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Math Help - algebraic manipulation for proof

  1. #1
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    algebraic manipulation for proof

    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)
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  2. #2
    MHF Contributor red_dog's Avatar
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    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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