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:

  1. Vertex
  2. Focus
  3. Axis
  4. Length of Latus Rectum
  5. Co-ordinates of end points of Latus Rectum
  6. 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:

enter image description here

  • $\begingroup$ this is not a parabola as you have $x$ and $y$ squared $\endgroup$ – Vasya Oct 4 '17 at 17:09
  • $\begingroup$ This question gives the equation of a parabola with directrix $lx+my+n=0$ and focus $P(x_P,y_P)$ as $\frac{(lx+my+n)^2}{{l}^2+{m}^2}=(x-x_P)^2+(y-y_P)^2$. $\endgroup$ – Jan-Magnus Økland Oct 4 '17 at 17:09
  • $\begingroup$ @Vasya please see the image. It might take a bit of time to load. $\endgroup$ – Wrichik Basu Oct 4 '17 at 18:03
  • $\begingroup$ @Jan-MagnusØkland that shows that the equation of a parabola can have both $x^2$ and $y^2$ terms. $\endgroup$ – Wrichik Basu Oct 4 '17 at 18:04
  • 1
    $\begingroup$ Since you know how to find these values for an axis-aligned parabola, try finding a coordinate transformation that aligns the parabola. Hint: such a transformation will eliminate the cross-term $2axy$. You might also find some clues to finding these properties in the derivation of this general equation. $\endgroup$ – amd Oct 4 '17 at 19:48

$y'=bx-ay$ 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 $$(\frac12 \frac{b c a^2-2 f a g+g^2 b+c b^3-f^2 b}{(a^2+b^2) (f a-g b)},-\frac12 \frac{c a^3+a f^2+a c b^2-g^2 a-2 f g b}{(a^2+b^2) (f a-g b)}).$$


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