Question 25

Mathematics Tangents and Normals Medium

Normal to the curve \(y=x^{3}-3 x+2\) at the point \((2,4)\) is

(A) \(9 x-y-14=0\)
(B) \(x-9 y+40=0\)
(C) \(x+9 y-38=0\)
(D) \(-9 x+y+22=0\)
View Dynamic Solution & Explanation
Correct Solution: Option C

Step-by-step Solution:

\(y=x^{3}-3 x+2\) 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)\[\]