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 (i.e. $a^2+b^2=1$), $P=(p,q)$ 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:
$\sqrt{(x-f)^2 + (y-g)^2} = |a(x-p) + b(y-q)| \implies$ $\left(x-f\right)^2 + \left(y-g\right)^2 = \left( a \left( x-p \right) + b \left( y-q \right) \right)^2 \implies$ $\tag{2} k_{x^2}x^2 + k_{y^2}y^2 + k_{xy}xy + k_{x}x + k_{y}y + k = 0$
Where $$ k_{x^2} = a^2-1 ;\\ k_{y^2} = b^2-1 ;\\ k_{xy} = 2ab ;\\ k_{x} = 2(f - a^2p - abq) ;\\ k_{y} = 2(g - abp - b^2q) ;\\ k = a^2p^2+2abpq+b^2q^2-f^2+g^2 $$
How can I derive necessary conditions for the constant coefficients, in order for the equation to be for a parabola?
I suppose we need to prove that not at least one of $k_x, k_y$ should be $\ne 0$, but I don't know how to prove that (by using the parabola definition in equation $1$ above).
We know that $N \ne (0, 0)$, so not both $a, b$ can be $0$. However, even proving at least one of $P, F$ has to be non-0 also is out of my reach for now. When $P=F=0$, I get $(a^2-1)x^2 - y^2 + 2a\sqrt {1-a^2} xy = 0$, and I'm stuck in proving that's not a parabola.
Thanks!
Context
Starting with a wrong premise that the sufficient condition for a parabola is $ab=c^2$, I started the question for the first part of the generic parabolic equation here.
Now, thanks to getting an answer to that question, I understood that that's neither sufficient nor even correct condition, and since the recommendation is to start a new question, that's what I'm doing here, with the wish to get to the generic parabolic equation, together with the sufficient conditions.
Given the parabola directrix
$$ p = p_0 + \lambda \vec v,\ \ \ \|\vec v\|=1 $$
and the focus $f$, with $q = (x,y)'$, the parabola has the property:
$$ \|f-q\|^2=\|q-p_0-((q-p_0)\cdot\vec v)\vec v\|^2 $$
so developing and grouping terms we arrive at
$$ q'\cdot(\vec v\cdot\vec v')\cdot q-2(f+(\vec v\cdot\vec v'-I_2)\cdot p_0)\cdot q+\|f\|^2-p_0'\cdot (\vec v\cdot\vec v'-I_2)\cdot p_0=0 $$
here
$$ \vec v\cdot\vec v'=\left(\matrix{v_x^2&v_xv_y\\ v_xv_y & v_y^2}\right)\\ I_2=\left(\matrix{1&0\\ 0 & 1}\right) $$
Now given a quadratic $q'\cdot M\cdot q+B\cdot q + C=0$ it is a parabola if $M$ can be written in the form
$$ M = \left(\matrix{a^2& a b\\ a b& b^2}\right),\ \ \ a^2+b^2\ne 0 $$
and also
$$ (f-p_0)\times\vec v\ne \vec 0 $$
NOTE
The involved parameters are
$$ \cases{ p_0\ \ \, (2)\\ f\ \ \ \ (2)\\ \vec v\ \ \ \ (2) } $$
and the conditions are
$$ \cases{ \|\vec v\| \ne 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\\ (f-p_0)\times\vec v \ne 0 \ \ (1) } $$
so we have $(6)-(2)=(4)$ characteristic parameters. Now regarding the form
$$ a x^2+b x y + c y^2+d x+ e y + g=0 $$
we have $(6)$ free parameters and $(2)$ conditions to establish a parabola, remaining $(4)$ characteristic parameters.