Mister Exam
8*x^2-4*x*y+5*y^2-52*x+22*y+53=0 canonical form
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2 2 53 - 52*x + 5*y + 8*x + 22*y - 4*x*y = 0
$$8 x^{2} - 4 x y - 52 x + 5 y^{2} + 22 y + 53 = 0$$
8*x^2 - 4*x*y - 52*x + 5*y^2 + 22*y + 53 = 0
Detail solution
Given line equation of 2-order:
$$8 x^{2} - 4 x y - 52 x + 5 y^{2} + 22 y + 53 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = -2$$
$$a_{13} = -26$$
$$a_{22} = 5$$
$$a_{23} = 11$$
$$a_{33} = 53$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}8 & -2\\-2 & 5\end{matrix}\right|$$
$$\Delta = 36$$
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
$$8 x_{0} - 2 y_{0} - 26 = 0$$
$$- 2 x_{0} + 5 y_{0} + 11 = 0$$
then
$$x_{0} = 3$$
$$y_{0} = -1$$
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
$$a'_{33} = - 26 x_{0} + 11 y_{0} + 53$$
$$a'_{33} = -36$$
then equation turns into
$$8 x'^{2} - 4 x' y' + 5 y'^{2} - 36 = 0$$
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
$$\cot{\left(2 \phi \right)} = - \frac{3}{4}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{3}{4} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{4}{5}$$
$$\cos{\left(2 \phi \right)} = \frac{3}{5}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{2 \sqrt{5}}{5}$$
$$\sin{\left(\phi \right)} = - \frac{\sqrt{5}}{5}$$
substitute coefficients
$$x' = \frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}$$
$$y' = - \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}$$
then the equation turns from
$$8 x'^{2} - 4 x' y' + 5 y'^{2} - 36 = 0$$
to
$$5 \left(- \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right)^{2} - 4 \left(- \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) \left(\frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}\right) + 8 \left(\frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}\right)^{2} - 36 = 0$$
simplify
$$9 \tilde x^{2} + 4 \tilde y^{2} - 36 = 0$$
Given equation is ellipse
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{2 \sqrt{5}}{5}, \ - \frac{\sqrt{5}}{5}\right)$$
$$\vec e_2 = \left( \frac{\sqrt{5}}{5}, \ \frac{2 \sqrt{5}}{5}\right)$$
$$8 x^{2} - 4 x y - 52 x + 5 y^{2} + 22 y + 53 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = -2$$
$$a_{13} = -26$$
$$a_{22} = 5$$
$$a_{23} = 11$$
$$a_{33} = 53$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}8 & -2\\-2 & 5\end{matrix}\right|$$
$$\Delta = 36$$
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
$$8 x_{0} - 2 y_{0} - 26 = 0$$
$$- 2 x_{0} + 5 y_{0} + 11 = 0$$
then
$$x_{0} = 3$$
$$y_{0} = -1$$
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
$$a'_{33} = - 26 x_{0} + 11 y_{0} + 53$$
$$a'_{33} = -36$$
then equation turns into
$$8 x'^{2} - 4 x' y' + 5 y'^{2} - 36 = 0$$
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
$$\cot{\left(2 \phi \right)} = - \frac{3}{4}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{3}{4} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{4}{5}$$
$$\cos{\left(2 \phi \right)} = \frac{3}{5}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{2 \sqrt{5}}{5}$$
$$\sin{\left(\phi \right)} = - \frac{\sqrt{5}}{5}$$
substitute coefficients
$$x' = \frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}$$
$$y' = - \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}$$
then the equation turns from
$$8 x'^{2} - 4 x' y' + 5 y'^{2} - 36 = 0$$
to
$$5 \left(- \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right)^{2} - 4 \left(- \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) \left(\frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}\right) + 8 \left(\frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}\right)^{2} - 36 = 0$$
simplify
$$9 \tilde x^{2} + 4 \tilde y^{2} - 36 = 0$$
Given equation is ellipse
2 2
\tilde x \tilde y
--------- + --------- = 1
2 2
/ 1 \ / 1 \
|-----| |-----|
\3*1/6/ \2*1/6/ - reduced to canonical form
The center of canonical coordinate system at point O
(3, -1)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{2 \sqrt{5}}{5}, \ - \frac{\sqrt{5}}{5}\right)$$
$$\vec e_2 = \left( \frac{\sqrt{5}}{5}, \ \frac{2 \sqrt{5}}{5}\right)$$
Invariants method
Given line equation of 2-order:
$$8 x^{2} - 4 x y - 52 x + 5 y^{2} + 22 y + 53 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = -2$$
$$a_{13} = -26$$
$$a_{22} = 5$$
$$a_{23} = 11$$
$$a_{33} = 53$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22}$$
$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12}\\a_{12} & a_{22} - \lambda\end{matrix}\right|$$
substitute coefficients
$$I_{1} = 13$$
$$I_{3} = \left|\begin{matrix}8 & -2 & -26\\-2 & 5 & 11\\-26 & 11 & 53\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}8 - \lambda & -2\\-2 & 5 - \lambda\end{matrix}\right|$$
$$I_{1} = 13$$
$$I_{2} = 36$$
$$I_{3} = -1296$$
$$I{\left(\lambda \right)} = \lambda^{2} - 13 \lambda + 36$$
$$K_{2} = -108$$
Because
$$I_{2} > 0 \wedge I_{1} I_{3} < 0$$
then by line type:
this equation is of type : ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 13 \lambda + 36 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = 4$$
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
$$9 \tilde x^{2} + 4 \tilde y^{2} - 36 = 0$$
- reduced to canonical form
$$8 x^{2} - 4 x y - 52 x + 5 y^{2} + 22 y + 53 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = -2$$
$$a_{13} = -26$$
$$a_{22} = 5$$
$$a_{23} = 11$$
$$a_{33} = 53$$
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
$$I_{1} = 13$$
|8 -2|
I2 = | |
|-2 5 |$$I_{3} = \left|\begin{matrix}8 & -2 & -26\\-2 & 5 & 11\\-26 & 11 & 53\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}8 - \lambda & -2\\-2 & 5 - \lambda\end{matrix}\right|$$
| 8 -26| |5 11|
K2 = | | + | |
|-26 53 | |11 53|$$I_{1} = 13$$
$$I_{2} = 36$$
$$I_{3} = -1296$$
$$I{\left(\lambda \right)} = \lambda^{2} - 13 \lambda + 36$$
$$K_{2} = -108$$
Because
$$I_{2} > 0 \wedge I_{1} I_{3} < 0$$
then by line type:
this equation is of type : ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 13 \lambda + 36 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = 4$$
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
$$9 \tilde x^{2} + 4 \tilde y^{2} - 36 = 0$$
2 2
\tilde x \tilde y
--------- + --------- = 1
2 2
/ 1 \ / 1 \
|-----| |-----|
\3*1/6/ \2*1/6/ - reduced to canonical form