# What is the equation of the line normal to  f(x)=xtanx at  x=pi/3?

Jul 28, 2016

$x + \left(\sqrt{3} + \frac{4 \pi}{9}\right) y - \frac{4 \pi}{3} \left(1 + \frac{\pi}{3 \sqrt{3}}\right) = 0$

#### Explanation:

Let us find the slope of the tangent, which should give us slope of normal (as it is perpendicular to tangent and hence product of their slopes would be $- 1$).

Slope of tangent is given by first derivative and this is

$f ' \left(x\right) = 1 \times \tan x + x \times {\sec}^{2} x$ and as we need slope of tangent at $x = \frac{\pi}{3}$,

f'([i/3)=tan(pi/3)+pi/3xxsec^2(pi/3)=sqrt3+pi/3xx(2/sqrt3)^2

= $\sqrt{3} + \frac{4 \pi}{9}$

and slope of normal is $- \frac{1}{\sqrt{3} + \frac{4 \pi}{9}}$

As normal is needed at $\left(x , y\right)$ with $x = \frac{\pi}{3}$, $\left(x , y\right)$ is $\left(\frac{\pi}{3} , \frac{\pi}{3} \times \tan \left(\frac{\pi}{3}\right)\right)$ or $\left(\frac{\pi}{3} , \frac{\pi}{\sqrt{3}}\right)$

Hence, using point slope form of equation $\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)$

Equation of normal is $\left(y - \frac{\pi}{\sqrt{3}}\right) = - \frac{1}{\sqrt{3} + \frac{4 \pi}{9}} \left(x - \frac{\pi}{3}\right)$ or

$\left(\sqrt{3} + \frac{4 \pi}{9}\right) \left(y - \frac{\pi}{\sqrt{3}}\right) = \frac{\pi}{3} - x$

$\sqrt{3} y - \pi + \frac{4 \pi y}{9} - \frac{4 {\pi}^{2}}{9 \sqrt{3}} - \frac{\pi}{3} + x = 0$ or

$x + \left(\sqrt{3} + \frac{4 \pi}{9}\right) y - \frac{4 \pi}{3} \left(1 + \frac{\pi}{3 \sqrt{3}}\right) = 0$