# algebraic manipulation for proof

• Feb 8th 2009, 09:28 AM
tombrownington
algebraic manipulation for proof
Given the four variables x, x', y and y', with initial conditions:

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

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

show that:

$\displaystyle a^2b^2(x^2+y^2+x'^2+y'^2) = (a^2+b^2)(x^2+y^2)(x'^2+y'^2)$
• Feb 8th 2009, 11:08 AM
red_dog
$\displaystyle xx'+yy'=0\Rightarrow yy'=-xx'$

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

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

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

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

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

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

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

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

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

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

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