# Find the equation of the normal to curve y^{2} = 4x at the point (1, 2)

**Solution:**

Using the concept of derivatives we can write down the equations of tangents and normals to arbitrary curves at arbitrary points

The equation of the given curve is

y^{2} = 4x

Differentiating with respect to x, we have:

2y dy/dx = 4

⇒ dy/dx = 4/2y

= dy/dx = 2/y

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

= 1

Now, the slope of the normal at point (1, 2) is

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

Equation of the normal at (1, 2) is

⇒ y - 2 = - x + 1

⇒ x + y - 3 = 0

Hence the required equation is x + y - 3 = 0

NCERT Solutions Class 12 Maths - Chapter 6 Exercise ME Question 4

## Find the equation of the normal to curve y^{2} = 4x at the point (1, 2)

**Summary:**

The equation of the normal to curve y^{2} = 4x at the point (1, 2) is x + y - 3 = 0. Using the concept of derivatives we can write down the equations of tangents and normals to arbitrary curves at arbitrary points

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