Question

Find the equation of the normal to the curve ay² = x³ at the point (am², am³).

Answer 1

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