^2-5^2}#
#=(16+5sqrt2)/103#.
