Question

Show that all curves for which the slope at any point (x, y) on it is \[\frac{x^2 + y^2}{2xy}\]  are rectangular hyperbola.

Answer 1

We have, 
\[\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}\]
\[\text{ Let }y = vx\]
\[\frac{dy}{dx} = v + x\frac{dv}{dx}\]
\[ \therefore v + x\frac{dv}{dx} = \frac{x^2 + v^2 x^2}{2v x^2}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{1 + v^2}{2v} - v\]
\[ \Rightarrow \frac{x dv}{dx} = \frac{1 + v^2 - 2 v^2}{2v}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{1 - v^2}{2v}\]
\[ \Rightarrow \frac{2v}{1 - v^2}dv = \frac{1}{x}dx\]
Integrating both sides, we get
\[\int\frac{2v}{1 - v^2}dy = \int\frac{1}{x}dx\]
\[ \Rightarrow - \log \left| 1 - v^2 \right| = \log \left| x \right| - \log \left| C \right|\]
\[ \Rightarrow \log \left| \frac{1 - v^2}{C} \right| = - \log \left| x \right|\]
\[ \Rightarrow 1 - v^2 = \frac{C}{x}\]
\[ \Rightarrow 1 - \left( \frac{y}{x} \right)^2 = \frac{C}{x}\]
\[ \Rightarrow \frac{x^2 - y^2}{x^2} = \frac{C}{x}\]
\[ \Rightarrow x^2 - y^2 = Cx\]
Thus,
\[x^2 - y^2 = Cx\] is the equation of the rectangular hyperbola.