Question Number 173163 by dragan91 last updated on 07/Jul/22
Answered by mr W last updated on 07/Jul/22
$${y}=\pm{x} \\ $$$$\left({x}−\mathrm{2}\left({a}−\mathrm{1}\right)\right)^{\mathrm{2}} +\left({x}+{a}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}\left({a}−\mathrm{2}\right){x}+\left({a}−\mathrm{1}\right)\left(\mathrm{5}{a}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\left({a}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}\left({a}−\mathrm{1}\right)\left(\mathrm{5}{a}−\mathrm{3}\right)>\mathrm{0} \\ $$$$\mathrm{9}{a}^{\mathrm{2}} −\mathrm{12}{a}+\mathrm{2}<\mathrm{0} \\ $$$$\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{3}}<{a}<\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$$$\left({x}−\mathrm{2}\left({a}−\mathrm{1}\right)\right)^{\mathrm{2}} +\left(−{x}+{a}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{3}{a}−\mathrm{2}\right){x}+\left({a}−\mathrm{1}\right)\left(\mathrm{5}{a}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\left(\mathrm{3}{a}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}\left({a}−\mathrm{1}\right)\left(\mathrm{5}{a}−\mathrm{3}\right)>\mathrm{0} \\ $$$${a}^{\mathrm{2}} −\mathrm{4}{a}+\mathrm{2}<\mathrm{0} \\ $$$$\mathrm{2}−\sqrt{\mathrm{2}}<{a}<\mathrm{2}+\sqrt{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow\mathrm{2}−\sqrt{\mathrm{2}}<{a}<\mathrm{1}\:\vee\:\mathrm{1}<{a}<\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$
Commented by mr W last updated on 07/Jul/22
$${read}\:{my}\:{solution}\:{again}! \\ $$$${both}\:{cases}\:{y}=\pm{x}\:{are}\:{considered}. \\ $$
Commented by dragan91 last updated on 08/Jul/22
$$\mathrm{Sorry}.\:\mathrm{Well}\:\mathrm{done} \\ $$