Mister Exam

# Canonical form of a parabola

For example, you have entered:
$$x^{2} - 2 x y - 10 x + y^{2} - 6 y + 25 = 0$$

### Detail solution

Given line equation of 2-order:
$$x^{2} - 2 x y - 10 x + y^{2} - 6 y + 25 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = -1$$
$$a_{13} = -5$$
$$a_{22} = 1$$
$$a_{23} = -3$$
$$a_{33} = 25$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & -1\\-1 & 1\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
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
$$\cot{\left(2 \phi \right)} = 0$$
then
$$\phi = \frac{\pi}{4}$$
$$\sin{\left(2 \phi \right)} = 1$$
$$\cos{\left(2 \phi \right)} = 0$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
substitute coefficients
$$x' = \frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}$$
$$y' = \frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}$$
then the equation turns from
$$x'^{2} - 2 x' y' - 10 x' + y'^{2} - 6 y' + 25 = 0$$
to
$$\left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right)^{2} - 2 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right) \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right) - 10 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right) + \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right)^{2} - 6 \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right) + 25 = 0$$
simplify
$$- 8 \sqrt{2} \tilde x + 2 \tilde y^{2} + 2 \sqrt{2} \tilde y + 25 = 0$$
$$\left(\sqrt{2} \tilde y + 1\right)^{2} = 8 \sqrt{2} \tilde x - 24$$
$$\left(\tilde y + \frac{\sqrt{2}}{2}\right)^{2} = 4 \sqrt{2} \left(\tilde x - \frac{3 \sqrt{2}}{2}\right)$$
$$\tilde y'^{2} = 4 \sqrt{2} \tilde x'$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \frac{\sqrt{2}}{2} + 0 \frac{\sqrt{2}}{2}$$
$$y_{0} = 0 \frac{\sqrt{2}}{2} + 0 \frac{\sqrt{2}}{2}$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
(0, 0)

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

### Invariants method

Given line equation of 2-order:
$$x^{2} - 2 x y - 10 x + y^{2} - 6 y + 25 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = -1$$
$$a_{13} = -5$$
$$a_{22} = 1$$
$$a_{23} = -3$$
$$a_{33} = 25$$
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
$$I_{1} = 2$$
     |1   -1|
I2 = |      |
|-1  1 |

$$I_{3} = \left|\begin{matrix}1 & -1 & -5\\-1 & 1 & -3\\-5 & -3 & 25\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & -1\\-1 & 1 - \lambda\end{matrix}\right|$$
     |1   -5|   |1   -3|
K2 = |      | + |      |
|-5  25|   |-3  25|

$$I_{1} = 2$$
$$I_{2} = 0$$
$$I_{3} = -64$$
$$I{\left(\lambda \right)} = \lambda^{2} - 2 \lambda$$
$$K_{2} = 16$$
Because
$$I_{2} = 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : parabola
$$I_{1} \tilde y^{2} + 2 \tilde x \sqrt{- \frac{I_{3}}{I_{1}}} = 0$$
or
$$8 \sqrt{2} \tilde x + 2 \tilde y^{2} = 0$$
$$\tilde y^{2} = 4 \sqrt{2} \tilde x$$
- reduced to canonical form