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Question Number 173163 by dragan91 last updated on 07/Jul/22

Answered by mr W last updated on 07/Jul/22

y=±x  (x−2(a−1))^2 +(x+a)^2 =1  2x^2 −2(a−2)x+(a−1)(5a−3)=0  (a−2)^2 −2(a−1)(5a−3)>0  9a^2 −12a+2<0  ((2−(√2))/3)<a<((2+(√2))/3)  (x−2(a−1))^2 +(−x+a)^2 =1  2x^2 −2(3a−2)x+(a−1)(5a−3)=0  (3a−2)^2 −2(a−1)(5a−3)>0  a^2 −4a+2<0  2−(√2)<a<2+(√2)    ⇒2−(√2)<a<1 ∨ 1<a<((2+(√2))/3)

$${y}=\pm{x} \\ $$$$\left({x}−\mathrm{2}\left({a}−\mathrm{1}\right)\right)^{\mathrm{2}} +\left({x}+{a}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}\left({a}−\mathrm{2}\right){x}+\left({a}−\mathrm{1}\right)\left(\mathrm{5}{a}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\left({a}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}\left({a}−\mathrm{1}\right)\left(\mathrm{5}{a}−\mathrm{3}\right)>\mathrm{0} \\ $$$$\mathrm{9}{a}^{\mathrm{2}} −\mathrm{12}{a}+\mathrm{2}<\mathrm{0} \\ $$$$\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{3}}<{a}<\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$$$\left({x}−\mathrm{2}\left({a}−\mathrm{1}\right)\right)^{\mathrm{2}} +\left(−{x}+{a}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{3}{a}−\mathrm{2}\right){x}+\left({a}−\mathrm{1}\right)\left(\mathrm{5}{a}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\left(\mathrm{3}{a}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}\left({a}−\mathrm{1}\right)\left(\mathrm{5}{a}−\mathrm{3}\right)>\mathrm{0} \\ $$$${a}^{\mathrm{2}} −\mathrm{4}{a}+\mathrm{2}<\mathrm{0} \\ $$$$\mathrm{2}−\sqrt{\mathrm{2}}<{a}<\mathrm{2}+\sqrt{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow\mathrm{2}−\sqrt{\mathrm{2}}<{a}<\mathrm{1}\:\vee\:\mathrm{1}<{a}<\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$

Commented by mr W last updated on 07/Jul/22

read my solution again!  both cases y=±x are considered.

$${read}\:{my}\:{solution}\:{again}! \\ $$$${both}\:{cases}\:{y}=\pm{x}\:{are}\:{considered}. \\ $$

Commented by dragan91 last updated on 08/Jul/22

Sorry. Well done

$$\mathrm{Sorry}.\:\mathrm{Well}\:\mathrm{done} \\ $$

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