Aspire Faculty ID #3629 · Topic: NIMCET 2012 · Just now
NIMCET 2012

Normal to the curve $y = x^3 - 3x + 2$ at the point $(2,4)$ is:

Solution

Differentiate: 
$\dfrac{dy}{dx} = 3x^2 - 3$ 
At $x = 2$: 
$\dfrac{dy}{dx} = 3(4) - 3 = 12 - 3 = 9$ 
So slope of tangent $m_t = 9$ 
Slope of normal: $m_n = -\dfrac{1}{9}$ 
Equation of normal at $(2,4)$: 
$y - 4 = -\dfrac{1}{9}(x - 2)$ 
Multiply by 9: 
$9y - 36 = -(x - 2)$ 
$9y - 36 = -x + 2$ 
$ x + 9y - 38 = 0$ 
Answer: (c) $x + 9y - 38 = 0$

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