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Question
Find the equation of the normal to the curve ay2 = x3 at the point (am2, am3) ?
Solution
\[a y^2 = x^3 \]
\[\text {Differentiating both sides w.r.t.x }, \]
\[2ay \frac{dy}{dx} = 3 x^2 \]
\[ \Rightarrow \frac{dy}{dx} = \frac{3 x^2}{2ay}\]
\[\text { Slope of tangent } = \left( \frac{dy}{dx} \right)_\left( a m^2 , a m^3 \right) =\frac{3 a^2 m^4}{2 a^2 m^3}=\frac{3m}{2}\]
\[\text { Given } \left( x_1 , y_1 \right) = \left( a m^2 , a m^3 \right)\]
\[\text { Equation of normal is },\]
\[y - y_1 = \frac{- 1}{m} \left( x - x_1 \right)\]
\[ \Rightarrow y - a m^3 = \frac{- 2}{3m} \left( x - a m^2 \right)\]
\[ \Rightarrow 3my - 3a m^4 = - 2x + 2a m^2 \]
\[ \Rightarrow 2x + 3my - a m^2 \left( 2 + 3 m^2 \right) = 0\]
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