Q.If \(\cfrac{x}{y} \propto (x + y)\) and \(\cfrac{y}{x} \propto (x - y)\), then show that \((x^2 - y^2)\) is constant.

\(\cfrac{x}{y} \propto x + y\) i.e., \(\cfrac{x}{y} = k_1(x + y)\) [where \(k_1\) is a non-zero constant] \(\cfrac{y}{x} \propto x - y\) i.e., \(\cfrac{y}{x} = k_2(x - y)\) [where \(k_2\) is a non-zero constant] \(\therefore k_1(x + y) \cdot k_2(x - y) = \cfrac{x}{y} \cdot \cfrac{y}{x}\) i.e., \(k_1k_2(x^2 - y^2) = 1\) i.e., \(x^2 - y^2 = \cfrac{1}{k_1k_2} =\) constant (since \(k_1, k_2\) are constants) (Proved)
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