x-a-y-b-1-a-x-b-y-1-a-b-R-

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

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

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

l e t X = \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

{A Y}^{2} + {B X}^{2} = X^{2} Y^{2}

{A B}^{4} (1 - \frac{2 X}{A^{2}} + \frac{X^{2}}{A^{4}}) + {B X}^{2} = X^{2} B^{4} (1 - \frac{2 X}{A^{2}} + \frac{X^{2}}{A^{4}})

{A B}^{4} (A^{4} - 2 A^{2} X + X^{2}) + A^{4} {B X}^{2} = X^{2} B^{4} (A^{4} - 2 A^{2} X + X^{2})

A^{5} B^{4} - 2 A^{3} B^{4} X + {A B}^{4} X^{2} + A^{4} {B X}^{2} = A^{4} B^{4} X^{2} - 2 A^{2} B^{4} X^{3} + B^{4} X^{4}

\Rightarrow X^{4} - 2 A^{2} X^{3} - A (\frac{A^{3}}{B^{3}} + 1 - A^{3}) X^{2} + 2 A^{3} X - A^{5} = 0

\Rightarrow X = \dots . .

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