The normal to the curve x² = 4y passing (1,2) is

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