A) \[2x+\sqrt{3}y=0\]
B) \[2x-\sqrt{3}x=0\]
C) \[2y+\sqrt{3}x=0\]
D) \[2x-\sqrt{3}y=0\] \[x=0\Rightarrow y=0\]
Correct Answer: A
Solution :
Differentiating we have \[\cos y\frac{dy}{dx}=x\cos \left( \frac{\pi }{3}+y \right)\frac{dy}{dx}+\sin \left( \frac{\pi }{3}+y \right)\] \[x=0\] \[y=0\]\[\Rightarrow \]\[\frac{dy}{dx}=\frac{\sqrt{3}}{2}\]\[\Rightarrow \]\[\frac{-dx}{dy}=\frac{-2}{\sqrt{3}}\] \[\therefore \]equation of normal is\[y=\frac{-2}{\sqrt{3}}x\] i.e., \[\sqrt{3}y=-2x\] \[2x+\sqrt{3}y=0\]
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