Eliminate the D and E terms of the general equation of a Conic section?

Question

We have the general equation of a conic section:

$$Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0$$

Where, the $D.x$ and $E.y$ terms determine the translation/offset of the conic section from the origin.

I have an equation with all these terms and am trying to eliminate these terms in order to translate my conic section equation to the origin:

$$Ax^{2} + Bxy + Cy^{2} + F = 0$$

Where $D=E=0$.

I tried understanding how the equation changes with such a translation by manually moving a conic to the origin on Geogebra:

Unfortunately, it seems the D and E terms aren't eliminated (at least on Geogebra) and instead, get offsetted by some amount and the F value also seems to change drastically.

How can I achieve the same result? That is, generate the equation for the conic being centred at the origin. With or without the D and E terms (the latter preferred). Thanks!

Answer 1

$2Ax+By+D=0,By+2Cy+E=0$ has solution $(x,y)=\left(\frac{2CD-BE}{B^2-4AC},-\frac{BD-2AE}{B^2-4AC}\right)$,

provided $B^2-4AC\not=0$, thus the conic should not be a parabola. Geometrically, the formula breaks down for a parabola because it intends to identify the center of the conic but the center of a parabola does not exist.

Then substituting $(x,y)\to \left(x+\frac{2CD-BE}{B^2-4AC},y-\frac{BD-2AE}{B^2-4AC}\right),$ we get $$A\left(x+\frac{2CD-BE}{B^2-4AC}\right)^2+B\left(x+\frac{2CD-BE}{B^2-4AC}\right)\left(y-\frac{BD-2AE}{B^2-4AC}\right)+C\left(y-\frac{BD-2AE}{B^2-4AC}\right)^2+D\left(x+\frac{2CD-BE}{B^2-4AC}\right)+E\left(y-\frac{BD-2AE}{B^2-4AC}\right)+F=0$$ which expands to $$Ax^2+Bxy+Cy^2+F+\frac{-AE^2+BDE-CD^2}{4AC-B^2}=0$$

Answer 2

By polynomials comparison

$$ Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = c_1(x-x_0)^2+c_2(x-x_0)(y-y_0)+c_3(y-y_0)^2+c_4 $$

we have the relationships

$$ \left\{ \begin{array}{rcl} A-c_1 & = & 0\\ B-c_2 & = & 0 \\ D+2 c_1 x_0+c_2 y_0 & = & 0\\ C-c_3 & = & 0 \\ E+c_2 x_0+2 c_3 y_0 & = & 0 \\ F-c_1 x_0^2-c_2 y_0 x_0-c_3 y_0^2-c_4 & = & 0 \\ \end{array} \right. $$

and after solving

$$ \left\{ \begin{array}{rcl} c_1&=&A \\ c_2&=&B \\ c_3&=&C \\ c_4&=&\frac{C D^2-B E D+A E^2}{B^2-4 A C}+F \\ x_0&=&\frac{2 C D-B E}{B^2-4 A C} \\ y_0&=&\frac{2 A E-B D}{B^2-4 A C} \\ \end{array} \right. $$

so this translation is possible as far as $B^2-4AC \ne 0$ which means that the conic cannot be a parabola.

Answer 3

Displacements $(h,k)$ are determined to find Central Conicoids (without $x,y$ terms) using Mathematica:

We expand and solve for (h,k) which are needed to rid of (x,y) terms:

Expand[aa (x+h)^2+bb(x+h)(y+k)+cc (y+k)^2+dd(x+h)+ee(y+k)+ff==0]
Solve[{dd+2 aa h+bb k==0,ee+bb h+2 cc k==0},{h,k}]
(* Plug in above solution*)
{aa (x+h)^2+bb(x+h)(y+k)+cc (y+k)^2+dd(x+h)+ee(y+k)+ff}/.{h->-((-2 cc dd+bb ee)/(bb^2-4 aa cc)),k->-((bb dd-2 aa ee)/(bb^2-4 aa cc))}
(* and simplify *)
FullSimplify[{ff+dd (-((-2 cc dd+bb ee)/(bb^2-4 aa cc))+x)+aa (-((-2 cc dd+bb ee)/(bb^2-4 aa cc))+x)^2+ee (-((bb dd-2 aa ee)/(bb^2-4 aa cc))+y)+bb (-((-2 cc dd+bb ee)/(bb^2-4 aa cc))+x) (-((bb dd-2 aa ee)/(bb^2-4 aa cc))+y)+cc (-((bb dd-2 aa ee)/(bb^2-4 aa cc))+y)^2}]

Next for the parabola:

Set the discriminant of a general/classical conicoid to zero. Instead of 5 constants you have 4 constants. If solved for one of the independent variables equation of a tilted parabola (axis of symmetry not parallel to either of the coordinate axes) can be expressed as:

$$ y= \sqrt{ax+b}\pm\sqrt{cx+d}\,$$

In polar coordinates only an implicit form of a parabola

$$ r \sin \theta = \sqrt{a \,r \cos \theta+b}\pm\sqrt{c\,r \cos \theta+d}\,$$

that is non-focal, is presently known to me.