Mister Exam
Other calculators
- Canonical form of a elliptical paraboloid
- Canonical form of a double hyperboloid
- Canonical form of a imaginary ellipsoid
- Canonical form of a degenerate ellipse
- Canonical form of a parabola
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How to use it?
- Canonical form:
- 4x^2+y^2/16=4
- 5x^2+6y^2+7z^2-4xy+4yz-10x+8y+14z-6=0
- 2(x+2)(x+6)
- 9x^2-y^2=9
- Plot:
- (|y|)=(|1-1/3*x|)
- ((|y|)+2)^2+((|x|)-4)^2=8
- y^2-4*x+x^2+16=0
-
y^2=x*y-2
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Identical expressions
- nine x^ two -y^ two =9
- 9x squared minus y squared equally 9
- nine x to the power of two minus y to the power of two equally 9
- 9x2-y2=9
- 9x²-y²=9
- 9x to the power of 2-y to the power of 2=9
-
Similar expressions
- 9x^2+y^2=9
9x^2-y^2=9 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given line equation of 2-order:
$$9 x^{2} - y^{2} - 9 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -1$$
$$a_{23} = 0$$
$$a_{33} = -9$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}9 & 0\\0 & -1\end{matrix}\right|$$
$$\Delta = -9$$
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
$$9 x_{0} = 0$$
$$- y_{0} = 0$$
then
$$x_{0} = 0$$
$$y_{0} = 0$$
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
$$a'_{33} = -9$$
$$a'_{33} = -9$$
then equation turns into
$$9 x'^{2} - y'^{2} - 9 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{1} - \frac{\tilde y^{2}}{9} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$9 x^{2} - y^{2} - 9 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -1$$
$$a_{23} = 0$$
$$a_{33} = -9$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}9 & 0\\0 & -1\end{matrix}\right|$$
$$\Delta = -9$$
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
$$9 x_{0} = 0$$
$$- y_{0} = 0$$
then
$$x_{0} = 0$$
$$y_{0} = 0$$
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
$$a'_{33} = -9$$
$$a'_{33} = -9$$
then equation turns into
$$9 x'^{2} - y'^{2} - 9 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{1} - \frac{\tilde y^{2}}{9} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(0, 0)
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:
$$9 x^{2} - y^{2} - 9 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -1$$
$$a_{23} = 0$$
$$a_{33} = -9$$
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
$$I_{1} = 8$$
$$I_{3} = \left|\begin{matrix}9 & 0 & 0\\0 & -1 & 0\\0 & 0 & -9\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & 0\\0 & - \lambda - 1\end{matrix}\right|$$
$$I_{1} = 8$$
$$I_{2} = -9$$
$$I_{3} = 81$$
$$I{\left(\lambda \right)} = \lambda^{2} - 8 \lambda - 9$$
$$K_{2} = -72$$
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
$$\lambda^{2} - 8 \lambda - 9 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = -1$$
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
$$9 \tilde x^{2} - \tilde y^{2} - 9 = 0$$
$$\frac{\tilde x^{2}}{1} - \frac{\tilde y^{2}}{9} = 1$$
- reduced to canonical form
$$9 x^{2} - y^{2} - 9 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -1$$
$$a_{23} = 0$$
$$a_{33} = -9$$
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
$$I_{1} = 8$$
|9 0 |
I2 = | |
|0 -1|$$I_{3} = \left|\begin{matrix}9 & 0 & 0\\0 & -1 & 0\\0 & 0 & -9\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & 0\\0 & - \lambda - 1\end{matrix}\right|$$
|9 0 | |-1 0 |
K2 = | | + | |
|0 -9| |0 -9|$$I_{1} = 8$$
$$I_{2} = -9$$
$$I_{3} = 81$$
$$I{\left(\lambda \right)} = \lambda^{2} - 8 \lambda - 9$$
$$K_{2} = -72$$
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
$$\lambda^{2} - 8 \lambda - 9 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = -1$$
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
$$9 \tilde x^{2} - \tilde y^{2} - 9 = 0$$
$$\frac{\tilde x^{2}}{1} - \frac{\tilde y^{2}}{9} = 1$$
- reduced to canonical form