# The normal to the

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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