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  • Answer please! - Limit , continuity and differentiability - JEE Main-10

Let x = t^2, y = t^3 represent a curve in parametric form, then slope of normal at  a point t = 3 is

  • Option 1)

    -\frac{1}{9}

  • Option 2)

    -\frac{4}{9}

  • Option 3)

    -\frac{2}{9}

  • Option 4)

    -\frac{5}{9}

 

Answers (1)

best_answer

As we have learned

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}}

 

 \frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{3t^{2}}{2t}= 3t/2

  \therefore slope of tangent at t= 3 is  3/2* 3=9/2

\therefore slope of normal = -1/9/2 = -2/9

 

 

 

 

 

 


Option 1)

-\frac{1}{9}

Option 2)

-\frac{4}{9}

Option 3)

-\frac{2}{9}

Option 4)

-\frac{5}{9}

Posted by

Himanshu

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