Find the equation of the normal
Question:

Find the equation of the normal to $y=2 x^{3}-x^{2}+3$ at $(1,4)$.

Solution:

$y=2 x^{3}-x^{2}+3$

Differentiating both sides w.r.t. $x$,

$\frac{d y}{d x}=6 x^{2}-2 x$

Slope of tangent $=\left(\frac{d y}{d x}\right)_{(1,4)}=6(1)^{2}-2(1)=4$

Slope of normal $=\frac{-1}{\text { Slope of tangent }}=\frac{-1}{4}$

Given $\left(x_{1}, y_{1}\right)=(1,4)$

Equation of normal is,

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-4=\frac{-1}{4}(x-1)$

$\Rightarrow 4 y-16=-x+1$

$\Rightarrow x+4 y=17$

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