Question

 Find the equation of the tangent and the normal to the following curve at the indicated point  x² = 4y at (2, 1) ?

Answer 1

\[x^2 = 4y\]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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