What is the equation of the normal line of #f(x)=(x-1)^2/(x-5)# at #x=4 #?

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Answer 1 · 2017-02-11T09:21:43

#x-15y-139=0#. See the normal-inclusive Socratic graph.

#y=-9#, at #x = 4#.

So, the foot of the normal is #P( 4, -9 )#.

By actual division,

#y = x+3+16/(x-5)#,

revealing asymptotes #y=x+3 and x =5#.

y'=1-16/(x-5)^2=-15, at x = 4.

The slope of the normal = -1/y'=1/15.

So, the equation to the normal at #P( 4, -9 )# is

#y+9=15(x-4)#, giving

#x-15y-139=0#

graph{((x-1)^2/(x-5)-y)(x-15y-139)((x-4)^2+(y+9)^2-1)=0 [-80, 80, -40, 40]}