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Question
Determine the equation(s) of tangent (s) line to the curve y = 4x3 − 3x + 5 which are perpendicular to the line 9y + x + 3 = 0 ?
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Solution
Let (x1, y1) be a point on the curve where we need to find the tangent(s).
Slope of the given line = \[\frac{- 1}{9}\]
Since, tangent is perpendicular to the given line,
Slope of the tangent = \[\frac{- 1}{\left( \frac{- 1}{9} \right)} = 9\]
\[\text { Let }\left( x_1 , y_1 \right)\text { be the point where the tangent is drawn to this curve }.\]
\[\text { Since, the point lies on the curve } . \]
\[\text { Hence }, y_1 = 4 {x_1}^3 - 3 x_1 + 5 \]
\[\text { Now }, y = 4 x^3 - 3x + 5\]
\[ \Rightarrow \frac{dy}{dx} = 12 x^2 - 3\]
\[\text { Slope of the tangent }= \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =12 {x_1}^2 - 3\]
\[\text { Given that },\]
\[\text { slope of the tangent = slope of the perpendicular line }\]
\[ \Rightarrow 12 {x_1}^2 - 3 = 9\]
\[ \Rightarrow 12 {x_1}^2 = 12\]
\[ \Rightarrow {x_1}^2 = 1\]
\[ \Rightarrow x_1 = \pm 1\]
\[\text { Case}-1: x_1 = 1\]
\[ y_1 = 4 {x_1}^3 - 3 x_1 + 5 = 4 - 3 + 5 = 6\]
\[ \therefore \left( x_1 , y_1 \right) = \left( 1, 6 \right)\]
\[\text { Slope of the tangent}=9\]
\[\text { Equation of tangent is},\]
\[y - y_1 = m \left( x - x_1 \right)\]
\[ \Rightarrow y - 6 = 9\left( x - 1 \right)\]
\[ \Rightarrow y - 6 = 9x - 9\]
\[ \Rightarrow 9x - y - 3 = 0\]
\[\text { Case }-2: x_1 = - 1\]
\[ y_1 = 4 {x_1}^3 - 3 x_1 + 5 = - 4 + 3 + 5 = 4\]
\[ \therefore \left( x_1 , y_1 \right) = \left( - 1, 4 \right)\]
\[\text { Slope of the tangent }=9\]
\[\text { Equation of tangent is },\]
\[y - y_1 = m \left( x - x_1 \right)\]
\[ \Rightarrow y - 4 = 9\left( x + 1 \right)\]
\[ \Rightarrow y - 4 = 9x + 9\]
\[ \Rightarrow 9x - y + 13 = 0\]
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