Question

Show that the family of curves for which the slope of the tangent at any point (x, y) on it is $\frac{x^2+y^2}{2xy}$ is given by x² − y² = Cx.

Answer 1

We have,
$$\begin{gathered} \frac{dy}{dx}=\frac{x^2+y^2}{2xy} \\ \mathrm{Let}y=vx \\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx} \\ \therefore v+x\frac{dv}{dx}=\frac{x^2+v^2{x}^2}{2v{x}^2} \\ \Rightarrow x\frac{dv}{dx}=\frac{1+v^2}{2v}-v \\ \Rightarrow x\frac{dv}{dx}=\frac{1+v^2-2v^2}{2v} \\ \Rightarrow x\frac{dv}{dx}=\frac{1-v^2}{2v} \\ \Rightarrow \frac{2v}{1-v^2}dv=\frac{1}{x}dx \end{gathered}$$

$$\begin{gathered} \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{2v}{1-v^2}dy=\int \frac{1}{x}dx \\ \Rightarrow -\log \left|1-v^2\right|=\log \left|x\right|-\log \left|C\right| \\ \Rightarrow \log \left|\frac{1-v^2}{C}\right|=-\log \left|x\right| \\ \Rightarrow 1-v^2=\frac{C}{x} \\ \Rightarrow 1-{\left(\frac{y}{x}\right)}^2=\frac{C}{x} \\ \Rightarrow \frac{x^2-y^2}{x^2}=\frac{C}{x} \\ \Rightarrow x^2-y^2=Cx \end{gathered}$$