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#