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

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

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