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  • Try this! - Limit , continuity and differentiability - JEE Main-14

The normal to the curve  x= a\left ( \cos \Theta +\Theta \sin \Theta \right ),

y= a\left ( \sin \Theta -\Theta \cos \Theta \right ), at any point \Theta  is such that

  • Option 1)

    it makes angle \frac{\pi }{2}+\Theta with x-axis

  • Option 2)

    it passes through the origin

  • Option 3)

    it is at a constant distance from the origin

  • Option 4)

    it passes through \left ( a\frac{\pi}{2},-a \right )

 

Answers (1)

best_answer

As we have learned

Slope of Normal -

M_{T}\times M_{N}=-1
 

\therefore \:M_{N}=\frac{-1}{M_T}
 

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

- wherein

Where ( x1, y1 ) is the point on the curve.

 

 

Slope of Normal for parametric form -

M_{N}=-\frac{f'_{x}}{f'_{y}}

and\:find\:M_{N}\:at(t)

- wherein

Where

M_{N}=\frac{-1}{M_{T}}

 

 

m_N= \frac{-1}{\frac{dy}{dx}} = \frac{-dx/d\theta }{dy/d\theta }= \frac{-a\left ( - \sin \theta +\theta \cos \theta +\sin \theta \right )}{a(\cos \theta + \theta \sin \theta - \cos \theta )} = - \left ( \frac{\theta \cos \theta }{\theta \sin \theta } \right )= - \cot \theta = \tan (\pi /2+\theta )

 


Option 1)

it makes angle \frac{\pi }{2}+\Theta with x-axis

Option 2)

it passes through the origin

Option 3)

it is at a constant distance from the origin

Option 4)

it passes through \left ( a\frac{\pi}{2},-a \right )

Posted by

Himanshu

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