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Question
Find the equation of the tangent and the normal to the following curve at the indicated point y2 = 4x at (1, 2) ?
Solution
\[y^2 = 4x\]
\[\text { Differentiating both sides w.r.t.x,} \]
\[2y \frac{dy}{dx} = 4\]
\[ \Rightarrow \frac{dy}{dx} = \frac{2}{y}\]
\[\text { Slope of tangent,}m= \left( \frac{dy}{dx} \right)_\left( 1, 2 \right) =\frac{2}{2}=1\]
\[\text { Given } \left( x_1 , y_1 \right) = \left( 1, 2 \right)\]
\[\text{ Equation of tangent is },\]
\[y - y_1 = m \left( x - x_1 \right)\]
\[ \Rightarrow y - 2 = 1\left( x - 1 \right)\]
\[ \Rightarrow y - 2 = x - 1\]
\[ \Rightarrow x - y + 1 = 0\]
\[\text { Equation of normal is},\]
\[y - y_1 = \frac{- 1}{m} \left( x - x_1 \right)\]
\[ \Rightarrow y - 2 = - 1\left( x - 1 \right)\]
\[ \Rightarrow y - 2 = - x + 1\]
\[ \Rightarrow x + y - 3 = 0\]
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