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Question-189684




Question Number 189684 by normans last updated on 20/Mar/23
Commented by a.lgnaoui last updated on 21/Mar/23
Answered by mr W last updated on 20/Mar/23
Commented by mr W last updated on 21/Mar/23
x^2 +y^2 =((√5)a)^2 =5a^2   (16−y−y)^2 +22^2 =(2(√5)a)^2   y^2 −16y+185−5a^2 =0   ...(i)  ⇒y=8−(√(5a^2 −121))  (22−x−x)^2 +16^2 =(4a)^2   x^2 −22x+185−4a^2 =0   ...(ii)  ⇒x=11−(√(4a^2 −64))  (i)+(ii):  11x+8y=185−2a^2   11(11−(√(4a^2 −64)))+8(8−(√(5a^2 −121)))=185−2a^2   11(√(4a^2 −64))=2a^2 −8(√(5a^2 −121))  a^2 −41=8(√(5a^2 −121))  a^4 −402a^2 +10705=0  ⇒a^2 =201−32(√(29))≈28.675
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\left(\sqrt{\mathrm{5}}{a}\right)^{\mathrm{2}} =\mathrm{5}{a}^{\mathrm{2}} \\ $$$$\left(\mathrm{16}−{y}−{y}\right)^{\mathrm{2}} +\mathrm{22}^{\mathrm{2}} =\left(\mathrm{2}\sqrt{\mathrm{5}}{a}\right)^{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} −\mathrm{16}{y}+\mathrm{185}−\mathrm{5}{a}^{\mathrm{2}} =\mathrm{0}\:\:\:…\left({i}\right) \\ $$$$\Rightarrow{y}=\mathrm{8}−\sqrt{\mathrm{5}{a}^{\mathrm{2}} −\mathrm{121}} \\ $$$$\left(\mathrm{22}−{x}−{x}\right)^{\mathrm{2}} +\mathrm{16}^{\mathrm{2}} =\left(\mathrm{4}{a}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{22}{x}+\mathrm{185}−\mathrm{4}{a}^{\mathrm{2}} =\mathrm{0}\:\:\:…\left({ii}\right) \\ $$$$\Rightarrow{x}=\mathrm{11}−\sqrt{\mathrm{4}{a}^{\mathrm{2}} −\mathrm{64}} \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$$\mathrm{11}{x}+\mathrm{8}{y}=\mathrm{185}−\mathrm{2}{a}^{\mathrm{2}} \\ $$$$\mathrm{11}\left(\mathrm{11}−\sqrt{\mathrm{4}{a}^{\mathrm{2}} −\mathrm{64}}\right)+\mathrm{8}\left(\mathrm{8}−\sqrt{\mathrm{5}{a}^{\mathrm{2}} −\mathrm{121}}\right)=\mathrm{185}−\mathrm{2}{a}^{\mathrm{2}} \\ $$$$\mathrm{11}\sqrt{\mathrm{4}{a}^{\mathrm{2}} −\mathrm{64}}=\mathrm{2}{a}^{\mathrm{2}} −\mathrm{8}\sqrt{\mathrm{5}{a}^{\mathrm{2}} −\mathrm{121}} \\ $$$${a}^{\mathrm{2}} −\mathrm{41}=\mathrm{8}\sqrt{\mathrm{5}{a}^{\mathrm{2}} −\mathrm{121}} \\ $$$${a}^{\mathrm{4}} −\mathrm{402}{a}^{\mathrm{2}} +\mathrm{10705}=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\mathrm{201}−\mathrm{32}\sqrt{\mathrm{29}}\approx\mathrm{28}.\mathrm{675} \\ $$

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