Question 147722
{{{x^2-4y^2+2x-32y=67}}}
Complete the square in both x and y.
{{{x^2+2x-4y^2-32y=67}}}
{{{(x^2+2x+1)-1-4(y^2+8y+16)+64=67}}}
{{{(x^2+2x+1)-4(y^2+8y+16)=67+1-64}}}
{{{(x+1)^2-4(y+4)^2=4}}}
{{{(x+1)^2/4-(y+4)^2=1}}}
This is an equation for a hyperbola which has a general form of
{{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}
{{{a=2}}}
{{{b=1}}}
Center of the hyperbola is (h,k) or in this case (-1,-4)
Distance between the foci is 2c where 
{{{c^2=a^2+b^2}}}
{{{c^2=4+1}}}
{{{c=sqrt(5)}}}
The eccentricity,e, is then
{{{e=c/a=sqrt(5)/2}}}
The foci of the hyperbola are located at
({{{h +- c}}},{{{k}}})=({{{-1+-sqrt(5)}}},{{{-4}}})
The vertices of the hyperbola are located at
({{{h +- a}}},{{{k}}})=({{{-1+-2}}},{{{-4}}})=(-3,-4) and (1,-4)
The equation of the asymptote lines is given by
{{{y=k +- (b/a)(x-h)}}}
{{{y=-4 +- (1/2)(x+1)}}}

The directrix of a hyperbola is given by
{{{x=h +- a^2/c}}}
{{{x=-1 +- 4/(sqrt(5)/2)}}}
{{{x=-1 +- 8/(sqrt(5))}}}