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

2 Answers
Jul 8, 2017

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.

Jul 8, 2017

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