The normal to the curve x² = 4y passing (1, 2) is
(A) x + y = 3  (B) x - y = 3  (C) x + y = 1  (D) x - y = 1

Solution:

The slope of a line is nothing but the change in y coordinate with respect to the change in x coordinate of that line

The equation of the given curve is x² = 4y

Differentiating with respect to x, we have:

2x = 4. dy/dx

⇒ dy/dx = x / 2

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

(- 1)/dy/dx]_{(h, k)} = - 2 / h

Hence, the equation of the normal to the given curve at (h, k) is given as:

y - k = - 2/h (x - h)

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

Therefore, we have:

2 - k = - 2 / h (1 - h)

⇒ k = 2 + 2 / h (1 - h)

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

we have h² = 4k

⇒ k = h²/4

From equation (1),

we have:

h²/4 = 2 + 2/h (1 - h)

h³/4 = 2h + 2 - 2h

h³/4 = 2

h³ = 8

h = 2

Therefore,

k = h²/4

On substituting the value of h, we get

k = 1

Hence, the equation of the normal is given as:

⇒ y - 1 = - 2/2 (x - 2)

⇒ y - 1 = - x + 2

⇒ x + y = 3

Thus, the correct option is A

---

NCERT Solutions Class 12 Maths - Chapter 6 Exercise ME Question 23

The normal to the curve x² = 4y passing (1, 2) is (A) x + y = 3 (B) x - y = 3 (C) x + y = 1 (D) x - y = 1

Summary:

The normal to the curve x² = 4y passing (1, 2) is  x + y = 3. Thus, the correct option is A