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

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

Solution

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

⇒ 3x= 12

⇒ x= 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

Concept: Tangents and Normals
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