Question 615643
My way:
{{{4x^2-16y^2+40x+128y-220=0}}} --> {{{(4x^2+40x)+(-16y^2+128y)=220}}} -->  {{{4(x^2+10x)-16(y^2-8y)=220}}}
Now, I add a term into each bracket to "complete the square" and (of course) add the same thing to the right side of the equal sign too.
{{{4(x^2+10x)-16(y^2-8y)=220}}} --> {{{4(x^2+10x+25)-16(y^2-8y+16)=220+4*25-16*16}}} -->  {{{4(x^2+10x+25)-16(y^2-8y+16)=220+100-256}}} -->  {{{4(x^2+10x-25)-16(y^2-8y+16)=64}}}
Next, I write the brackets as squares:
{{{4(x^2+10x+25)-16(y^2-8y+16)=64}}} -->  {{{4(x+5)^2-16(y-4)^2=64}}}
Finally, I divide both sides by the value on right side of the equal sign to get the equation in the form I like:
{{{4(x+5)^2-16(y-4)^2=64}}} --> {{{4(x+5)^2/64-16(y-4)^2/64=64/64}}} --> {{{(x+5)^2/16-(y-4)^2/4=1}}}
Or even better, I can write those denominators as squares, as in:
{{{(x+5)^2/4^2-(y-4)^2/2^2=1}}}
That matches the general equation for a hyperbola with a horizontal transverse axis, centered at (h,k):
{{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}
From that equation, I get the coordinates of the center and the vertices.
The center is (-5,4), from the numbers subtracted from x and y. (You can see that the curve is symmetrical with respect to x=-5 and symmetrical with respect to y=4).
When y=4, {{{(x+5)^2/4^2=1}}} --> {{{(x+5)^2=4^2}}} and that is the smallest possible value for {{{(x+5)^2}}}. For all other values of y,
{{{(x+5)^2/4^2=1+(y-4)^2/2^2>1}}} and {{{(x+5)^2>4^2}}}
So {{{abs(x+5)>=4}}}
That is as close to the center as the hyperbola can get, and it gives you the coordinates of the vertices:
{{{(x+5)^2=4^2}}} --> {{{x+5=4}}} or {{{x+5=4}}} --> {{{x=-5 +- 4}}} --> {{{highlight(x=-9)}}} or {{{highlight(x=-1)}}}
So the vertices are at (-9,4) and (-1,4).
The foci are on the line conecting the vertices (the transverse axis), at points 
at a distance c from the center, with c calculated as
{{{c=sqrt(a^2+b^2)}}}, so {{{c=sqrt(4^2+2^2)=sqrt(16+4)=sqrt(20)=2sqrt(5)}}}
So the foci are at ({{{-5-2sqrt(5)}}},4) and ({{{-5+2sqrt(5)}}},4).