How do you find the equation of the tangent and normal line to the curve #y=lnx# at x=e?

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Answer 1 · 2016-09-09T10:38:28

Tangent: #y=x/e#
Normal: #y=e(e-x)+1#

#y=lnx#
#:. y=1# at #x=e#

#y' = 1/x-> # the slope of #f(x) = 1/x#

The equation of a straight line with slope #(m)# passing throught point #(x_1,y_1)# is;

#(y-y_1) = m(x-x_1)#

Since #m# at #x=e# is #1/e#

The equation of tangent to #f(x)# at #(e,1)# is
#(y-1) = 1/e(x-e)#

#y= x/e -1 +1 #

#y=x/e#

And equation of normal to #f(x)# at #(e,1)# is
#(y-1) = -e(x-e)# (Since slope of normal #=-1/m#)

#y= -ex+e^2+1#

#y=e(e-x)+1#