Question

Find the points on the curve 2a²y = x³ − 3ax² where the tangent is parallel to x-axis.

Answer 1

Let (x₁, y₁) represent the required points.
The slope of the x-axis is 0.
Here,
$$\begin{gathered} 2a^{2}y=x^3-3a{x}^2 \\ \mathrm{Since},\mathrm{the}\mathrm{point}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{curve}. \\ \mathrm{Hence},2a^2{y}_1={x_1}^3-3a{x_1}^2...\left(1\right) \\ \mathrm{Now},2a^{2}y=x^3-3a{x}^2 \\ \text{On differentiating both sides w.r.t.}\mathit{\text{x}}\text{, we get} \\ 2a^2\frac{dy}{dx}=3x^2-6ax \\ \Rightarrow \frac{dy}{dx}=\frac{3x^2-6ax}{2a^2} \\ \text{Slope of the tangent at}\left(x_1,y_1\right)\text{=}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\text{=}\frac{3{x_1}^2-6a{x}_1}{2a^2} \\ \text{Given:} \\ \text{Slope of the tangent at}\left(x_1,y_1\right)\text{= Slope of the}\mathit{\text{x}}\text{-axis} \\ \Rightarrow \frac{3{x_1}^2-6a{x}_1}{2a^2}=0 \\ \Rightarrow 3{x_1}^2-6a{x}_1=0 \\ \Rightarrow x_1\left(3x_1-6a\right)=0 \\ \Rightarrow x_1=0\text{or}{x}_1=2a \\ \mathrm{Also}, \\ 2a^2{y}_1=0\text{or}2a^2{y}_1=8a^3-12a^3\text{[From eq. (1)]} \\ \Rightarrow y_1=0\text{or}{y}_1=-2a \\ \mathrm{Thus}\text{, the required points are}\left(0,0\right)\text{and}\left(2a,-2a\right)\text{.} \end{gathered}$$