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