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

##### 1 Answer
Oct 5, 2016

Slope of normal is

$\left(y - \frac{2}{\sqrt{2 + \sqrt{2}}}\right) = - \frac{\sqrt{2 + \sqrt{2}}}{2 \left(\sqrt{2} - 1\right)} \left(x - \frac{\pi}{8}\right)$

#### Explanation:

At $x = \frac{\pi}{8}$, $f \left(x\right) = \sec x = \frac{2}{\sqrt{2 + \sqrt{2}}}$

Hence we are seeking tangent at $\left(\frac{\pi}{8} , \frac{2}{\sqrt{2 + \sqrt{2}}}\right)$

As $\frac{\mathrm{df}}{\mathrm{dx}} = \sec x \tan x$

at $x = \frac{\pi}{8}$, slope of tangent is $\sec \left(\frac{\pi}{8}\right) \tan \left(\frac{\pi}{8}\right)$

= $\frac{2}{\sqrt{2 + \sqrt{2}}} \times \left(\sqrt{2} - 1\right)$

= $\frac{2 \left(\sqrt{2} - 1\right)}{\sqrt{2 + \sqrt{2}}}$

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

As given slope $m$ equation of line passing through $\left({x}_{1} , {y}_{1}\right)$ is

$\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)$

the equation of tangent with slope $\frac{2 \left(\sqrt{2} - 1\right)}{\sqrt{2 + \sqrt{2}}}$ and passing through $\left(\frac{\pi}{8} , \frac{2}{\sqrt{2 + \sqrt{2}}}\right)$ is

$\left(y - \frac{2}{\sqrt{2 + \sqrt{2}}}\right) = - \frac{\sqrt{2 + \sqrt{2}}}{2 \left(\sqrt{2} - 1\right)} \left(x - \frac{\pi}{8}\right)$