# Find the equation of the normals to the curve y = x^{3} + 2x + 6 which are parallel to the line x + 14y + 4 = 0

**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

y = x^{3} + 2x + 6

The slope of the tangent to the given curve at any point (x, y) is given by,

dy/dx = 3x^{2} + 2

Therefore, the slope of the normal to the given curve is,

- 1 / slope of the tangent

= - 1/(3x^{2} + 2)

The equation of the given line is x + 14 y + 4 = 0

⇒ y = - 1/14 x - 4/14,

which is the form of y = mx + c

Hence,

- 1/(3x^{2} + 2) = - 1/14

3x^{2} + 2 = 14

⇒ 3x^{2} = 12

⇒ x^{2} = 4

⇒ x = ± 2

When, x = 2, ⇒ y = 8 + 4 + 6 = 18

When, x = - 2, ⇒ y = - 8 - 4 + 6 = - 6

Therefore, there are two normal to the given curve with slope - 1/14 and passing through the points (2, 18) and (- 2, - 6).

Thus, the equation of the normal through (2,18) is

y - 18 = - 1/14 (x - 2)

⇒ 14y - 252 = x + 2

⇒ x + 14y - 254 = 0

And the equation of the normal through (- 2, - 6) is

y - (- 6) = - 1/14 (x - (- 2))

⇒ 14y + 84 = - x - 2

⇒ x + 14y + 86 = 0

Hence, the equations of the normal to the given curve are x + 14y - 254 = 0 and x + 14y + 86 = 0

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.3 Question 21

## Find the equation of the normal to the curve y = x^{3} + 2x + 6 which are parallel to the line x + 14y + 4 = 0.

**Summary:**

The equation of the normal to the curve y = x^{3} + 2x + 6 which are parallel to the line x + 14y + 4 = 0 are x + 14y - 254 = 0 and x + 14y + 86 = 0

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