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?

  • 1
    $\begingroup$ That equation is for any conic. First thing I would do is to reduce this general equation into one which only applies to parabolas. $\endgroup$
    – Peter Webb
    Commented Feb 2, 2014 at 11:40
  • $\begingroup$ Finding the vertex of a parabola. $\endgroup$ Commented Feb 2, 2014 at 12:01
  • $\begingroup$ @AnonSubmitter85, that video only handles the case $y=ax^2+bx+c$. It doesn't apply with xis non-vertical. $\endgroup$ Commented Feb 2, 2014 at 12:34
  • $\begingroup$ @GerryMyerson You are absolutely right. I added an answer that takes rotation into account. $\endgroup$ Commented Feb 2, 2014 at 14:33

3 Answers 3


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


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 $x,y$

  • $\begingroup$ Nice method. Very clean. Very easy. $\endgroup$ Commented Feb 2, 2014 at 14:47
  • 1
    $\begingroup$ @AnonSubmitter85, nice to hear that. $\endgroup$ Commented Feb 2, 2014 at 14:48
  • $\begingroup$ I think your solution is not correct, for example, with parabola $$x^2 + 2xy + y^2 + 2x + y + 1 = 0 $$ . Using your method, we get vertex at (-1,1). But Wolfram calculate vertex at (-13/16, 1/16). Here's the link: wolframalpha.com/input/?i=x^2%2B2xy%2By^2%2B2x%2By%2B1%3D0+parabola $\endgroup$
    – it4rb
    Commented Feb 3, 2014 at 11:14
  • $\begingroup$ @it4rb, could you pinpoint the mistake and in the meantime I shall try to look into this $\endgroup$ Commented Feb 3, 2014 at 11:42
  • $\begingroup$ I think the problem is that X and Y are not independent variables, but its just my guess. $\endgroup$
    – it4rb
    Commented Feb 4, 2014 at 10:03


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.


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 $(x,y)$.

  • $\begingroup$ I tried your method, but it didn't work. Consider parabola $x^2 + 2xy + y^2 + 2x + y + 1 = 0$, so we have A=1, B=2, C=1, D=2, E=1, F=1. Use your equation, we get a=A+C=2. Then because $c_\theta s_\theta = B/2a = 1/2$, so $c_\theta$ and $s_\theta$ must have same sign. Assume $c_\theta = \sqrt{\frac Aa} = \frac{\sqrt2}2$ and $s_\theta = \frac{\sqrt2}2$, then 2 equations $D = b c_\theta + s_\theta$ and $E=b s_\theta - c_\theta$ can not be satisfied. The same if we choose $c_\theta = -\frac{\sqrt2}2$ and $s_\theta = -\frac{\sqrt2}2$. $\endgroup$
    – it4rb
    Commented Feb 3, 2014 at 14:06
  • $\begingroup$ @it4rb Yeah, I realized my mistake write after I posted that comment. I'm double checking things. The method should work. $\endgroup$ Commented Feb 3, 2014 at 14:22
  • $\begingroup$ @it4rb I apologize but I cannot find the mistake in the above though I agree that something is wrong. I am still of the mind that the methodology should work, but cannot explain why it leads to an unsolvable system. When I plot the example you gave, I get a parabola whose axis is rotated by $\pi/4$, so the values of $c_\theta$ and $s_\theta$ are correct. Other than that, the above appears to be useless. $\endgroup$ Commented Feb 3, 2014 at 17:04
  • $\begingroup$ I found 1 article on Wiki mention about axes of conic, include parabola link. We can solve equation to get 2 eigenvalues, 0 and A+C. Therefore, the axis of parabola has form $y=\frac{A+C}{2}x + p$. From this we can get $cos(\theta)$ and $sin(\theta)$. I'm trying on this with your solution about rotation. $\endgroup$
    – it4rb
    Commented Feb 4, 2014 at 10:09
  • $\begingroup$ Oh I'm sorry, it should be $y=\frac{2C}{B}x+p$ $\endgroup$
    – it4rb
    Commented Feb 4, 2014 at 10:58

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 $m_1 m_2 = -1$


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