Find the equation of the normal to the curve 2y=x^2 ,which passes through the point (2, 1).

but how ????? i too know the answer but wht is d prodedure?

But the answer is coming out to be x + 2y = 4.
Can anyone kindly explain this answer please?? Thanks.

We obtain the slope of tangent by differentiating the curve
   2 * dy/dx = 2* x
slope of tangent = x
now, slope of normal = -1/slope of tangent 
                                   =  -1/x
  eqtn of normal (y-y_​1)=(-1/x₁)*(x-x₁)
                           y-1  =(-1/2)*(x-2)
                          2y-2 =  -x +2
thus required eqtn     x+2y =4

Sorry, there is a mistake in my previous ans . The correct answer is as follows
let the normal be at point (x₁,y₁) to the curve 2y=x²

slope of tangent  = dy/dx = x
hence slope of normal at point (x₁,y₁ ) = -1/​slope of tangent = -1/x₁
so eqn of normal  (y-y₁) = (-1/x₁)*(x-x₁)
since point (2,1) lies on the curve
       (1-y₁)=(​-1/x₁)​(2-x₁)            ......(1)
         x_{1 *} y₁ =2                           ......(2)
since ​(x₁,y₁) lies on the curve
        y₁= (1/2)*(x₁)²
putting value of y₁  in eqn (2)
           x₁*(1/2)*(x₁)²=2
  thus x₁ = 2^2/3
   putting value of x₁  in eqn (2)​
we get y₁ = 2^1/3
finally we put value of x₁ and y₁ in the eqn(1) of normal
       ​(y-2^1/3) = (-1/2^2/3)*(x-2^2/3)
       2^2/3y - 2= -x +​2^2/3
thus  x + ​  2^2/3y = 2 +​  2^2/3  is the required equation.
Hope this helps.