How do I find details of a parabola from its general two-degree equation? This general equation in two-degree represents a parabola: $$(ax + by)^2 + 2gx + 2fy + c = 0$$
How do I find the following from this equation:


*

*Vertex

*Focus

*Axis

*Length of Latus Rectum

*Co-ordinates of end points of Latus Rectum

*Equation of Directrix


I know how to find these if a parabola has its axis parallel to either of the co-ordinate axes. But I don't know how to derive these from the general equation.
N.B.: w.r.t. a comment by @Vasya, I produce this picture from my book, where it says that the general form of the equation of a parabola is the above equation. The book gives the equation, but doesn't derive any of the above from the equation:

Hence the general form of the equation of a parabola is
   $$ (ax + by)^2 + 2gx + 2fy + c = 0. $$



$y'=bx-ay$ is parallel to the directrix and $x'=ax+by$ parallel to the axis. This we can see from comparing to the form I mentioned in the comments. So let these be the new coordinates. The inverse is $x=\frac{ax'+by'}{a^2+b^2}$, $\frac{bx'-ay'}{a^2+b^2}$, transforming your equation into $$x'^2+2g\frac{ax'+by'}{a^2+b^2}+2f\frac{bx'-ay'}{a^2+b^2}+c=0,$$ solving for $y'$ we get $$y'=\frac{a^2+b^2}{2(fa-gb)}x'^2+\frac{ga+fb}{fa-gb}x'+c\frac{a^2+b^2}{2(fa-gb)}=Ax'^2+Bx'+C,$$ making (from the formulas given here) $x'=-\frac{B}{2A}$ or $ax+by=-\frac{ga+fb}{a^2+b^2}$ the axis and $y'=\frac{4AC-B^2-1}{4A}$ or $bx-ay=\frac{c(a^2+b^2)-g^2-f^2}{2(fa-gb)}$ the directrix. The semi latus rectum is $p=\frac1{2A}=\frac{fa-gb}{a^2+b^2}$. For the vertex and focus to undo the transformation we transform the point back. The vertex in the rotated coordinates is $(-\frac{B}{2A},\frac{4AC-B^2}{4A})$ making the vertex in the first coordinates $$(\frac12 \frac{-2 g a^3 f+g^2 a^2 b-2 f^2 b a^2+b c a^4+2 c a^2 b^3+c b^5-f^2 b^3}{(a^2+b^2)^2 (f a-g b)},-\frac12 \frac{-2 g^2 a b^2+f^2 b^2 a-2 f b^3 g+c a^5+2 c a^3 b^2+a c b^4-g^2 a^3}{(a^2+b^2)^2 (f a-g b)})$$ and the focus is $(-\frac{B}{2A},\frac{4AC-B^2+1}{4A})$, using the inverse transformation the focus is  $$(\frac12 \frac{b c a^2-2 f a g+g^2 b+c b^3-f^2 b}{(a^2+b^2) (f a-g b)},-\frac12 \frac{c a^3+a f^2+a c b^2-g^2 a-2 f g b}{(a^2+b^2) (f a-g b)}).$$
