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  • I have a doubt, kindly clarify. - Limit , continuity and differentiability - JEE Main-13

Length of normal to the curve y = x^3 at (2, 8) on it will be ?

  • Option 1)

    8\sqrt{45}

  • Option 2)

    6\sqrt{145}

  • Option 3)

    7\sqrt{145}

  • Option 4)

    8\sqrt{145}

 

Answers (1)

best_answer

As we have learned

Length of Normal -

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

- wherein

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

 

 dy/dx= 3x^{2}\Rightarrow dy/dx   at (2,8) = 3*2^{2} =12 

\therefore  Length of normal = 8\sqrt {1+144}= 8\sqrt{145}

 

 

 

 

 


Option 1)

8\sqrt{45}

Option 2)

6\sqrt{145}

Option 3)

7\sqrt{145}

Option 4)

8\sqrt{145}

Posted by

Himanshu

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