TITLE: How to convert the general form of ellipse equation to the standard form?
QUESTION [5 upvotes]: How to convert the general form of ellipse equation to the standard form?
$$-x+2y+x^2+xy+y^2=0$$

REPLY [4 votes]: $% Predefined Typography
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$
If you have a conic section characterized by : 
$$f(x, y) = A_0x^2 + 2B_0xy + C_0y^2 + 2D_0x + 2E_0y + F_0 = 0$$
Then the center will occur where $\indp{f}{x} = 0$ and $\indp{f}{y} = 0$, and solving the linear equations (2x2 matrix) you get the center $c$ at:
$$c_x = \frac{C_0D_0 - B_0E_0}{B_0^2 - A_0C_0}$$
$$c_y = \frac{A_0E_0 - B_0D_0}{B_0^2 - A_0C_0}$$
So the conic section shifted to be at the center will occur at 
$$g(x, y) = f(x + x_c, y + y_c) = Ax^2 + 2Bxy + Cy^2 + F = 0$$
for some new coefficients $A,~ B,~ C,~ \text{and } F$:
$$\begin{align}
A &= A_0(A_0C_0 - B_0^2) \\
B &= B_0(A_0C_0 - B_0^2) \\
C &= C_0(A_0C_0 - B_0^2) \\
F &= F_0(A_0C_0-B_0^2) -A_0E_0^2 - C_0D_0^2 +2B_0D_0E_0 \\
\end{align}$$
Now you must rotate the values of $x, y$ to align the axises, so you need to introduce two new variables $u, v$ under the constraint:
$$u = cx - sy$$
$$v = sy + cx$$
$$s^2 + c^2 = 1$$
Where $s$ and $c$ represent the sine and cosine of the rotation (although we will avoid trig functions like the plague).
Solving the transform for $x$ and $y$ and plugging into the definition for $g$ we get
$$h(u, v) = (Ac^2 -2Bcs+ Cs^2)u^2
          + 2(Acs + Bc^2 -Bs^2 -Ccs )uv
          + (As^2 -2Bcs+ Cc^2)v^2
          + F = 0$$
To obtain axial alignment, we need to choose $c$ and $s$ in a way that the coefficient of $uv$ is $0$ while maintaining $s^2 + c^2 = 1$.
$$Acs + Bc^2 -Bs^2 -Ccs = 0$$
$$c^2 + s^2 = 1$$
which solves to 
$$c^2 = \frac{\pm(C - A)\sqrt{M} + M}{2M}$$
$$s^2 = \frac{\mp(C - A)\sqrt{M} + M}{2M}$$
$$M = A^2 + C^2 + 4B^2 - 2AC$$
which is enough information to back substitute and get either an analytical expression or computational computation for a centered axial aligned conic section.

For your example:
$$f(x, y) = -x+2y+x^2+xy+y^2=0$$
So $A_0 = 1$, $B_0 = 1/2$, $C _0= 1$, $D_0 = -1/2$, $E_0 = 1$, $F_0 = 0$.
So the center occurs at 
$$c = \begin{bmatrix} \frac 43 \\ -\frac 53 \end{bmatrix}$$
and the origin centered shifted conic section is:
$$g(x, y) = \frac 34 x^2 + \frac 34 y^2 + \frac 34 xy - \frac 74 = 0$$
with the new $A = \frac 34$, $B = \frac 38$, $C = \frac 34$, $F = -\frac 74$.
So with yours, $M = \frac {9}{16}$, and $c = s = \pm \frac{1}{\sqrt{2}}$ (your conic section was rotated by $\frac 18$ of a revolution).
So the centered axially aligned transform of your example is:
$$9v^2 + 3u^2 - 14=0$$