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  • Can someone help me with this, The normal to the curve, x2 + 2xy - 3y2 = 0, at (1,1) :

The normal to the curve, x2 + 2xy - 3y2 = 0, at (1,1) :

  • Option 1)

    does not meet the curve again.

  • Option 2)

    meets the curve again in the second quadrant.

  • Option 3)

    meets the curve again in the third quadrant.

  • Option 4)

    meets the curve again in the fourth quadrant.

 

Answers (1)

As we learnt in 

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})

-

 

 Move to Tangent Normal

2x+2.1y+2x\frac{dy}{dx}-6y\frac{dy}{dx}=0

2+2+2\frac{dy}{dx}-6\frac{dy}{dx}=0

4-4\frac{dy}{dx}=0

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

m_{N}=-1

\therefore y-1=-1(x-1)=-x+1

\therefore x+y=2 solve from x & y

\therefore x=3, y=-1

 


Option 1)

does not meet the curve again.

This option is incorrect

Option 2)

meets the curve again in the second quadrant.

This option is incorrect

Option 3)

meets the curve again in the third quadrant.

This option is incorrect

Option 4)

meets the curve again in the fourth quadrant.

This option is correct

Posted by

Sabhrant Ambastha

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