NIMCET 2012 Question Paper with solutions

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Question : 5 of 120

Marks: +1, -0

Normal to the curve

y = x^3 - 3x + 2

at the point

(2 , 4)

is

9x - y - 14 = 0
x - 9y + 40 = 0
x + 9y - 38 = 0
- 9x + y + 22 = 0

Solution:

Given curve ,

y = x^3 - 3x + 2

Now ,

{dy}/{dx}

= 3x^2 - 3

⇒

{dy}/{dx}_{\at (2,4)}

= 3(2)^2 - 3 = 12 - 3 = 9

∴ Slope of normal = -

1/9

Hence, the equation of normal at point (2, 4)

⇒ (y - 4) = - 1/9 (x - 2)

⇒ 9y - 36 = - x + 2

⇒ x + 9y = 38

⇒ x + 9y - 38 = 0

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