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

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Answer 1 · 2015-11-15T22:16:54

#y=1/(3e^9)x-1/e^9+e^9#

If tangent line is #y=kx+n# then #k=f'(x_0)# where #x_0=3#.

#f'(x)=e^(x^2/(x-2)) * (2x(x-2)-x^2)/(x-2)^2#

#k = f'(3)=-3e^9#

Equation of the normal is #y-y_0 = k_1(x-x_0)# where:

#x_0=3 => y_0=f(x_0)=f(3)=e^9# and

#k*k_1=-1 => k_1=-1/k = 1/(3e^9)#

#y-e^9=1/(3e^9)(x-3)#

#y=1/(3e^9)(x-3)+e^9#

#y=1/(3e^9)x-1/e^9+e^9#