Question-220246

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Question Number 220246 by Spillover last updated on 10/May/25

Answered by Spillover last updated on 11/May/25

Answered by mr W last updated on 10/May/25

Commented by mr W last updated on 10/May/25

c=a+b y^2 =((√3)c)^2 +a^2 =3c^2 +a^2 x^2 =c^2 +(c+b)^2 −c(c+b)=c^2 +b^2 +bc (y^2 /x^2 )=((3c^2 +a^2 )/(c^2 +b^2 +bc))=(2/1) 3c^2 +a^2 =2c^2 +2b^2 +2bc (a+b)^2 +a^2 =2b^2 +2b(a+b) 2a^2 =3b^2 ⇒(a^2 /b^2 )=(3/2) ✓

$${c}={a}+{b} \\ $$$${y}^{\mathrm{2}} =\left(\sqrt{\mathrm{3}}{c}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} =\mathrm{3}{c}^{\mathrm{2}} +{a}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} ={c}^{\mathrm{2}} +\left({c}+{b}\right)^{\mathrm{2}} −{c}\left({c}+{b}\right)={c}^{\mathrm{2}} +{b}^{\mathrm{2}} +{bc} \\ $$$$\frac{{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }=\frac{\mathrm{3}{c}^{\mathrm{2}} +{a}^{\mathrm{2}} }{{c}^{\mathrm{2}} +{b}^{\mathrm{2}} +{bc}}=\frac{\mathrm{2}}{\mathrm{1}} \\ $$$$\mathrm{3}{c}^{\mathrm{2}} +{a}^{\mathrm{2}} =\mathrm{2}{c}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{bc} \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} =\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{b}\left({a}+{b}\right) \\ $$$$\mathrm{2}{a}^{\mathrm{2}} =\mathrm{3}{b}^{\mathrm{2}} \\ $$$$\Rightarrow\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\frac{\mathrm{3}}{\mathrm{2}}\:\checkmark \\ $$

Commented by Spillover last updated on 10/May/25