What is the equation of the line normal to  f(x)=tan^2x-3x at  x=pi/3?

Mar 13, 2016

Equation of normal at $x = \frac{\pi}{3}$ is $\left(3 \sqrt{3} - 8\right) y - \sqrt{3} x - 2 \sqrt{3} \pi + 8 \frac{\pi}{3} + 3 \sqrt{3} = 0$

Explanation:

If $f \left(x\right) = {\tan}^{2} x - 3 x$,

$f \left(\frac{\pi}{3}\right) = {\tan}^{2} \left(\frac{\pi}{3}\right) - 3 \times \frac{\pi}{3} = {\left(\sqrt{3}\right)}^{2} - \pi = 3 - \pi$.

Hence normal should pass through $\left(3 - \pi , \frac{\pi}{3}\right)$

Slope of the curve at $x$ is given by differential of $f \left(x\right)$ i.e.

$f ' \left(x\right) = 2 \tan x \times {\sec}^{2} x - 3$.

Hence slope of tangent at $x = \frac{\pi}{3}$ is

$f ' \left(\frac{\pi}{3}\right) = 2 \tan \left(\frac{\pi}{3}\right) \times {\sec}^{2} \left(\frac{\pi}{3}\right) - 3 = 2 \sqrt{3} \times {\left(\frac{2}{\sqrt{3}}\right)}^{2} - 3 = \frac{8}{\sqrt{3}} - 3 = \frac{8 - 3 \sqrt{3}}{\sqrt{3}}$ and as product of slopes of tangent and normal should be $- 1$

Hence, slope of normal is $\frac{\sqrt{3}}{3 \sqrt{3} - 8}$

Hence, equation of normal using point slope form is

$\left(y - \frac{\pi}{3}\right) = \frac{\sqrt{3}}{3 \sqrt{3} - 8} \left(x - 3 + \pi\right)$ or

$\left(3 \sqrt{3} - 8\right) y - \sqrt{3} x - \sqrt{3} \pi + 8 \frac{\pi}{3} + 3 \sqrt{3} - \sqrt{3} \pi = 0$

or $\left(3 \sqrt{3} - 8\right) y - \sqrt{3} x - 2 \sqrt{3} \pi + 8 \frac{\pi}{3} + 3 \sqrt{3} = 0$