Find the equation of tangent and normal to the curve $y = x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5$ at $(1, 3)$

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Solution

The equation of the curve is $y = x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5$
On differentiating with respect to $x$ , we get
$\begin{matrix} \frac{d y}{d x} = 4 x^{3} - 18 x^{2} + 26 x - 10 {\frac{d y}{d x}]}_{(1, 3)} = 4 - 18 + 26 - 10 = 2 \end{matrix}$
Thus, the slope of the tangent at $(1, 3)$ is $2$ . The equation of the tangent is given as:
$\begin{matrix} y - 3 = 2 (x - 1) \Rightarrow y - 3 = 2 x - 2 \Rightarrow y = 2 x + 1 \end{matrix}$
The slope of the normal at $(1, 3)$ is $\frac{1}{s l o p e o f t h e tan g e n t a t (1, 3)} = \frac{- 1}{2}$
Therefore, the equation of the normal at $(1, 3)$ is given as;
$\begin{matrix} y - 3 = - \frac{1}{2} (x - 1) \Rightarrow 2 y - 6 = - x + 1 \Rightarrow x + 2 y - 7 = 0 \end{matrix}$ is the required equation.

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