Question

The equation of a normal to the curve, $\sin y=x\sin (\frac{\pi }{3}+y)$ 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

The correct option is D $2x+\sqrt{3}y=0$

$\sin y=x\sin (\frac{\pi }{3}+y)..\left(1\right)$

Putting $x=0\Rightarrow \sin y=0\Rightarrow y=0$

Now differentiating eqn (1) w.r.t $x$

$\cos y\frac{dy}{dx}=\sin (\frac{\pi }{3}+y)+x\cos (\frac{\pi }{3}+y)\frac{dy}{dx}$

Putting $x=0$ and $y=0$ to get the slope of tangent at these points,

$\frac{dy}{dx}=\sin \left(\frac{\pi }{3}\right)=\frac{\sqrt{3}}{2}=m$(say)

Thus slope of normal at $(0,0)$ is $=-\frac{1}{m}=-\frac{2}{\sqrt{3}}$

Hence required equation of normal is,

$y-0=-\frac{2}{\sqrt{3}}(x-0)$

$2x+\sqrt{3}y=0$