Question

Find an equation of normal line to the curve y = x³ + 2x + 6 which is parallel to the line x+ 14y + 4 = 0 ?

Answer 1

Let (x₁, y₁) be a point on the curve where we need to find the normal.
Slope of the given line = \[\frac{- 1}{14}\]

\[\text { Since, the point lies on the curve } . \]

\[\text { Hence}, y_1 = {x_1}^3 + 2 x_1 + 6 \]

\[\text{ Now,} y = x^3 + 2x + 6\]

\[ \Rightarrow \frac{dy}{dx} = 3 x^2 + 2\]

\[\text { Slope of the tangent }= \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =3 {x_1}^2 +2\]

\[\text { Slope of the normal }=\frac{- 1}{\text { slope of the tangent}}= = \frac{- 1}{3 {x_1}^2 + 2}\]

\[\text { Given that},\]

\[\text{ slope of the normal=slope of the given line }\]

\[ \Rightarrow \frac{- 1}{3 {x_1}^2 + 2} = \frac{- 1}{14}\]

\[ \Rightarrow 3 {x_1}^2 + 2 = 14\]

\[ \Rightarrow 3 {x_1}^2 = 12\]

\[ \Rightarrow {x_1}^2 = 4\]

\[ \Rightarrow x_1 = \pm 2\]

\[\text { Case }-1: x_1 = 2\]

\[ y_1 = {x_1}^3 + 2 x_1 + 6 = 8 + 4 + 6 = 18\]

\[ \therefore \left( x_1 , y_1 \right) = \left( 2, 18 \right)\]

\[\text{ Slope of the normal },m=\frac{- 1}{14}\]

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

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

\[ \Rightarrow y - 18 = \frac{- 1}{14}\left( x - 2 \right)\]

\[ \Rightarrow 14y - 252 = - x + 2\]

\[ \Rightarrow x + 14y - 254 = 0\]

\[\text { Case }-2: x_1 = - 2\]

\[ y_1 = {x_1}^3 + 2 x_1 + 6 = - 8 - 4 + 6 = - 6\]

\[ \therefore \left( x_1 , y_1 \right) = \left( - 2, - 6 \right)\]

\[\text { Slope of the normal},m=\frac{- 1}{14}\]

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

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

\[ \Rightarrow y + 6 = \frac{- 1}{14}\left( x + 2 \right)\]

\[ \Rightarrow 14y + 84 = - x - 2\]

\[ \Rightarrow x + 14y + 86 = 0\]