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

$$\{\begin{array}{l}\frac{\sqrt{\mathbf{x}}}{\mathbf{a}}+\frac{\sqrt{\mathbf{y}}}{\mathbf{b}}=1\\ \frac{\sqrt{\mathbf{a}}}{\mathbf{x}}+\frac{\sqrt{\mathbf{b}}}{\mathbf{y}}=1\end{array}\mathbf{a},\mathbf{b}\in {\mathrm{R}}^{+}$$

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

$$letX=\sqrt{x},Y=\sqrt{y},A=\sqrt{a},B=\sqrt{b}$$$$\frac{X}{A^2}+\frac{Y}{B^2}=1$$$$\Rightarrow Y=B^2(1-\frac{X}{A^2})$$$$\frac{A}{X^2}+\frac{B}{Y^2}=1$$$${AY}^2+{BX}^2=X^2{Y}^2$$$${AB}^4(1-\frac{2X}{A^2}+\frac{X^2}{A^4})+{BX}^2=X^2{B}^4(1-\frac{2X}{A^2}+\frac{X^2}{A^4})$$$${AB}^4(A^4-2A^{2}X+X^2)+A^4{BX}^2=X^2{B}^4(A^4-2A^{2}X+X^2)$$$$A^5{B}^4-2A^3{B}^{4}X+{AB}^4{X}^2+A^4{BX}^2=A^4{B}^4{X}^2-2A^2{B}^4{X}^3+B^4{X}^4$$$$\Rightarrow X^4-2A^2{X}^3-A(\frac{A^3}{B^3}+1-A^3)X^2+2A^{3}X-A^5=0$$$$\Rightarrow X=.....$$

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