Question

Find the equation of tangents to the curve y= x³ + 2x – 4, which are perpendicular to line x + 14y + 3 = 0.

Answer 1

Consider the given equation,

y = x³+2x-4

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

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

⇒  m₁ = 3x²+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,

m₁ x m₂ = -1

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

⇒ 3x² + 2 = 14

⇒ 3x² = 12

⇒ x² = 4

⇒ x = ±2

if x =2, y=x³ + 2x -4

⇒ y = 2³ + 2 x 2 - 4

⇒ y = 8

if x = -2, y =x³ + 2x -4

⇒ y = (-2)³+ 2 x (-2) - 4

⇒ y = -16

Equation of the tangent having slope m at the point (x₁,y₁) is (y-y₁)=m(x-x₁)

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