\(2x^2 + 3y^2 \propto xy\) Or, \(2x^2 + 3y^2 = k \cdot xy\) Or, \(\frac{2x^2 + 3y^2}{xy} = k\) Or, \(\frac{2x^2 + 3y^2}{2\sqrt{6} \cdot xy} = \frac{k}{2\sqrt{6}}\) \[ \text{[Dividing both sides by } 2\sqrt{6} \text{]} \] Or, \(\frac{2x^2 + 3y^2}{2\sqrt{6} \cdot xy} = p\) [Let \(\frac{k}{2\sqrt{6}} = p\)] Or, \(\frac{2x^2 + 3y^2 + 2\sqrt{6}xy}{2x^2 + 3y^2 - 2\sqrt{6}xy} = \frac{p + 1}{p - 1}\) \[ \text{[Using componendo and dividendo]} \] Or, \(\frac{(\sqrt{2}x + \sqrt{3}y)^2}{(\sqrt{2}x - \sqrt{3}y)^2} = \frac{p + 1}{p - 1}\) Or, \(\frac{\sqrt{2}x + \sqrt{3}y}{\sqrt{2}x - \sqrt{3}y} = \sqrt{\frac{p + 1}{p - 1}}\) Or, \(\frac{\sqrt{2}x + \sqrt{3}y}{\sqrt{2}x - \sqrt{3}y} = q\) \[ \text{[Let } \sqrt{\frac{p + 1}{p - 1}} = q \text{]} \] Or, \(\frac{\sqrt{2}x + \sqrt{3}y + \sqrt{2}x - \sqrt{3}y}{\sqrt{2}x + \sqrt{3}y - \sqrt{2}x + \sqrt{3}y} = \frac{q + 1}{q - 1}\) \[ \text{[Using componendo and dividendo]} \] Or, \(\frac{2\sqrt{2}x}{2\sqrt{3}y} = \frac{q + 1}{q - 1}\) Or, \(\frac{x}{y} = \frac{q + 1}{q - 1} \cdot \frac{\sqrt{3}}{\sqrt{2}} =\) constant ∴ \(x \propto y\) (Proved)