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  • Give answer! Equation of curve for which , normal at its any points passes through (1,1) is

Equation of curve for which , normal at its any points passes through (1,1) is 

  • Option 1)

    x^2+y^2 =c

  • Option 2)

    x^2+ (y-1)^2=C

  • Option 3)

    (x-1)^2+ (y-1)^2= c

  • Option 4)

    (x-1)^2 + (y+1)^2=C

 

Answers (1)

best_answer

As we have learned

Evation of normal at a point (x,y) -

(y-x)= \frac{-1}{\left ( \frac{dy}{dx} \right )}\left ( X-x \right )

- wherein

\frac{dy}{dx} is slop of  tangent at (x,y) on the curve.

 

 Equation of nbormal at (x,y) will be 

Y-y = -\frac{dx}{dy}(X-x)

\because  it passes through (1,1) so 

1-y = -\frac{dx}{dy}(1-x)

\Rightarrow (1-y) dy+(1-x){dx}=0

\Rightarrow (y-1) dy+(x-1){dx}=0  

on integrating we get ,

\frac{(y-1)^2}{2}+ \frac{(x-1)^2}{2}= C\: \: or\: \: (x-1)^2+(y-1)^2= 2C

WHich can also be written as (x-1)^2+(y-1)^2= C

 

 

 

 

 

 


Option 1)

x^2+y^2 =c

Option 2)

x^2+ (y-1)^2=C

Option 3)

(x-1)^2+ (y-1)^2= c

Option 4)

(x-1)^2 + (y+1)^2=C

Posted by

Himanshu

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