I fail to derive the parabola equation from here, which states that the general equation of any conic section is
$\tag{0} ax^2+by^2+cxy+dx+ey+f=0$
where for a parabola we have $ab-c^2=0$ [2].
I operate with the parabola definition $\tag{1} \|X-F\| = distance(X, L)$, where $F=(f, g)$ is the focus point, $X=(x, y)$ is a point on the parabola, $L = \{Y | (Y-P)N=0\}$ is the directrix line, $N = (a, b)$ is unit-vector normal to the directrix, $P=(p,q)$ is a point on the directrix.
$distance(X,L) = |(X-P)N| = |(x-p, y-q)(a,b)| = |a(x-p) + b(y-q)|$
If I plug that back into $1$, I get:
$\sqrt{(x-f)^2 + (y-g)^2} = |a(x-p) + b(y-q)| \implies$
$(x-f)^2 + (y-g)^2 = (a(x-p) + b(y-q))^2 \implies$
$\tag{2} x^2 (a^2-1) + y^2(b^2-1) + 2abxy + k_1x + k_2y + k_3 = 0$
Where $k_1, k_2, k_3$ are some constants, which are not important for this discussion.
If we take an example parabola: $x^{2}-4xy+4y^{2}+40x+20y=100$, it would not fit into the equation $2$, but it also does not satisfy the initial condition of $1\cdot4 – 4^2=0$
What I'm doing wrong here?
[2]: The correct condition is $4ab-c^2=0$, but I left the original text of the question for posterity, even though after reading the answers here I realized that the referenced answer was wrong.
Best Answer
As I wrote in the comments, $x^2-4xy+4y^2+40x+20y=100$ is a parabola and there is no problem here, since the correct condition is $\color{red}4ab=c^2$.
On the other hand, note that\begin{multline}x^2-4xy+4y^2+40x+20y-100=\\=-5\left(\left(\left(\frac2{\sqrt5}\right)^2-1\right)x^2+2\frac2{\sqrt5}\frac1{\sqrt5}xy+\left(\left(\frac1{\sqrt5}\right)^2-1\right)y^2-8x-4y+20\right).\end{multline}Since $\left(\frac2{\sqrt5}\right)^2+\left(\frac1{\sqrt5}\right)^2=1$, this is indeed a parabola. You did not take into consideration the possibility of multiplying the expression of a parabola by a constant.