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Question
Find the equation of the normal to y = 2x3 − x2 + 3 at (1, 4) ?
Solution
\[y = 2 x^3 - x^2 + 3\]
\[\text { Differentiating both sides w.r.t.x,} \]
\[\frac{dy}{dx} = 6 x^2 - 2x\]
\[\text { Slope of tangent } = \left( \frac{dy}{dx} \right)_\left( 1, 4 \right) =6 \left( 1 \right)^2 -2\left( 1 \right)=4\]
\[\text { Slope of normal } =\frac{- 1}{\text { Slope of tangent}}=\frac{- 1}{4}\]
\[\text { Given }\left( x_1 , y_1 \right) = \left( 1, 4 \right)\]
\[\text { Equation of normal is},\]
\[y - y_1 = m \left( x - x_1 \right)\]
\[ \Rightarrow y - 4 = \frac{- 1}{4} \left( x - 1 \right)\]
\[ \Rightarrow 4y - 16 = - x + 1\]
\[ \Rightarrow x + 4y = 17\]
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