Normal to the curve \(y=x^{3}-3 x+2\) at the point \((2,4)\) is
\(y=x^{3}-3 x+2\) On differentiating \[\] \(\frac{d y}{d x}=3 x^{2}-3 \Rightarrow \frac{d y}{d x}=12-3=9\) \[\] Slope of normal \(=-\frac{1}{9}\) \[\] Equation of normal at \((2,4)\) is \(y-4=-\frac{1}{9}(x-2)\) \[\] \(9 y-36=-x+2 \Rightarrow x+9 y=38\) \[\] Choice (C)\[\]
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