Mister Exam

# Canonical form of a degenerate ellipse

For example, you have entered (calculator here):
$$5 x^{2} + 4 x y - 6 x + y^{2} - 2 y + 2 = 0$$

### Detail solution

Given line equation of 2-order:
$$5 x^{2} + 4 x y - 6 x + y^{2} - 2 y + 2 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 5$$
$$a_{12} = 2$$
$$a_{13} = -3$$
$$a_{22} = 1$$
$$a_{23} = -1$$
$$a_{33} = 2$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}5 & 2\\2 & 1\end{matrix}\right|$$
$$\Delta = 1$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} x_{0} + a_{12} y_{0} + a_{13} = 0$$
$$a_{12} x_{0} + a_{22} y_{0} + a_{23} = 0$$
substitute coefficients
$$5 x_{0} + 2 y_{0} - 3 = 0$$
$$2 x_{0} + y_{0} - 1 = 0$$
then
$$x_{0} = 1$$
$$y_{0} = -1$$
Thus, we have the equation in the coordinate system O'x'y'
$$a'_{33} + a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = - 3 x_{0} - y_{0} + 2$$
$$a'_{33} = 0$$
then equation turns into
$$5 x'^{2} + 4 x' y' + y'^{2} = 0$$
Rotate the resulting coordinate system by an angle φ
$$x' = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y' = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
φ - determined from the formula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
substitute coefficients
$$\cot{\left(2 \phi \right)} = 1$$
then
$$\phi = \frac{\pi}{8}$$
$$\sin{\left(2 \phi \right)} = \frac{\sqrt{2}}{2}$$
$$\cos{\left(2 \phi \right)} = \frac{\sqrt{2}}{2}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}$$
$$y' = \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} + \tilde y \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}$$
then the equation turns from
$$5 x'^{2} + 4 x' y' + y'^{2} = 0$$
to
$$\left(\tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} + \tilde y \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}\right)^{2} + 4 \left(\tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} + \tilde y \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}\right) + 5 \left(\tilde x \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}\right)^{2} = 0$$
simplify
$$\sqrt{2} \tilde x^{2} + 4 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} + 3 \tilde x^{2} - 8 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} + 2 \sqrt{2} \tilde x \tilde y - \sqrt{2} \tilde y^{2} - 4 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} + 3 \tilde y^{2} = 0$$
$$2 \sqrt{2} \tilde x^{2} + 3 \tilde x^{2} - 2 \sqrt{2} \tilde y^{2} + 3 \tilde y^{2} = 0$$
Given equation is degenerate ellipse
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{2 \sqrt{2} + 3}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{3 - 2 \sqrt{2}}}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
(1, -1)

Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}, \ \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}\right)$$
$$\vec e_2 = \left( - \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}, \ \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}\right)$$

### Invariants method

Given line equation of 2-order:
$$5 x^{2} + 4 x y - 6 x + y^{2} - 2 y + 2 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 5$$
$$a_{12} = 2$$
$$a_{13} = -3$$
$$a_{22} = 1$$
$$a_{23} = -1$$
$$a_{33} = 2$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22}$$
     |a11  a12|
I2 = |        |
|a12  a22|

$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12}\\a_{12} & a_{22} - \lambda\end{matrix}\right|$$
     |a11  a13|   |a22  a23|
K2 = |        | + |        |
|a13  a33|   |a23  a33|

substitute coefficients
$$I_{1} = 6$$
     |5  2|
I2 = |    |
|2  1|

$$I_{3} = \left|\begin{matrix}5 & 2 & -3\\2 & 1 & -1\\-3 & -1 & 2\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}5 - \lambda & 2\\2 & 1 - \lambda\end{matrix}\right|$$
     |5   -3|   |1   -1|
K2 = |      | + |      |
|-3  2 |   |-1  2 |

$$I_{1} = 6$$
$$I_{2} = 1$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 6 \lambda + 1$$
$$K_{2} = 2$$
Because
$$I_{3} = 0 \wedge I_{2} > 0$$
then by line type:
this equation is of type : degenerate ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 6 \lambda + 1 = 0$$
$$\lambda_{1} = 3 - 2 \sqrt{2}$$
$$\lambda_{2} = 2 \sqrt{2} + 3$$
then the canonical form of the equation will be
$$\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2} + \frac{I_{3}}{I_{2}} = 0$$
or
$$\tilde x^{2} \left(3 - 2 \sqrt{2}\right) + \tilde y^{2} \left(2 \sqrt{2} + 3\right) = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{3 - 2 \sqrt{2}}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{2 \sqrt{2} + 3}}\right)^{2}} = 0$$
- reduced to canonical form