Question

Find the points on the curve y² = 2x³ at which the slope of the tangent is 3.

Answer 1

Let (x₁, y₁) be the required point.
Given:
$$\begin{gathered} y^2=2x^3 \\ \mathrm{Since}\left(x_1{y}_1\right)\mathrm{lies}\mathrm{on}\mathrm{a}\mathrm{curve},{y_1}^2=2{x_1}^3....\left(1\right) \\ \Rightarrow 2y\frac{dy}{dx}=6x^2 \\ \Rightarrow \frac{dy}{dx}=\frac{6x^2}{2y}=\frac{3x^2}{y} \\ \text{Slope of the tangent at}\left(x,y\right)\text{=}\frac{3{x_1}^2}{y_1} \\ \mathrm{S}\text{lope of the tangent=3 [Given]} \\ \therefore \frac{3{x_1}^2}{y_1}=3....\left(2\right) \\ \Rightarrow y_1={x_1}^2 \\ \text{On substituting the value of}{y}_1\text{in eq. (1), we get} \\ {x_1}^4=2{x_1}^3 \\ \Rightarrow {x_1}^3\left(x_1-2\right)=0 \\ \Rightarrow x_1=0,2 \\ \text{Case 1} \\ \text{When}{x}_1=0,y_1=x^2=0.\text{Thus, we get the point}\left(0,0\right)\text{. But, it does not satisfy eq. (2).} \\ \text{So, we can ignore (0, 0).} \\ \text{Case 2} \\ \text{When}{x}_1=2,y_1={x_1}^2=4.\text{Thus, we get the point}\left(2,4\right)\text{.} \end{gathered}$$