What is the equation of the normal line of #f(x)= secx# at #x = pi/8#?

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Answer 1 · 2016-10-05T10:24:08

Slope of normal is

#(y-2/sqrt(2+sqrt2))=-(sqrt(2+sqrt2))/(2(sqrt2-1))(x-pi/8)#

At #x=pi/8#, #f(x)=secx=2/sqrt(2+sqrt2)#

Hence we are seeking tangent at #(pi/8,2/sqrt(2+sqrt2))#

As #(df)/(dx)=secxtanx#

at #x=pi/8#, slope of tangent is #sec(pi/8)tan(pi/8)#

= #2/sqrt(2+sqrt2)xx(sqrt2-1)#

= #(2(sqrt2-1))/sqrt(2+sqrt2)#

and slope of normal is #-sqrt(2+sqrt2)/(2(sqrt2-1)#

As given slope #m# equation of line passing through #(x_1,y_1)# is

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

the equation of tangent with slope #(2(sqrt2-1))/sqrt(2+sqrt2)# and passing through #(pi/8,2/sqrt(2+sqrt2))# is

#(y-2/sqrt(2+sqrt2))=-(sqrt(2+sqrt2))/(2(sqrt2-1))(x-pi/8)#