Question

Find the equation of the tangent and the normal to the following curve at the indicated point  y² = 4x at (1, 2)  ?

Answer 1

\[y^2 = 4x\]

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

\[2y \frac{dy}{dx} = 4\]

\[ \Rightarrow \frac{dy}{dx} = \frac{2}{y}\]

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

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

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

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

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

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

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

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

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

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

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

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