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Question
Find the equation of the tangent and the normal to the following curve at the indicated point y = x2 + 4x + 1 at x = 3 ?
Solution
\[y = x^2 + 4x + 1\]
\[\text { Differentiating both sides w.r.t.x, } \]
\[\frac{dy}{dx} = 2x + 4\]
\[\text { When x}=3,y = 9 + 12 + 1 = 22\]
\[\text { So }, \left( x_1 , y_1 \right) = \left( 3, 22 \right)\]
\[\text { Slope of tangent },m= \left( \frac{dy}{dx} \right)_{x = 3} =10\]
\[\text { Equation of tangent is },\]
\[y - y_1 = m \left( x - x_1 \right)\]
\[ \Rightarrow y - 22 = 10\left( x - 3 \right)\]
\[ \Rightarrow y - 22 = 10x - 30\]
\[ \Rightarrow 10x - y - 8 = 0\]
\[\text { Equation of normal is },\]
\[y - y_1 = \frac{- 1}{m} \left( x - x_1 \right)\]
\[ \Rightarrow y - 22 = \frac{- 1}{10} \left( x - 3 \right)\]
\[ \Rightarrow 10y - 220 = - x + 3\]
\[ \Rightarrow x + 10y - 223 = 0\]
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