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  • The equation of a normal to the curve,   <img alt="\sin y= x\sin \left ( \frac{\pi }{3} +y\right )" src="https://learn.careers360.com/latex-image/?%5Csin%20y%3D%20x%5Csin%20%5Cleft%2

The equation of a normal to the curve,

  \sin y= x\sin \left ( \frac{\pi }{3} +y\right )at x = 0, is :

Option: 1

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


Option: 2

2y-\sqrt{3}\: x= 0


Option: 3

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


Option: 4

2x-\sqrt{3}\: y= 0


Answers (1)

best_answer

Use the concept of

Equation of Normal -

Equation of normal to the curve  y = f(x) at the point  P(x1, y1) on the curve having a slope  MN  is 

(y-y_{1})=M_{N}(x-x_{1})


=\frac{-1}{\frac{dy}{dx}_{(x_{1},y_{1})}}(x-x_{1})

-

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

Put x = 0

\sin y=0

y=\sin^{-1}0=0

\cos y \frac{d y}{d x}=1 . \sin (\pi / 3+y)+x \cdot \cos (\pi / 3+y) \times \frac{d y}{d x}

\cos 0 \frac{d y}{d x}=1 \cdot \sin (\pi / 3+0)=\frac{\sqrt{3}}{2}

\therefore \frac{d y}{d x}=\frac{\sqrt{3}}{2}

\therefore \text { equation of normal }

y=\frac{-2 x}{\sqrt{3}}

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

Posted by

vinayak

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