Question

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

Answer 1

Please see below.

Explanation:

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

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`