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

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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.

Explanation:

As $a x + b y$ $ax+by$ varies as $\sqrt{x y}$ $sqrt(xy)$

$(a x + b y) = k \sqrt{x y}$ $(ax+by)=ksqrt(xy)$ , where $k$ $k$ is a constant.

Squaring this we get

${(a x + b y)}^{2} = k^{2} x y$ $(ax+by)^2=k^2xy$

or ${(a x)}^{2} + {(b y)}^{2} + 2 a b x y = k^{2} x y$ $(ax)^2+(by)^2+2abxy=k^2xy$

or ${(a x)}^{2} + {(b y)}^{2} = (k^{2} - 2 a b) x y$ $(ax)^2+(by)^2=(k^2-2ab)xy$

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

Hence ${(a x)}^{2} + {(b y)}^{2}$ $(ax)^2+(by)^2$ varies directly as $x y$ $xy$ .

Answer 2 · 2017-07-08T07:35:41

Given

$a x + b y \propto \sqrt{x y}$ $ax+bypropsqrt(xy)$

$\Rightarrow a x + b y = k \sqrt{x y}$ $=>ax+by=ksqrt(xy)$ , where k = proportionality constant

$\Rightarrow \frac{\sqrt{x y}}{a x + b y} = \frac{1}{k}$ $=>sqrt(xy)/(ax+by)=1/k$

$\Rightarrow \frac{x y}{{(a x + b y)}^{2}} = \frac{1}{k^{2}}$ $=>(xy)/(ax+by)^2=1/k^2$

$\Rightarrow \frac{4 a b x y}{{(a x + b y)}^{2}} = \frac{4 a b}{k^{2}}$ $=>(4abxy)/(ax+by)^2=(4ab)/k^2$

$\Rightarrow 1 - \frac{4 a b x y}{{(a x + b y)}^{2}} = 1 - \frac{4 a b}{k^{2}}$ $=>1-(4abxy)/(ax+by)^2=1-(4ab)/k^2$

$\Rightarrow \frac{{(a x + b y)}^{2} - 4 a b x y}{{(a x + b y)}^{2}} = \frac{k^{2} - 4 a b}{k^{2}}$ $=>((ax+by)^2-4abxy)/(ax+by)^2=(k^2-4ab)/k^2$

$\Rightarrow \frac{{(a x - b y)}^{2}}{{(a x + b y)}^{2}} = \frac{k^{2} - 4 a b}{k^{2}} = m^{2} (say)$ $=>(ax-by)^2/(ax+by)^2=(k^2-4ab)/k^2=m^2 "(say)"$ , where m= constant

$\Rightarrow \frac{a x - b y}{a x + b y} = m$ $=>(ax-by)/(ax+by)=m$

$\Rightarrow \frac{a x + b y}{a x - b y} = \frac{1}{m}$ $=>(ax+by)/(ax-by)=1/m$

Now by componendo and dividendo we get

$\Rightarrow \frac{2 a x}{2 b y} = \frac{1 + m}{1 - m}$ $=>(2ax)/(2by)=(1+m)/(1-m)$

$\Rightarrow x = \frac{1 + m}{1 - m} \times \frac{b}{a} \times y$ $=>x=(1+m)/(1-m)xxb/axxy$

$\Rightarrow x = n \times y$ $=>x=nxxy$ ,

where $\frac{1 + m}{1 - m} \times \frac{b}{a} = n \to another constant$ $(1+m)/(1-m)xxb/a=n-> "another constant"$

Now

$\frac{a x^{2} + b y^{2}}{x y}$ $(ax^2+by^2)/(xy)$

$= \frac{a n^{2} y^{2} + b y^{2}}{n y^{2}}$ $=(an^2y^2+by^2)/(ny^2)$

$= \frac{a n^{2} + b}{n} \to A CONSTANT$ $=(an^2+b)/n-> " A CONSTANT"$

Hence

$(a x^{2} + b y^{2}) \propto x y$ $(ax^2+by^2)propxy$

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

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