What is the equation of the line that is normal to #f(x)= sqry(x-2) # at # x=5 #?

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Answer 1 · 2017-05-23T14:52:53

Equation of tangent is #x-2sqrt3y+1=0#

We are seeking equation at #x=5# on the curve #f(x)=sqrt(x-2)# i.e. at #(5,sqrt3)# as #f(5)=sqrt(5-2)=sqrt3#

As slope of tangent at #f(5)# is #f'(5)# let us first find #f'(x)#

#f'(x)=1/(2sqrt(x-2))# and hence slope of tangent is

#1/(2sqrt(5-2))=1/(2sqrt3)#

and equation of tangent is #y-sqrt3=1/(2sqrt3)(x-5)#

or #x-2sqrt3y+1=0#

graph{(x-2sqrt3y+1)(y-sqrt(x-2))=0 [0.81, 10.81, -0.56, 4.44]}