Question 982534
{{{4y^2 - x^2 + 8y - 6x - 41 = 0}}}


{{{(4y^2 + 8y)- (x^2 + 6x) -41 = 0}}}....complete square


{{{4(y^2 + 2y+b^2)-4b^2 -(x^2 + 6x+b^2)-(-b^2) -41 = 0}}}


{{{4(y^2 + 2y+1^2)-4*1^2 -(x^2 + 6x+3^2)+3^2 -41 = 0}}}


{{{4(y + 1)^2-4 -(x + 3)^2+9 -41 = 0}}}


{{{4(y + 1)^2- (x + 3)^2 -36 = 0}}}


{{{4(y + 1)^2- (x + 3)^2 =36 }}}

{{{4(y + 1)^2/36 - (x + 3)^2/36=36/36 }}}


{{{(y + 1)^2/9- (x + 3)^2/36=1 }}}

{{{(y + 1)^2/3^2- (x + 3)^2/6^2=1 }}}

The standard form of an hyperbola with the transverse axis being vertical.

{{{( y - k )^2/a^2 - ( x - h )^2/b^2 = 1}}}

The center of the hyperbola is at the point ( {{{h}}},{{{k}}}) or  ({{{-3}}},{{{-1}}})

{{{a=3}}} and {{{b=6}}}, so
semimajor axis length {{{3}}}
semiminor axis length {{{ 6}}}

The foci are {{{c}}} units from the center of the hyperbola and is calculated from {{{c^2 = a^2 + b^2}}}.
{{{c^2 = 9 + 36 =45}}}=>{{{c = sqrt (45)=sqrt (9*5)}}}=>{{{c=3sqrt(5)}}}

The foci are therefore, ( {{{h}}}, {{{k+c}}} ) or ( {{{h}}}, {{{k-c}}} )

or  ({{{-3}}},{{{ -1-3sqrt(5)}}})  and ({{{-3}}}, ({{{-1+3sqrt(5)}}})
=>({{{-3}}},{{{ -7.7}}})  and ({{{-3}}}, {{{5.7}}})

The vertices are a units from the center of the hyperbola, ( {{{h}}}, {{{k +a}}} ) and ( {{{h}}}, {{{k -a}}} )
or ( {{{-3}}}, {{{-4}}} ) and ( {{{-3}}}, {{{2}}} )


asymptotes: 
{{{y = -x/2-5/2 }}} 
{{{ y = x/2+1/2}}}


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(-3,-1,.12),locate(-3,-1,C(-3,-1)),
circle(-3,-4,.12),locate(-3,-4,v(-3,-4)),
circle(-3,2,.12),locate(-3,2,v(-3,2)),
circle(-3,-7.7,.12),locate(-3,-7.7,F(-3,-7.7)),
circle(-3,5.7,.12),locate(-3,5.7,F(-3,5.7)),
 graph( 600, 600, -10, 10, -10, 10, x/2+1/2,-x/2-5/2,-sqrt(9((x + 3)^2/36+1))-1 ,sqrt(9((x + 3)^2/36+1))-1)) }}}