Question

Find the equation of a curve passing through the point (−2, 3), given that the slope of the tangent to the curve at any point (x, y) is $\frac{2x}{y^2}$.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}=\frac{2x}{y^2} \\ \Rightarrow y^{2}dy=2xdx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int y^{2}dy=2\int xdx \\ \Rightarrow \frac{y^3}{3}=x^2+C.....\left(1\right) \\ \mathrm{Now}\mathrm{the}\mathrm{given}\mathrm{curve}\mathrm{passes}\mathrm{theough}\left(-2,3\right) \\ \mathrm{Therefore},\mathrm{when}x=-2,y=3 \\ \mathrm{Substituting}x=-2\mathrm{and}y=3\mathrm{in}\left(1\right)\mathrm{we}\mathrm{get} \\ \frac{3^3}{3}={\left(-2\right)}^2+C \\ \Rightarrow 9=4+C \\ \Rightarrow C=5 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \frac{y^3}{3}=x^2+5 \\ \Rightarrow y^3=3x^2+15 \\ \Rightarrow y={\left(3x^2+15\right)}^{\frac{1}{3}} \end{gathered}$$