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Question Number 62276 by behi83417@gmail.com last updated on 18/Jun/19

{ ((((√x)/a)+((√y)/b)=1)),((((√a)/x)+((√b)/y)=1)) :} a,b∈R^+

$$\begin{cases}{\frac{\sqrt{\boldsymbol{\mathrm{x}}}}{\boldsymbol{\mathrm{a}}}+\frac{\sqrt{\boldsymbol{\mathrm{y}}}}{\boldsymbol{\mathrm{b}}}=\mathrm{1}}\\{\frac{\sqrt{\boldsymbol{\mathrm{a}}}}{\boldsymbol{\mathrm{x}}}+\frac{\sqrt{\boldsymbol{\mathrm{b}}}}{\boldsymbol{\mathrm{y}}}=\mathrm{1}}\end{cases}\:\:\:\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}}\in\mathrm{R}^{+} \\ $$

Answered by mr W last updated on 19/Jun/19

let X=(√x), Y=(√y), A=(√a), B=(√b) (X/A^2 )+(Y/B^2 )=1 ⇒Y=B^2 (1−(X/A^2 )) (A/X^2 )+(B/Y^2 )=1 AY^2 +BX^2 =X^2 Y^2 AB^4 (1−((2X)/A^2 )+(X^2 /A^4 ))+BX^2 =X^2 B^4 (1−((2X)/A^2 )+(X^2 /A^4 )) AB^4 (A^4 −2A^2 X+X^2 )+A^4 BX^2 =X^2 B^4 (A^4 −2A^2 X+X^2 ) A^5 B^4 −2A^3 B^4 X+AB^4 X^2 +A^4 BX^2 =A^4 B^4 X^2 −2A^2 B^4 X^3 +B^4 X^4 ⇒X^4 −2A^2 X^3 −A((A^3 /B^3 )+1−A^3 )X^2 +2A^3 X−A^5 =0 ⇒X=.....

$${let}\:{X}=\sqrt{{x}},\:{Y}=\sqrt{{y}},\:{A}=\sqrt{{a}},\:{B}=\sqrt{{b}} \\ $$$$\frac{{X}}{{A}^{\mathrm{2}} }+\frac{{Y}}{{B}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow{Y}={B}^{\mathrm{2}} \left(\mathrm{1}−\frac{{X}}{{A}^{\mathrm{2}} }\right) \\ $$$$\frac{{A}}{{X}^{\mathrm{2}} }+\frac{{B}}{{Y}^{\mathrm{2}} }=\mathrm{1} \\ $$$${AY}^{\mathrm{2}} +{BX}^{\mathrm{2}} ={X}^{\mathrm{2}} {Y}^{\mathrm{2}} \\ $$$${AB}^{\mathrm{4}} \left(\mathrm{1}−\frac{\mathrm{2}{X}}{{A}^{\mathrm{2}} }+\frac{{X}^{\mathrm{2}} }{{A}^{\mathrm{4}} }\right)+{BX}^{\mathrm{2}} ={X}^{\mathrm{2}} {B}^{\mathrm{4}} \left(\mathrm{1}−\frac{\mathrm{2}{X}}{{A}^{\mathrm{2}} }+\frac{{X}^{\mathrm{2}} }{{A}^{\mathrm{4}} }\right) \\ $$$${AB}^{\mathrm{4}} \left({A}^{\mathrm{4}} −\mathrm{2}{A}^{\mathrm{2}} {X}+{X}^{\mathrm{2}} \right)+{A}^{\mathrm{4}} {BX}^{\mathrm{2}} ={X}^{\mathrm{2}} {B}^{\mathrm{4}} \left({A}^{\mathrm{4}} −\mathrm{2}{A}^{\mathrm{2}} {X}+{X}^{\mathrm{2}} \right) \\ $$$${A}^{\mathrm{5}} {B}^{\mathrm{4}} −\mathrm{2}{A}^{\mathrm{3}} {B}^{\mathrm{4}} {X}+{AB}^{\mathrm{4}} {X}^{\mathrm{2}} +{A}^{\mathrm{4}} {BX}^{\mathrm{2}} ={A}^{\mathrm{4}} {B}^{\mathrm{4}} {X}^{\mathrm{2}} −\mathrm{2}{A}^{\mathrm{2}} {B}^{\mathrm{4}} {X}^{\mathrm{3}} +{B}^{\mathrm{4}} {X}^{\mathrm{4}} \\ $$$$\Rightarrow{X}^{\mathrm{4}} −\mathrm{2}{A}^{\mathrm{2}} {X}^{\mathrm{3}} −{A}\left(\frac{{A}^{\mathrm{3}} }{{B}^{\mathrm{3}} }+\mathrm{1}−{A}^{\mathrm{3}} \right){X}^{\mathrm{2}} +\mathrm{2}{A}^{\mathrm{3}} {X}−{A}^{\mathrm{5}} =\mathrm{0} \\ $$$$\Rightarrow{X}=..... \\ $$

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