If A.M. and G.M. of x and y are in the ratio | vedclass

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(B) Given that $\frac{A.M.}{G.M.} = \frac{p}{q}$,where $A.M. = \frac{x+y}{2}$ and $G.M. = \sqrt{xy}$.So,$\frac{x+y}{2\sqrt{xy}} = \frac{p}{q} \dots (i

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$p + \sqrt{p^2 - q^2} : p - \sqrt{p^2 - q^2}$

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(B) Given that $\frac{A.M.}{G.M.} = \frac{p}{q}$,where $A.M. = \frac{x+y}{2}$ and $G.M. = \sqrt{xy}$.So,$\frac{x+y}{2\sqrt{xy}} = \frac{p}{q} \dots (i)$.Using the property of componendo and dividendo,let $x = u^2$ and $y = v^2$. Then $\frac{u^2+v^2}{2uv} = \frac{p}{q}$.$\frac{u^2+v^2+2uv}{u^2+v^2-2uv} = \frac{p+q}{p-q} \implies \frac{(u+v)^2}{(u-v)^2} = \frac{p+q}{p-q}$.Alternatively,from $(i)$,$\frac{x+y}{2\sqrt{xy}} = \frac{p}{q} \implies \frac{x+y}{2\sqrt{xy}} + 1 = \frac{p}{q} + 1$ and $\frac{x+y}{2\sqrt{xy}} - 1 = \frac{p}{q} - 1$.This leads to $\frac{(\sqrt{x}+\sqrt{y})^2}{2\sqrt{xy}} = \frac{p+q}{q}$ and $\frac{(\sqrt{x}-\sqrt{y})^2}{2\sqrt{xy}} = \frac{p-q}{q}$.Dividing these two,we get $\frac{(\sqrt{x}+\sqrt{y})^2}{(\sqrt{x}-\sqrt{y})^2} = \frac{p+q}{p-q}$.Taking the square root,$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \sqrt{\frac{p+q}{p-q}}$.Applying componendo and dividendo again,$\frac{\sqrt{x}}{\sqrt{y}} = \frac{\sqrt{p+q} + \sqrt{p-q}}{\sqrt{p+q} - \sqrt{p-q}}$.Squaring both sides,$\frac{x}{y} = \frac{(p+q) + (p-q) + 2\sqrt{p^2-q^2}}{(p+q) + (p-q) - 2\sqrt{p^2-q^2}} = \frac{2p + 2\sqrt{p^2-q^2}}{2p - 2\sqrt{p^2-q^2}} = \frac{p + \sqrt{p^2-q^2}}{p - \sqrt{p^2-q^2}}$.

If $A.M.$ and $G.M.$ of $x$ and $y$ are in the ratio $p : q$,then $x : y$ is

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