Answer :

It is given that the equation of curve is x2 = 4y


Differentiating w.r.t. x, we get,




The slope of the normal to the given curve at point (h,k) is



Then, the equation of the normal to the curve at (h,k) is


y – k =


Now, it is given that the normal passes through the point (1,2)


Thus, we get,


2 – k =


k = ………………(1)


Since (h,k) lies on the curve x2 = 4y, we have h2 = 4k


k =


Now putting the value of of k in (1), we get




h3 = 8


h = 2


Therefore, the equation of the normal is given as:


y – 1 =


y - 1 = - (x - 2)


x + y = 3

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