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  • Need clarity, kindly explain! - Co-ordinate geometry - JEE Main-2

The normal to the curve x=a(1+\cos \Theta ),y=a\sin \Theta\; at\; \Theta  always passes through the fixed point

  • Option 1)

    (0,0)\;

  • Option 2)

    \; (0,a)\;

  • Option 3)

    \; (a,0)\;

  • Option 4)

    \; (a,a)

 

Answers (1)

As we learnt in

Slope – point from of a straight line -

y-y_{1}=m(x-x_{1})

- wherein

m\rightarrow slope

\left ( x_{1},y_{1} \right )\rightarrow point through which line passes

 

 \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{a\cos \theta }{-a\sin \theta }=-\cot \theta

Slope of normal=\tan \theta

Hence equation of normal by point-slope method 

(y-a\sin \theta )=\frac{\sin \theta }{\cos \theta }(x-a\cos \theta -a)

Clearly this line passes through (a,0) 


Option 1)

(0,0)\;

This option is incorrect 

Option 2)

\; (0,a)\;

This option is incorrect 

Option 3)

\; (a,0)\;

This option is correct 

Option 4)

\; (a,a)

This option is incorrect 

Posted by

Vakul

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