Question 840431
The intersection of a line and a hyperbola,
1.{{{x^2-y^2=-8}}}
2.{{{y=2x+1}}}
Subsstitute eq. 2 into eq. 1,
{{{x^2-(2x+1)^2=-8}}}
{{{x^2-(4x^2+4x+1)=-8}}}
{{{-3x^2-4x-1=-8}}}
{{{-3x^2-4x+7=0}}}
{{{3x^2+4x-7=0}}}
{{{(3x+7)(x-1)=0}}}
Two solutions:
{{{3x+7=0}}}
{{{3x=-7}}}
{{{x=-7/3}}}
Then
{{{y=2(-7/3)+1}}}
{{{y=-14/3+3/3}}}
{{{y=-11/3}}}
.
.
.
{{{x-1=0}}}
{{{x=1}}}
Then
{{{y=2x+1}}}
{{{y=2(1)+1}}}
{{{y=3}}}
(-7/3,-11/3) and (1,3)
.
.
{{{drawing(300,300,-5,5,-5,5,grid(1),circle(1,3,0.15),circle(-7/3,-11/3,0.15),graph(300,300,-5,5,-5,5,sqrt(x^2+8),-sqrt(x^2+8),2x+1))}}}