The most general conic section has an equation of the form
$$Ax^2 + By^2 + Cxy + Dx + E y + F = 0.$$
So far we have taken $C=D=E=0$, except
for parabolas. Here we'll see how to adjust for nonzero values of $D$
and $E$. We'll still assume that $C=0$, which means that our curves
will point in the direction of the coordinate axes. $C \ne 0$
describes rotated conic sections.
If $A \ne 0$, then we can always absorb the $Dx$ term by completing
the square:
\begin{eqnarray*} Ax^2 + Dx & = & A \left (x^2 + \frac{D}{A}x \right)
\cr \cr & = & A\left (x + \frac{D}{2A}\right )^2 
\frac{D^2}{4A}\end{eqnarray*}
Likewise, if $B \ne 0$ then we can absorb
the $Ey$ term:
$$\displaystyle{By^2 + Ey = B\left (y+\frac{E}{2B}\right )^2\frac{E^2}{4B}}.$$
If $B=0$ then we cannot absorb the $Ey$ term, and the equation of a
parabola winds up looking like $\displaystyle{yk = \frac{A}{E} (xh)^2}$.
Example : Identify the center, foci and
eccentricity of the ellipse $$9x^2 + 25y^2 18 x + 50y  191 = 0.$$
Solution: Rewrite $9x^2 18x$ as $9(x1)^2 9$, and rewrite $25 y^2 +
50 y$ as $25 (y+1)^2 25$. This makes the equation
$$9(x1)^2 + 25(y+1)^2 225=0,$$

or equivalently
$$\frac{(x1)^2}{25}+\frac{(y+1)^2}{9}=1.$$
This is an ellipse centered
at $(1,1)$ with $a=5$ and $b=3$, hence $c=\sqrt{5^23^2}=4$. The foci
are 4 units the the right and left of the center, at $(5,1)$ and
$(3,1)$. This ellipse has eccentricity $c/a=4/5=0.8$.

