If ax+by varies as √(xy), Can you prove that ax^2+by^2 varies as xy?

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Answer 1 · 2017-07-08T07:30:29

Please see below.

As $ax+by$ varies as $sqrt(xy)$

$(ax+by)=ksqrt(xy)$ , where $k$ is a constant.

Squaring this we get

$(ax+by)^2=k^2xy$

or $(ax)^2+(by)^2+2abxy=k^2xy$

or $(ax)^2+(by)^2=(k^2-2ab)xy$

and as $k,a,b$ are constant $k^2-2ab$ is another constant

Hence $(ax)^2+(by)^2$ varies directly as $xy$ .

Answer 2 · 2017-07-08T07:35:41

Given

$ax+bypropsqrt(xy)$

$=>ax+by=ksqrt(xy)$ , where k = proportionality constant

$=>sqrt(xy)/(ax+by)=1/k$

$=>(xy)/(ax+by)^2=1/k^2$

$=>(4abxy)/(ax+by)^2=(4ab)/k^2$

$=>1-(4abxy)/(ax+by)^2=1-(4ab)/k^2$

$=>((ax+by)^2-4abxy)/(ax+by)^2=(k^2-4ab)/k^2$

$=>(ax-by)^2/(ax+by)^2=(k^2-4ab)/k^2=m^2 "(say)"$ , where m= constant

$=>(ax-by)/(ax+by)=m$

$=>(ax+by)/(ax-by)=1/m$

Now by componendo and dividendo we get

$=>(2ax)/(2by)=(1+m)/(1-m)$

$=>x=(1+m)/(1-m)xxb/axxy$

$=>x=nxxy$ ,

where $(1+m)/(1-m)xxb/a=n-> "another constant"$

Now

$(ax^2+by^2)/(xy)$

$=(an^2y^2+by^2)/(ny^2)$

$=(an^2+b)/n-> " A CONSTANT"$

Hence

$(ax^2+by^2)propxy$