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  • Help me solve this - Limit , continuity and differentiability - JEE Main-8

The angle made by normal drawn at (0, 0) to the curve y^3 + y + x = 0 with positive x-axis is 

  • Option 1)

    \frac{\pi}{4}

  • Option 2)

    \frac{\pi}{}3

  • Option 3)

    \frac{\pi}{}2

  • Option 4)

    \frac{3\pi}{4}

 

Answers (1)

best_answer

As we  have learned

Slope of Normal -

M_{T}\times M_{N}=-1
 

\therefore \:M_{N}=\frac{-1}{MT}
 

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

- wherein

Where ( x, y ) is the point on the curve.

 

 \frac{dy}{dx} at (0,0) =  value of slope of tangent at (0,0)

\Rightarrow (3y^{2}+1) \frac{dy}{dx} =-1\Rightarrow \frac{dy}{dx} =-1/3y^{2}+1

now , \frac{dy}{dx} at (0,0) =-1/ 0+1 =-1=  slope of tangent  

\therefore  slope of normal = -1/-1=1= tan \theta 

\Rightarrow \theta =\pi /4

 

 

 

 


Option 1)

\frac{\pi}{4}

Option 2)

\frac{\pi}{}3

Option 3)

\frac{\pi}{}2

Option 4)

\frac{3\pi}{4}

Posted by

Himanshu

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