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