One
general format of an ellipse is ax^{2}
+ by^{2} + cx + dy + e = 0.
But the more useful form looks quite different:

...where
the point (h,
k) is the center
of the ellipse, and the focal points and the axis lengths of the ellipse
can be found from the values of a
and b.
But how do you convert
from the general form to the useful form? By completing the square.

Find
the focus equation of the ellipse given by 4x^{2}
+ 9y^{2} – 48x + 72y + 144 = 0.

Move
the loose number over to the other side, and group the x-stuff
and y-stuff
together.

Factor
out whatever is on the squared terms.

Divide
through by whatever you factored out of the x-stuff.

Divide
through by whatever you factored out of the y-stuff.

Leave
space in the x-stuff,
the y-stuff,
and on the right side. Whatever numbers are multiplied on the
x-stuff
and the y-stuff,
put them on the spaces on the right.

Take half of the
coefficients of the first-degree terms (and don't forget their
signs!), square them, and add the squares into the appropriate
spaces on both sides of the equation.

Simplify
on the right, and convert to squared form on the left. (This is
where you use those signs you kept track of earlier.)

Whatever
value you get on the right side, divide through by it, so the
right side becomes "=
1".

The
process for hyperbolas is the same, except that the signs on the x-squared
and y-squared
terms will be opposite; that is, while both the x-squared
and y-squared
terms are added in the case of ellipses (and circles), one or the other
will be subtracted in the case of hyperbolas.