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  • Solve this problem - Limit , continuity and differentiability - JEE Main-22

Equation of normal to the curve y^2 = 8x at (2, 4) on it is ?

  • Option 1)

    x + y = 2

  • Option 2)

    x + y = 4

  • Option 3)

    x + y = 6

  • Option 4)

    x - y = -2

 

Answers (1)

best_answer

As we have learned

Equation of Normal -

Equation of normal to the curve  y = f(x) at the point  P(x1, y1) on the curve having a slope  MN  is 

(y-y_{1})=M_{N}(x-x_{1})


=\frac{-1}{\frac{dy}{dx}_{(x_{1},y_{1})}}(x-x_{1})

-

 

slope of normal = -1/ slope of tangent = -1/   \frac{dy}{dx}/(2,4) 

Diffrentiating both sides  y^{2}=8x , we get 

\frac{2ydy}{dx}=8\Rightarrow \frac{dy}{dx}=\frac{4}{y}

\therefore \frac{dy}{dx}=4/4=1

slope of normal = -1

equation will be : y-4=-1(x-2)\Rightarrow x+y=6

 

 

 

 


Option 1)

x + y = 2

Option 2)

x + y = 4

Option 3)

x + y = 6

Option 4)

x - y = -2

Posted by

Himanshu

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