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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 \[\left( \frac{a}{m^2}, \frac{2a}{m} \right)\] ?
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Solution
\[y^2 =4ax\]
\[\text { Differentiating both sides w.r.t.x,} \]
\[2y \frac{dy}{dx} = 4a\]
\[ \Rightarrow \frac{dy}{dx} = \frac{2a}{y}\]
\[\text { Given } \left( x_1 , y_1 \right) = \left( \frac{a}{m^2}, \frac{2a}{m} \right)\]
\[\text { Slope of tangent }= \left( \frac{dy}{dx} \right)_\left( \frac{a}{m^2}, \frac{2a}{m} \right) =\frac{2a}{\left( \frac{2a}{m} \right)}=m\]
\[\text { Equation of tangent is, }\]
\[y - y_1 = m \left( x - x_1 \right)\]
\[ \Rightarrow y - \frac{2a}{m} = m \left( x - \frac{a}{m^2} \right)\]
\[ \Rightarrow \frac{my - 2a}{m} = m\left( \frac{m^2 x - a}{m^2} \right)\]
\[ \Rightarrow my - 2a = m^2 x - a\]
\[ \Rightarrow m^2 x - my + a = 0\]
\[\text { Equation of normal is},\]
\[y - y_1 = \frac{1}{\text { Slope of tangent}} \left( x - x_1 \right)\]
\[ \Rightarrow y - \frac{2a}{m} = \frac{- 1}{m}\left( x - \frac{a}{m^2} \right)\]
\[ \Rightarrow \frac{my - 2a}{m} = \frac{- 1}{m}\left( \frac{m^2 x - a}{m^2} \right)\]
\[ \Rightarrow m^3 y - 2a m^2 = - m^2 x + a\]
\[ \Rightarrow m^2 x + m^3 y - 2a m^2 - a = 0\]
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