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) ?

Answer 1

\[y=2 x^2 -3x-1\]

\[\text { Differentiating both sides w.r.t.x,} \]

\[\frac{dy}{dx} = 4x - 3\]

\[\text { Given } \left( x_1 , y_1 \right) = \left( 1, - 2 \right)\]

\[\text { Slope of tangent },m= \left( \frac{dy}{dx} \right)_\left( 1, - 2 \right) =4-3=1\]

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

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

\[ \Rightarrow y + 2 = 1 \left( x - 1 \right)\]

\[ \Rightarrow y + 2 = x - 1\]

\[ \Rightarrow x - y - 3 = 0\]

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

\[y - y_1 = \frac{- 1}{m} \left( x - x_1 \right)\]

\[ \Rightarrow y + 2 = - 1 \left( x - 1 \right)\]

\[ \Rightarrow y + 2 = - x + 1\]

\[ \Rightarrow x + y + 1 = 0\]