algebraic manipulation for proof

**tombrownington** · February 8th 2009, 09:28 AM

Given the four variables x, x', y and y', with initial conditions:

$xx'+yy' = 0$
$\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$
$\frac{x'^2}{a^2}+\frac{y'^2}{b^2} = 1$

where a and b are some non-zero arbitrary constants,

show that:

$a^2b^2(x^2+y^2+x'^2+y'^2) = (a^2+b^2)(x^2+y^2)(x'^2+y'^2)$

**red_dog** · February 8th 2009, 11:08 AM

$xx'+yy'=0\Rightarrow yy'=-xx'$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\Rightarrow b^2x^2+a^2y^2=a^2b^2$

$\frac{x'^2}{a^2}+\frac{y'^2}{b^2}=1\Rightarrow b^2x'^2+a^2y'^2=a^2b^2$

$a^2b^2(x^2+y^2+x'^2+y'^2)=$

$=a^2b^2(x^2+y^2)+a^2b^2(x'^2+y'^2)=$

$=(b^2x'^2+a^2y'^2)(x^2+y^2)+(b^2x^2+a^2y^2)(x'^2+y '^2)$

$=2b^2x^2x'^2+2a^2y^2y'^2+b^2x'^2y^2+a^2x^2y'^2+b^2 x^2y'^2+a^2x'^2y^2=$

$2b^2x^2x'^2+2a^2(-xx')^2+(a^2+b^2)(x'^2y^2+x^2y'^2)=$

$=(a^2+b^2)(2x^2x'^2+x'^2y^2+x^2y'^2)=$

$=(a^2+b^2)(x^2x'^2+y^2y'^2+x'^2y^2+x^2y'^2)=$

$=(a^2+b^2)(x^2(x'^2+y'^2)+y^2(x'^2+y'^2))=$

$=(a^2+b^2)(x^2+y^2)(x'^2+y'^2)$

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