Mister Exam
3x^2-3y^2-16x+6y+24=0 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
2 2 24 - 16*x - 3*y + 3*x + 6*y = 0
$$3 x^{2} - 16 x - 3 y^{2} + 6 y + 24 = 0$$
3*x^2 - 16*x - 3*y^2 + 6*y + 24 = 0
Detail solution
Given line equation of 2-order:
$$3 x^{2} - 16 x - 3 y^{2} + 6 y + 24 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = -8$$
$$a_{22} = -3$$
$$a_{23} = 3$$
$$a_{33} = 24$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}3 & 0\\0 & -3\end{matrix}\right|$$
$$\Delta = -9$$
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
$$3 x_{0} - 8 = 0$$
$$3 - 3 y_{0} = 0$$
then
$$x_{0} = \frac{8}{3}$$
$$y_{0} = 1$$
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
$$a'_{33} = - 8 x_{0} + 3 y_{0} + 24$$
$$a'_{33} = \frac{17}{3}$$
then equation turns into
$$3 x'^{2} - 3 y'^{2} + \frac{17}{3} = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{\frac{17}{9}} - \frac{\tilde y^{2}}{\frac{17}{9}} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$3 x^{2} - 16 x - 3 y^{2} + 6 y + 24 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = -8$$
$$a_{22} = -3$$
$$a_{23} = 3$$
$$a_{33} = 24$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}3 & 0\\0 & -3\end{matrix}\right|$$
$$\Delta = -9$$
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
$$3 x_{0} - 8 = 0$$
$$3 - 3 y_{0} = 0$$
then
$$x_{0} = \frac{8}{3}$$
$$y_{0} = 1$$
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
$$a'_{33} = - 8 x_{0} + 3 y_{0} + 24$$
$$a'_{33} = \frac{17}{3}$$
then equation turns into
$$3 x'^{2} - 3 y'^{2} + \frac{17}{3} = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{\frac{17}{9}} - \frac{\tilde y^{2}}{\frac{17}{9}} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(8/3, 1)
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:
$$3 x^{2} - 16 x - 3 y^{2} + 6 y + 24 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = -8$$
$$a_{22} = -3$$
$$a_{23} = 3$$
$$a_{33} = 24$$
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
$$I_{1} = 0$$
$$I_{3} = \left|\begin{matrix}3 & 0 & -8\\0 & -3 & 3\\-8 & 3 & 24\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & 0\\0 & - \lambda - 3\end{matrix}\right|$$
$$I_{1} = 0$$
$$I_{2} = -9$$
$$I_{3} = -51$$
$$I{\left(\lambda \right)} = \lambda^{2} - 9$$
$$K_{2} = -73$$
Because
$$I_{2} < 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 9 = 0$$
$$\lambda_{1} = -3$$
$$\lambda_{2} = 3$$
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
$$- 3 \tilde x^{2} + 3 \tilde y^{2} + \frac{17}{3} = 0$$
$$\frac{\tilde x^{2}}{\frac{17}{9}} - \frac{\tilde y^{2}}{\frac{17}{9}} = 1$$
- reduced to canonical form
$$3 x^{2} - 16 x - 3 y^{2} + 6 y + 24 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = -8$$
$$a_{22} = -3$$
$$a_{23} = 3$$
$$a_{33} = 24$$
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
$$I_{1} = 0$$
|3 0 |
I2 = | |
|0 -3|$$I_{3} = \left|\begin{matrix}3 & 0 & -8\\0 & -3 & 3\\-8 & 3 & 24\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & 0\\0 & - \lambda - 3\end{matrix}\right|$$
|3 -8| |-3 3 |
K2 = | | + | |
|-8 24| |3 24|$$I_{1} = 0$$
$$I_{2} = -9$$
$$I_{3} = -51$$
$$I{\left(\lambda \right)} = \lambda^{2} - 9$$
$$K_{2} = -73$$
Because
$$I_{2} < 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 9 = 0$$
$$\lambda_{1} = -3$$
$$\lambda_{2} = 3$$
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
$$- 3 \tilde x^{2} + 3 \tilde y^{2} + \frac{17}{3} = 0$$
$$\frac{\tilde x^{2}}{\frac{17}{9}} - \frac{\tilde y^{2}}{\frac{17}{9}} = 1$$
- reduced to canonical form