NCERT Solutions
Class 12 Maths
Application of Derivatives

Ex.6.3 Q.21
Find the equation of the normals to the curve y = x3 + 2x + 6 which are parallel to the line
x + 14y + 4 = 0.
The equation of the given curve is y = x3 + 2x + 6
The slope of the tangent to the given curve at any point (x, y) is given by,
= 3x2 + 2
So, the slope of the normal to the given curve at any point (x, y)
= at the point (x, y)
=
The equation of the given line is x + 14y + 4 = 0
⟹ (which is of the form y = mx + c)
So, the slope of the given line =
If the normal is parallel to the line, then we must have the slope of the normal being equal to
the slope of the line.
So,
⟹ 3x2 + 2 = 14
⟹ 3x2 = 12
⟹ x2 = 4
⟹ x = ±2
When x = 2, y = 8 + 4 + 6 = 18
When x = −2, y = − 8 − 4 + 6 = −6
Therefore, there are two normals to the given curve with slope and passing through the
points (2, 18) and (−2, −6).
Thus, the equation of the normal through (2, 18) is given by,
y – 18 = (x - 2)
⟹ 14y – 252 = x – 2
⟹ x + 14y – 254 = 0
And, the equation of the normal through (−2, −6) is given by,
y – (-6) = [x – (-2)]
⟹ y + 6 = (x + 2)
⟹ 14y + 84 = -x - 2
⟹ x + 14y + 86 = 0
Hence, the equations of the normals to the given curve (which are parallel to the given line)
are x + 14y - 254 = 0 and x + 14y + 86 = 0