Question

If normal to $y=f\left(x\right)$ makes an angle $\frac{2\pi }{3}$ with positive $x-$axis at $(2,6)$, then $f^{\prime }\left(2\right)$ is

• A. $-\sqrt{3}$
• B. $\sqrt{3}$
• C. $-\frac{1}{\sqrt{3}}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

The correct option is D $\frac{1}{\sqrt{3}}$

Given : slope of normal
$m=\tan \left(\frac{2\pi }{3}\right)=-\sqrt{3}$
$\Rightarrow$ slope of tangent
$\frac{dy}{dx}=f^{\prime }\left(x\right)$
At $(2,6)$
$\frac{dy}{dx}{|}_{(2,6)}=f^{\prime }\left(2\right)$
$\Rightarrow f^{\prime }\left(2\right)=-\frac{1}{m}$
$\because (\text{slope of tangent}=-\frac{1}{\text{slope of normal}})$

$\therefore f^{\prime }\left(2\right)=\frac{1}{\sqrt{3}}$