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  • Help me solve this - Limit , continuity and differentiability - JEE Main-9

Equation of tangent to the curve x = t^2, \; y = t^3 at t = 2 is

  • Option 1)

    y = 3x + 4

  • Option 2)

    y =3x-4

  • Option 3)

    x + y =12

  • Option 4)

    2x - y = 0

 

Answers (1)

best_answer

As we have learned

Slope of tangent for parametric form -

M_{T}=\frac{dy}{dt}.\frac{dt}{dx}

      
        =\frac{f'_{y}}{f'_{x}}

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

- wherein

Where\:\:y=f(t)\:\:\:and\:\:\:x=f(t)

 

 Slope of tangent = \frac{dy}{dx} =\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{3t^{2}}{2t} =\frac{3t}{2}

\therefore slope at t =2 is 3/2*2=3

and t= 2 will give x= 4 , y= 8

\therefore Equation of tangent : y-8 =3(x-4)

\Rightarrow = 3x-4

 

 

 

 

 

 


Option 1)

y = 3x + 4

Option 2)

y =3x-4

Option 3)

x + y =12

Option 4)

2x - y = 0

Posted by

Himanshu

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