Question 337121
1.{{{x^2/4-y^2 = 1}}} 
2.{{{x=y^2 + 1}}} 
From eq. 2,
{{{y^2=x-1}}}
Substitute into eq. 1,
{{{x^2/4-(x-1)=1}}}
{{{x^2-4x+4=4}}}
{{{x^2-4x=0}}}
{{{x(x-4)=0}}}
Two solutions:
{{{x=0}}}
Then from eq. 2,
{{{y^2=-1}}}
No solution
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{{{x-4=0}}}
{{{x=4}}}
Then from eq. 2,
{{{y^2=4-1}}}
{{{y^2=3}}}
{{{y=0 +- sqrt(3)}}}
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({{{4}}},{{{-sqrt(3)}}}),({{{4}}},{{{sqrt(3)}}})


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circle(4,-sqrt(3),0.3),
circle(4,sqrt(3),0.3),
graph(300,300,-10,10,-10,10,sqrt(x-1),sqrt(x^2/4-1)),
graph(300,300,-10,10,-10,10,-sqrt(x-1),-sqrt(x^2/4-1)))}}}