If \((u^2 + v^2) \propto (x^2 + y^2)\) and\(uv \propto xy\), then show that\((u + v) \propto (x + y)\) when \(\frac{u}{v} + \frac{v}{u} = \frac{x}{y} + \frac{y}{x}\)."

Q.If \((u^2 + v^2) \propto (x^2 + y^2)\) and\(uv \propto xy\), then show that\((u + v) \propto (x + y)\) when \(\frac{u}{v} + \frac{v}{u} = \frac{x}{y} + \frac{y}{x}\)."

Since \((u^2 + v^2) \propto (x^2 + y^2)\), Therefore \((u^2 + v^2) = k(x^2 + y^2)\) ——— (i) [where \(k\) is a non-zero constant] And since \(uv \propto xy\), Therefore \(uv = m \cdot xy\) ——— (ii) [where \(m\) is a non-zero constant] Therefore, \(\frac{u^2 + v^2}{uv} = \frac{k}{m} \cdot \frac{x^2 + y^2}{xy}\) [Dividing equation (i) by equation (ii)] Or, \(\frac{u}{v} + \frac{v}{u} = \frac{k}{m} \left( \frac{x}{y} + \frac{y}{x} \right)\) Again, \(\frac{u}{v} + \frac{v}{u} = \frac{x}{y} + \frac{y}{x}\) Therefore, \(\frac{k}{m} = 1\) So, \(k = m\) Therefore, \((u + v)^2 = u^2 + v^2 + 2uv\) \(= k(x^2 + y^2) + 2mxy\) [from equations (i) and (ii)] \(= k(x^2 + y^2) + 2kxy\) [since \(k = m\)] \(= k(x^2 + y^2 + 2xy)\) \(= k(x + y)^2\) Therefore, \(u + v = \sqrt{k}(x + y)\) Therefore, \((u + v) \propto (x + y)\)