Question

Find the equation of the normal to y = 2x³ − x² + 3 at (1, 4) ?

Answer 1

\[y = 2 x^3 - x^2 + 3\]

\[\text { Differentiating both sides w.r.t.x,} \]

\[\frac{dy}{dx} = 6 x^2 - 2x\]

\[\text { Slope of tangent } = \left( \frac{dy}{dx} \right)_\left( 1, 4 \right) =6 \left( 1 \right)^2 -2\left( 1 \right)=4\]

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

\[\text { Given }\left( x_1 , y_1 \right) = \left( 1, 4 \right)\]

\[\text { Equation of normal is},\]

\[y - y_1 = m \left( x - x_1 \right)\]

\[ \Rightarrow y - 4 = \frac{- 1}{4} \left( x - 1 \right)\]

\[ \Rightarrow 4y - 16 = - x + 1\]

\[ \Rightarrow x + 4y = 17\]