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