how to obtain parabola general expression or Parametric equation from quadratic matrix made up by five points on a parabola curve?

Question

first please be noted that it is a general expression ,because it may be translated and rotated. I have read Matrix representation of conic sections

from Wiki and I make it plot the hyperbola or ellipse depending on five points input by ginput() using matlab. I think it is wonderful ,because it utilize the matrix properties such as det, eigenvalue ,and eigenvector to extract the curve feature, which can be directly used to do construct the curve.

But I don't know how to construct the parabola ,as you know there are some exception, the det of matrix $A33$ is zero, and it only has one eigenvalue. one of eigenvector is the direction that the parabola opens at.

I suppose the only eigenvalue is the $1/4a^2$ value in the expression $y=x^2/(4a^2)$. after a simple test , this is not always right. after only rotate the curve,yes, the only eigenvalue is the $4a^2$ value. but after only Translation, it is not.

is there anyone can tell me why and how to construct the parabola ,I would appreciate

Answer 1

The equation of the parabola having line $ax+by+c=0$ as directrix and point $F=(f_1,f_2)$ as focus can be written as: $$ (a^2 + b^2)\big((x - f_1)^2 + (y - f_2)^2\big) - (a x + b y + c)^2=0. $$ Comparing that with the generic equation of a conic: $$ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0, $$ one can find $a$, $b$, $c$, $f_1$, $f_2$ in terms of $A$, $C$, $D$, $E$, $F$ and the sign of $B$, taking into account that $B^2=4AC$.

Notice also that the generic equation can be rewritten as: $$ (\sqrt{A}x \pm \sqrt{C}y)^2 + Dx + Ey + F = 0, $$ and you may then consider one of the alternate methods given here to find axis and vertex of the parabola.