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Question
Find an equation of normal line to the curve y = x3 + 2x + 6 which is parallel to the line x+ 14y + 4 = 0 ?
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Solution
Let (x1, y1) be a point on the curve where we need to find the normal.
Slope of the given line = \[\frac{- 1}{14}\]
\[\text { Since, the point lies on the curve } . \]
\[\text { Hence}, y_1 = {x_1}^3 + 2 x_1 + 6 \]
\[\text{ Now,} y = x^3 + 2x + 6\]
\[ \Rightarrow \frac{dy}{dx} = 3 x^2 + 2\]
\[\text { Slope of the tangent }= \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =3 {x_1}^2 +2\]
\[\text { Slope of the normal }=\frac{- 1}{\text { slope of the tangent}}= = \frac{- 1}{3 {x_1}^2 + 2}\]
\[\text { Given that},\]
\[\text{ slope of the normal=slope of the given line }\]
\[ \Rightarrow \frac{- 1}{3 {x_1}^2 + 2} = \frac{- 1}{14}\]
\[ \Rightarrow 3 {x_1}^2 + 2 = 14\]
\[ \Rightarrow 3 {x_1}^2 = 12\]
\[ \Rightarrow {x_1}^2 = 4\]
\[ \Rightarrow x_1 = \pm 2\]
\[\text { Case }-1: x_1 = 2\]
\[ y_1 = {x_1}^3 + 2 x_1 + 6 = 8 + 4 + 6 = 18\]
\[ \therefore \left( x_1 , y_1 \right) = \left( 2, 18 \right)\]
\[\text{ Slope of the normal },m=\frac{- 1}{14}\]
\[\text { Equation of normal is },\]
\[y - y_1 = m \left( x - x_1 \right)\]
\[ \Rightarrow y - 18 = \frac{- 1}{14}\left( x - 2 \right)\]
\[ \Rightarrow 14y - 252 = - x + 2\]
\[ \Rightarrow x + 14y - 254 = 0\]
\[\text { Case }-2: x_1 = - 2\]
\[ y_1 = {x_1}^3 + 2 x_1 + 6 = - 8 - 4 + 6 = - 6\]
\[ \therefore \left( x_1 , y_1 \right) = \left( - 2, - 6 \right)\]
\[\text { Slope of the normal},m=\frac{- 1}{14}\]
\[\text { Equation of normal is },\]
\[y - y_1 = m \left( x - x_1 \right)\]
\[ \Rightarrow y + 6 = \frac{- 1}{14}\left( x + 2 \right)\]
\[ \Rightarrow 14y + 84 = - x - 2\]
\[ \Rightarrow x + 14y + 86 = 0\]
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