How to find vertex of a parabola from its second degree equation Given a parabola with second degree equation as
$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 $$ 
assume that this isn't degenerate case, and $B^2-4AC=0$
How can I find its vertex position?
 A: If $\displaystyle  A$ or $C=0, B=0,$  the derivation is trivial  
Else $\displaystyle C=\frac{B^2}{4A}$
Replacing this value of $C$ in the given equation 
$$(2Ax+By)^2=-4A(Dx+Ey+F)$$
Now comparing with one of standard forms $\displaystyle Y^2=-4AX$ whose vertex is $(0,0)$ in $X-Y$ coordinate 
we have $ \displaystyle Y=2Ax+By\  \ \ \ (1), X=Dx+Ey+F\  \ \ \ (2)$
Set $X=Y=0$ and solve for $x,y$
A: EDIT:
Apparently this method has been tried and found not to work. I'm leaving the answer because the idea seems solid enough. There exists a rotated version of the $(x,y)$ coordinate system, call it $(u,v)$, such that the axis of symmetry of the parabola is parallel to the $v$-axis. In the $(u,v)$ system, the parabola will have a standard form. Since we can freely switch coordinate systems, there should exist a way to write $Ax + Bxy + \cdots$ in terms of $(u,v)$ such that the vertex can be found by one of the standard methods. However, my attempt at this has been found lacking.
ORIGINAL:
I'm going to assume that if we can get the parabola back to the standard form of $v = a u^2 + b u + c$, where $u$ and $v$ are rotated versions of $x$ and $y$, respectively, that you can then easily find the vertex.
The $(u,v)$ coordinate system is a rotated version of the $(x,y)$ system, so
$$
\left[
\begin{array}{cc}
\cos(\theta) & \sin(\theta) \\
-\sin(\theta) & \cos(\theta)
\end{array}
\right]
\left[
\begin{array}{c}
x \\ y
\end{array}
\right]
=
\left[
\begin{array}{cc}
c_\theta & s_\theta \\
-s_\theta & c_\theta
\end{array}
\right]
\left[
\begin{array}{c}
x \\ y
\end{array}
\right]
=
\left[
\begin{array}{c}
u \\ v
\end{array}
\right],
$$
which means that $u = c_\theta x + s_\theta y$ and $v = -s_\theta x + c_\theta y$. When we substitute these expressions into $v = a u^2 + b u + c$ and group them according to $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$, we get that
$$
\begin{array}{ccl}
A & = & a c_\theta^2 \\
B & = & 2 a c_\theta s_\theta \\
C & = & a s_\theta^2 \\
D & = & b c_\theta + s_\theta \\
E & = & b s_\theta - c_\theta \\
F & = & c
\end{array}
$$
Looking at the above we note that $A + C = a c_\theta^2 + a s_\theta^2 = a$. Knowing $a$, you can then find $c_\theta$ and $s_\theta$, which can then be used to find $b$. A tricky part might (or might not) be choosing the correct signs of $c_\theta$ and $s_\theta$, however, you should be able to figure it out from the equation for $B$ and noting that $A+B+C=a(c_\theta + s_\theta)^2$. Also, using $D$ and $E$ might remove the ambiguity. I am not sure as I have not investigated it enough, but I'm sure it's possible.
Once you have $a$, $b$, and $c$, you can easily find the vertex in $(u,v)$ coordinates and use $c_\theta$ and $s_\theta$ to rotate the answer back to $(x,y)$.
A: Rearrange to get in form $(ax+by)^2=-c-2gx-2fy$
If you got back to the eccentricity definition and how we got this standard second degree curve for parabola, We simplified,
$ \frac{PFocus}{PDirectrix}=1 $
$(x − h)^2 + (y − k)^2 = (lx + my + n)^2 / (l^2 + m^2)$
$(mx − ly)^2 + 2gx + 2fy + d = 0$
$ax+by+k$ with arbitrary k is parallel to the axis and Normal to Parabola at Vertex, Now use this to find Vertex. At one point inthe parabola, the above line will be perpendicular to tangent at that point (differentiate and equate $m_1 m_2 = -1$
