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  • I need help with - Limit , continuity and differentiability - JEE Main-12

Length of normal drawn to the curve xy = 16 at its point (4, 4) equals ?

  • Option 1)

    6\sqrt2

  • Option 2)

    5\sqrt2

  • Option 3)

    4\sqrt2

  • Option 4)

    3\sqrt2

 

Answers (1)

best_answer

As we have learned

Length of Normal -

L_{N}=y\sqrt{1+y'^{2}}

- wherein

Where\:\:\:y'=\frac{dy}{dx}

 

 Length of normal = y\sqrt {1+(y')^{2}}

xy =16\Rightarrow y=\frac{16}{x}\Rightarrow y'= -16/x^{2}\Rightarrow y'  at (4,4)= -1 

\therefore  Length of normal = 4 \sqrt {1+1}= 4\sqrt2

 

 

 

 

 


Option 1)

6\sqrt2

Option 2)

5\sqrt2

Option 3)

4\sqrt2

Option 4)

3\sqrt2

Posted by

Himanshu

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