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Question
Find the equation of the tangent and the normal to the following curve at the indicated point y2 = 4ax at (x1, y1)?
Solution
\[y^2 =4ax\]
\[\text { Differentiating both sides w.r.t.x}, \]
\[2y\frac{dy}{dx} = 4a\]
\[ \Rightarrow \frac{dy}{dx} = \frac{2a}{y}\]
\[\text { At }\left( x_1 , y_1 \right)\]
\[\text { Slope of tangent } = \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =\frac{2a}{y_1}=m\]
\[\text { Equation of tangent is },\]
\[y - y_1 = m\left( x - x_1 \right)\]
\[ \Rightarrow y - y_1 = \frac{2a\left( x - x_1 \right)}{y_1}\]
\[ \Rightarrow y y_1 - {y_1}^2 = 2ax - 2a x_1 \]
\[ \Rightarrow y y_1 - 4a x_1 = 2ax - 2a x_1 \]
\[ \Rightarrow y y_1 = 2ax + 2a x_1 \]
\[ \Rightarrow y y_1 = 2a\left( x + x_1 \right)\]
\[\text { Equation of normal is,}\]
\[y - y_1 = \frac{1}{\text { Slope of tangen}t} \left( x - x_1 \right)\]
\[ \Rightarrow y - y_1 = \frac{- 1}{m}\left( x - x_1 \right)\]
\[ \Rightarrow y - y_1 = \frac{- y_1}{2a}\left( x - x_1 \right)\]
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