Question

The equation of a normal to the curve, $\sin y=x\sin \left(\frac{\pi }{3}+y\right)$ at $x=0$, is:

• A. $2x+\sqrt{3}y=0$
• B. $2y-\sqrt{3}x=0$
• C. $2y+\sqrt{3}x=0$
• D. $2x-\sqrt{3}y=0$

Answer 1

Answer: (A)

Given value curve is
$\sin y=x\sin \left(\frac{\pi }{3}+y\right)$
$\Rightarrow \cos y\frac{dy}{dx}=x\cos \left(\frac{\pi }{3}+y\right)\frac{dy}{dx}+\sin \left(\frac{\pi }{3}+y\right)$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{x=0}=\frac{\sqrt{3}}{2}$
$\Rightarrow {\left(\frac{-dx}{dy}\right)}_{x=0}=\frac{-2}{\sqrt{3}}$
$\therefore$ Equation of normal at $(0,0)$ is $(y-0)=\frac{-2}{\sqrt{3}}(x-0)$
$\Rightarrow y=\frac{-2}{\sqrt{3}}x$ or $2x+\sqrt{3}y=0$