The general equation of a parabola is $$\left(\frac{ax+by+c}{\sqrt{a^2+b^2}} \right) ^2=(Latus Rectum).\frac{bx-ay+c'}{\sqrt{a^2+b^2}}$$ To get to this from $y^2=4ax$, and similarly for other conics, it seems sufficient to prove that the equations of conics are coordinate invariant under rotations and shifting. If I can assume so, I can work out the coefficients for the axis and tangent-at-latus-rectum equations because they're always going to be perpendicular. Thus if I take $a, b,$ and $c$ for one, the other could have coefficients $\pm b, \mp a,$ and $c'$,as determined by the sign of the Latus Rectum term.
How can I prove the coordinate invariance?
And if this method to arrive at the equations is wrong (ignore this if it isn't), how do I work out an intuition for the same?
I would say that $$L_1^2=4A L_2~~~~(1)$$ represents a parabola if two lines $L_1=0$ and $L_2=0$ are non parallel. The condition of $H^2=AB$ for a parabola turns the well known quadratic to the form (1) as $$(\sqrt{a} x+\sqrt{b} y)^2=-2gx-2fy-c~~~~(2).$$
But if $L_1$ and $L_2$ are perpendicular and normalized, then $4A$ gives the latus rectum, the axis of the paralola: $L_1=\frac{ax+by+c}{\sqrt{a^2+b^2}}=0$, the tangent at vertex as $L_2=\frac{bx-ay+c'}{\sqrt{a^2+b^2}}=0.$ The focus given by the point of intersection of $L_1=0, L_1=A$, the equation of the directrix is $L_2=-A$.