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  • Try this! - Limit , continuity and differentiability - JEE Main-13

Equation of normal drawn at (3, 4) on the curve x^2 + y^2 - 4x -6y + 11 = 0 is 

  • Option 1)

    y = x +1

  • Option 2)

    x + y = 7

  • Option 3)

    2x - y =2

  • Option 4)

    x + 2 y =11

 

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})

-

 

diffrentiating we have \rightarrow 2x+2y\frac{dy}{dx}-4-6\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}= \frac{2-x}{y-3}

slope of tangent at (3,4) = \frac{2-3}{4-3}=-1

\therefore slope of normal = 1 

\therefore equation of normal : (y-4)=1(x-3)\Rightarrow y=x+1

 

 

 


Option 1)

y = x +1

Option 2)

x + y = 7

Option 3)

2x - y =2

Option 4)

x + 2 y =11

Posted by

Himanshu

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