How to find vertex of a parabola from its second degree equation

Question

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?

Answer 1

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 $x,y$

Answer 2

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 $(x,y)$.