Find the equation of tangents to the curve y= x^{3} + 2x – 4, which are perpendicular to line x + 14y + 3 = 0.

#### Solution

Consider the given equation,

y = x^{3}+2x-4

Differentiating the above function with respect to x, we have,

`dy/dx=3x^2+2`

⇒ m_{1} = 3x^{2}+2

Given that the tangents to the given curve are perpendicular to the line x+14y+3=0

Slope of this line, `m_2=(-1)/14`

Since the given line and the tangents to the given curve are perpendicular, we have,

m_{1} x m_{2} = -1

`=>(3x^2+2)((-1)/14)=-1`

⇒ 3x^{2} + 2 = 14

⇒ 3x^{2 }= 12

⇒ x^{2 }= 4

⇒ x = ±2

if x =2, y=x^{3} + 2x -4

⇒ y = 2^{3 }+^{ }2 x 2 - 4

⇒ y = 8

if x = -2, y =x^{3} + 2x -4

⇒ y = (-2)^{3}+ 2 x (-2) - 4

⇒ y = -16

Equation of the tangent having slope m at the point (x_{1},y_{1}) is (y-y_{1})=m(x-x_{1})

Equation of the tangent at P(2,8) with slope 14

(y-8)=14(x-2)

⇒ y -8 = 14x -28

⇒ 14x -y= 20

Equation of the tangent at P(-2,-16) with slope 14

(y+1=6) = 14(x+2)

⇒ y +16 = 14x +28

⇒ 14x - y = -12