SOLUTION: How do you find the equation, center, vertices, and foci of the following problem: 4x^2-16y^2+40x+128y-220=0

Question 615643: How do you find the equation, center, vertices, and foci of the following problem:
4x^2-16y^2+40x+128y-220=0
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
My way:
--> -->
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.
--> --> -->
Next, I write the brackets as squares:
-->
Finally, I divide both sides by the value on right side of the equal sign to get the equation in the form I like:
--> -->
Or even better, I can write those denominators as squares, as in:

That matches the general equation for a hyperbola with a horizontal transverse axis, centered at (h,k):

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, --> and that is the smallest possible value for . For all other values of y,
and
So
That is as close to the center as the hyperbola can get, and it gives you the coordinates of the vertices:
--> or --> --> or
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
, so
So the foci are at (,4) and (,4).
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