Question 971232
{{{4x^2-y^2+8x+2y-1=0}}}.......group {{{x}}}'s and {{{y}}}'s

{{{(4x^2+8x)-(y^2-2y-1)=0}}}.........complete squares

{{{4(x^2+2x)-(y-1)^2=0}}}

{{{4(x^2+2x+1^2)-4*1^2-(y-1)^2=0}}}

{{{4(x+1)^2-4-(y-1)^2=0}}}

{{{4(x+1)^2-(y-1)^2=4}}}..........divide by {{{4}}}

{{{4(x+1)^2/4-(y-1)^2/4=4/4}}}

{{{(x+1)^2- (y-1)^2/4 = 1}}} ..........you have a hyperbola

compare it to the standard formula {{{(x-h)^2/a^2-(y-k)^2/b^2 = 1}}}

as you can see, {{{h=-1}}}, {{{k=1}}}, {{{a=1}}}, and {{{b=2}}}

then {{{c=sqrt(1^2+2^2)=sqrt(5)}}}
so, 
the center is at ({{{-1}}},{{{1}}}
foci:
(({{{h-c}}}, {{{a}}})  and  ({{{h+c}}},{{{ a}}}))
(({{{-1-sqrt(5)}}}, {{{1}}})  and ({{{-1+sqrt(5)}}}, {{{1}}}))
or, approximately (({{{-3.24}}},{{{ 1}}})  and  ({{{1.24}}},{{{ 1}}}))

vertices: | ({{{-2}}}, {{{1}}})  and  ({{{0}}}, {{{1}}})

semi-major axis length: {{{1}}}
semi-minor axis length: {{{2}}}

asymptotes: {{{y = 2x+3}}}  and  {{{y = -2x-1}}}



{{{drawing(600, 600, -5,5, -10, 10, 
circle(-3.24,1,.05),locate(-3.24,1,f(-3.24,1)),
circle(1.24,1,.05),locate(1.24,1,f(1.24,1)),
circle(-1,1,.05),locate(-1,1.5,C(-1,1)),
circle(-2,1.2,.05),locate(-2,1.2,V(-2,1)),
circle(0,1,.05),locate(0,1,V(0,1)),

 graph(600, 600, -5,5, -10, 10,-2x-1 ,sqrt(((x+1)^2- 1)4)+1,2x+3,-sqrt(((x+1)^2- 1)4)+1) ) }}}