Mister Exam
(x-3)^2/25+(y+2)^2/49=1 canonical form
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The solution
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2 2
(-3 + x) (2 + y)
-1 + --------- + -------- = 0
25 49
$$\frac{\left(x - 3\right)^{2}}{25} + \frac{\left(y + 2\right)^{2}}{49} - 1 = 0$$
(x - 3)^2/25 + (y + 2)^2/49 - 1 = 0
Detail solution
Given line equation of 2-order:
$$\frac{\left(x - 3\right)^{2}}{25} + \frac{\left(y + 2\right)^{2}}{49} - 1 = 0$$
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
$$a_{11} = \frac{1}{25}$$
$$a_{12} = 0$$
$$a_{13} = - \frac{3}{25}$$
$$a_{22} = \frac{1}{49}$$
$$a_{23} = \frac{2}{49}$$
$$a_{33} = - \frac{684}{1225}$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}\frac{1}{25} & 0\\0 & \frac{1}{49}\end{matrix}\right|$$
$$\Delta = \frac{1}{1225}$$
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
$$\frac{x_{0}}{25} - \frac{3}{25} = 0$$
$$\frac{y_{0}}{49} + \frac{2}{49} = 0$$
then
$$x_{0} = 3$$
$$y_{0} = -2$$
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
$$a'_{33} = - \frac{3 x_{0}}{25} + \frac{2 y_{0}}{49} - \frac{684}{1225}$$
$$a'_{33} = -1$$
then equation turns into
$$\frac{x'^{2}}{25} + \frac{y'^{2}}{49} - 1 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{5}{1}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{7}{1}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$\frac{\left(x - 3\right)^{2}}{25} + \frac{\left(y + 2\right)^{2}}{49} - 1 = 0$$
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
$$a_{11} = \frac{1}{25}$$
$$a_{12} = 0$$
$$a_{13} = - \frac{3}{25}$$
$$a_{22} = \frac{1}{49}$$
$$a_{23} = \frac{2}{49}$$
$$a_{33} = - \frac{684}{1225}$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}\frac{1}{25} & 0\\0 & \frac{1}{49}\end{matrix}\right|$$
$$\Delta = \frac{1}{1225}$$
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
$$\frac{x_{0}}{25} - \frac{3}{25} = 0$$
$$\frac{y_{0}}{49} + \frac{2}{49} = 0$$
then
$$x_{0} = 3$$
$$y_{0} = -2$$
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
$$a'_{33} = - \frac{3 x_{0}}{25} + \frac{2 y_{0}}{49} - \frac{684}{1225}$$
$$a'_{33} = -1$$
then equation turns into
$$\frac{x'^{2}}{25} + \frac{y'^{2}}{49} - 1 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{5}{1}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{7}{1}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(3, -2)
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
Invariants method
Given line equation of 2-order:
$$\frac{\left(x - 3\right)^{2}}{25} + \frac{\left(y + 2\right)^{2}}{49} - 1 = 0$$
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
$$a_{11} = \frac{1}{25}$$
$$a_{12} = 0$$
$$a_{13} = - \frac{3}{25}$$
$$a_{22} = \frac{1}{49}$$
$$a_{23} = \frac{2}{49}$$
$$a_{33} = - \frac{684}{1225}$$
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
$$I_{1} = \frac{74}{1225}$$
$$I_{3} = \left|\begin{matrix}\frac{1}{25} & 0 & - \frac{3}{25}\\0 & \frac{1}{49} & \frac{2}{49}\\- \frac{3}{25} & \frac{2}{49} & - \frac{684}{1225}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}\frac{1}{25} - \lambda & 0\\0 & \frac{1}{49} - \lambda\end{matrix}\right|$$
$$I_{1} = \frac{74}{1225}$$
$$I_{2} = \frac{1}{1225}$$
$$I_{3} = - \frac{1}{1225}$$
$$I{\left(\lambda \right)} = \lambda^{2} - \frac{74 \lambda}{1225} + \frac{1}{1225}$$
$$K_{2} = - \frac{61}{1225}$$
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
$$\lambda^{2} - \frac{74 \lambda}{1225} + \frac{1}{1225} = 0$$
$$\lambda_{1} = \frac{1}{25}$$
$$\lambda_{2} = \frac{1}{49}$$
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
$$\frac{\tilde x^{2}}{25} + \frac{\tilde y^{2}}{49} - 1 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{5}{1}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{7}{1}\right)^{2}} = 1$$
- reduced to canonical form
$$\frac{\left(x - 3\right)^{2}}{25} + \frac{\left(y + 2\right)^{2}}{49} - 1 = 0$$
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
$$a_{11} = \frac{1}{25}$$
$$a_{12} = 0$$
$$a_{13} = - \frac{3}{25}$$
$$a_{22} = \frac{1}{49}$$
$$a_{23} = \frac{2}{49}$$
$$a_{33} = - \frac{684}{1225}$$
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
$$I_{1} = \frac{74}{1225}$$
|1/25 0 |
I2 = | |
| 0 1/49|$$I_{3} = \left|\begin{matrix}\frac{1}{25} & 0 & - \frac{3}{25}\\0 & \frac{1}{49} & \frac{2}{49}\\- \frac{3}{25} & \frac{2}{49} & - \frac{684}{1225}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}\frac{1}{25} - \lambda & 0\\0 & \frac{1}{49} - \lambda\end{matrix}\right|$$
|1/25 -3/25| |1/49 2/49 |
| | | |
K2 = | -684 | + | -684 |
|-3/25 -----| |2/49 -----|
| 1225| | 1225|$$I_{1} = \frac{74}{1225}$$
$$I_{2} = \frac{1}{1225}$$
$$I_{3} = - \frac{1}{1225}$$
$$I{\left(\lambda \right)} = \lambda^{2} - \frac{74 \lambda}{1225} + \frac{1}{1225}$$
$$K_{2} = - \frac{61}{1225}$$
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
$$\lambda^{2} - \frac{74 \lambda}{1225} + \frac{1}{1225} = 0$$
$$\lambda_{1} = \frac{1}{25}$$
$$\lambda_{2} = \frac{1}{49}$$
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
$$\frac{\tilde x^{2}}{25} + \frac{\tilde y^{2}}{49} - 1 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{5}{1}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{7}{1}\right)^{2}} = 1$$
- reduced to canonical form