Question Number 62276 by behi83417@gmail.com last updated on 18/Jun/19
$$\begin{cases}{\frac{\sqrt{\boldsymbol{\mathrm{x}}}}{\boldsymbol{\mathrm{a}}}+\frac{\sqrt{\boldsymbol{\mathrm{y}}}}{\boldsymbol{\mathrm{b}}}=\mathrm{1}}\\{\frac{\sqrt{\boldsymbol{\mathrm{a}}}}{\boldsymbol{\mathrm{x}}}+\frac{\sqrt{\boldsymbol{\mathrm{b}}}}{\boldsymbol{\mathrm{y}}}=\mathrm{1}}\end{cases}\:\:\:\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}}\in\mathrm{R}^{+} \\ $$
Answered by mr W last updated on 19/Jun/19
$${let}\:{X}=\sqrt{{x}},\:{Y}=\sqrt{{y}},\:{A}=\sqrt{{a}},\:{B}=\sqrt{{b}} \\ $$$$\frac{{X}}{{A}^{\mathrm{2}} }+\frac{{Y}}{{B}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow{Y}={B}^{\mathrm{2}} \left(\mathrm{1}−\frac{{X}}{{A}^{\mathrm{2}} }\right) \\ $$$$\frac{{A}}{{X}^{\mathrm{2}} }+\frac{{B}}{{Y}^{\mathrm{2}} }=\mathrm{1} \\ $$$${AY}^{\mathrm{2}} +{BX}^{\mathrm{2}} ={X}^{\mathrm{2}} {Y}^{\mathrm{2}} \\ $$$${AB}^{\mathrm{4}} \left(\mathrm{1}−\frac{\mathrm{2}{X}}{{A}^{\mathrm{2}} }+\frac{{X}^{\mathrm{2}} }{{A}^{\mathrm{4}} }\right)+{BX}^{\mathrm{2}} ={X}^{\mathrm{2}} {B}^{\mathrm{4}} \left(\mathrm{1}−\frac{\mathrm{2}{X}}{{A}^{\mathrm{2}} }+\frac{{X}^{\mathrm{2}} }{{A}^{\mathrm{4}} }\right) \\ $$$${AB}^{\mathrm{4}} \left({A}^{\mathrm{4}} −\mathrm{2}{A}^{\mathrm{2}} {X}+{X}^{\mathrm{2}} \right)+{A}^{\mathrm{4}} {BX}^{\mathrm{2}} ={X}^{\mathrm{2}} {B}^{\mathrm{4}} \left({A}^{\mathrm{4}} −\mathrm{2}{A}^{\mathrm{2}} {X}+{X}^{\mathrm{2}} \right) \\ $$$${A}^{\mathrm{5}} {B}^{\mathrm{4}} −\mathrm{2}{A}^{\mathrm{3}} {B}^{\mathrm{4}} {X}+{AB}^{\mathrm{4}} {X}^{\mathrm{2}} +{A}^{\mathrm{4}} {BX}^{\mathrm{2}} ={A}^{\mathrm{4}} {B}^{\mathrm{4}} {X}^{\mathrm{2}} −\mathrm{2}{A}^{\mathrm{2}} {B}^{\mathrm{4}} {X}^{\mathrm{3}} +{B}^{\mathrm{4}} {X}^{\mathrm{4}} \\ $$$$\Rightarrow{X}^{\mathrm{4}} −\mathrm{2}{A}^{\mathrm{2}} {X}^{\mathrm{3}} −{A}\left(\frac{{A}^{\mathrm{3}} }{{B}^{\mathrm{3}} }+\mathrm{1}−{A}^{\mathrm{3}} \right){X}^{\mathrm{2}} +\mathrm{2}{A}^{\mathrm{3}} {X}−{A}^{\mathrm{5}} =\mathrm{0} \\ $$$$\Rightarrow{X}=….. \\ $$