Question Number 209520 by Ismoiljon_008 last updated on 12/Jul/24
$$ \\ $$$$\:\:\:{At}\:{what}\:{value}\:{of}\:\:{a}\:{does}\:{the}\:{system} \\ $$$$\:\:\:{of}\:{equations}\:{have}\:\mathrm{4}\:{solutions}\:? \\ $$$$\:\:\:\left\{\:{x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{0}\right. \\ $$$$\:\:\:\left\{\left({x}−{a}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1}\right. \\ $$$$ \\ $$
Commented by Ismoiljon_008 last updated on 12/Jul/24
$$\:\:\:{help}\:{please} \\ $$
Answered by Rasheed.Sindhi last updated on 12/Jul/24
$$\begin{cases}{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{0}\Rightarrow{x}^{\mathrm{2}} ={y}^{\mathrm{2}} }\\{\left({x}−{a}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1}}\end{cases}\: \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{ax}+{a}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{ax}+{a}^{\mathrm{2}} +{x}^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{ax}+{a}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{2}{a}\pm\sqrt{\mathrm{4}{a}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{2}\right)\left({a}^{\mathrm{2}} −\mathrm{1}\right)}}{\mathrm{4}} \\ $$$${x}=\frac{\mathrm{2}{a}\pm\mathrm{2}\sqrt{\mathrm{2}−{a}^{\mathrm{2}} }}{\mathrm{4}} \\ $$$${x}=\frac{{a}\pm\sqrt{\mathrm{2}−{a}^{\mathrm{2}} }}{\mathrm{2}}\Rightarrow\mathrm{2}−{a}^{\mathrm{2}} \neq\mathrm{0}\Rightarrow{a}\neq\pm\sqrt{\mathrm{2}} \\ $$$${For}\:{all}\:{other}\:{values}\:{of}\:{a} \\ $$$${x}\:{has}\:{two}\:{values} \\ $$$${y}^{\mathrm{2}} ={x}^{\mathrm{2}} \Rightarrow{y}=\pm\mid{x}\mid \\ $$$$\left(\frac{{a}+\sqrt{\mathrm{2}−{a}^{\mathrm{2}} }}{\mathrm{2}}\:,\pm\frac{{a}+\sqrt{\mathrm{2}−{a}^{\mathrm{2}} }}{\mathrm{2}}\right) \\ $$$$\left(\frac{{a}−\sqrt{\mathrm{2}−{a}^{\mathrm{2}} }}{\mathrm{2}}\:,\pm\frac{{a}−\sqrt{\mathrm{2}−{a}^{\mathrm{2}} }}{\mathrm{2}}\right) \\ $$$${a}\in\:\mathbb{R}−\left\{\pm\sqrt{\mathrm{2}}\:\right\} \\ $$
Commented by Ismoiljon_008 last updated on 12/Jul/24
$$\:\:\:{thank}\:{you}\:{very}\:{much}\: \\ $$