Q.If \( \cfrac{1}{y} - \cfrac{1}{x} \propto \cfrac{1}{x - y} \), then show that \( x \propto y \).


\(\cfrac{1}{y} - \cfrac{1}{x} \propto \cfrac{1}{x - y}\)
Or, \(\cfrac{x - y}{xy} = \cfrac{k}{x - y}\) [\( k \) is a nonzero constant]
Or, \(\cfrac{(x - y)^2}{xy} = k\)
Or, \(\cfrac{x^2 + y^2 - 2xy}{xy} = k\)
Or, \(\cfrac{x^2 + y^2}{xy} - 2 = k\)
Or, \(\cfrac{x^2 + y^2}{xy} = k + 2\)
Or, \(\cfrac{x^2 + y^2}{2xy} = \cfrac{k + 2}{2} = m\)

[Let, \(\cfrac{k + 2}{2} = m\)]




Or, \(\cfrac{x^2 + y^2 + 2xy}{x^2 + y^2 - 2xy} = \cfrac{m + 1}{m - 1}\)

[Using addition and division process]



Or, \(\cfrac{(x + y)^2}{(x - y)^2} = n\)

[Let, \(\cfrac{m + 1}{m - 1} = n\)]


Or, \(\cfrac{(x + y)}{(x - y)} = \sqrt{n}\)
Or, \(\cfrac{(x + y + x - y)}{(x + y - x + y)} = \cfrac{\sqrt{n} + 1}{\sqrt{n} - 1}\)

[Using addition and division process]



Or, \(\cfrac{\cancel2x}{\cancel2y} = \cfrac{\sqrt{n} + 1}{\sqrt{n} - 1} =\) constant
\(\therefore x \propto y\) [Proved]
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