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
-
How to use it?
- Canonical form:
- x^2+y^2+z^2=2x+2y-1
- x^2-y^2-2x-4=0
- x^2-2x+10y+21=0
- 9x^2+4xy+6y^2+40x+20y-50=0
- Plot:
-
y^2=x(1-x^2)^3
- y=-x3+(17/2)*x2-20*x+(25/2)
- y=-2*sqrt(|x|)+4
-
y=(x^2-1)^1/2
-
Identical expressions
- 9x^ two -16y^ two - twelve = zero
- 9x squared minus 16y squared minus 12 equally 0
- 9x to the power of two minus 16y to the power of two minus twelve equally zero
- 9x2-16y2-12=0
- 9x²-16y²-12=0
- 9x to the power of 2-16y to the power of 2-12=0
- 9x^2-16y^2-12=O
-
Similar expressions
- 9x^2+16y^2-12=0
- 9x^2-16y^2+12=0
9x^2-16y^2-12=0 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
2 2 -12 - 16*y + 9*x = 0
$$9 x^{2} - 16 y^{2} - 12 = 0$$
9*x^2 - 16*y^2 - 12 = 0
Detail solution
Given line equation of 2-order:
$$9 x^{2} - 16 y^{2} - 12 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -16$$
$$a_{23} = 0$$
$$a_{33} = -12$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}9 & 0\\0 & -16\end{matrix}\right|$$
$$\Delta = -144$$
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
$$9 x_{0} = 0$$
$$- 16 y_{0} = 0$$
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
$$a'_{33} = -12$$
$$a'_{33} = -12$$
then equation turns into
$$9 x'^{2} - 16 y'^{2} - 12 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{\frac{4}{3}} - \frac{\tilde y^{2}}{\frac{3}{4}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$9 x^{2} - 16 y^{2} - 12 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -16$$
$$a_{23} = 0$$
$$a_{33} = -12$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}9 & 0\\0 & -16\end{matrix}\right|$$
$$\Delta = -144$$
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
$$9 x_{0} = 0$$
$$- 16 y_{0} = 0$$
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
$$a'_{33} = -12$$
$$a'_{33} = -12$$
then equation turns into
$$9 x'^{2} - 16 y'^{2} - 12 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{\frac{4}{3}} - \frac{\tilde y^{2}}{\frac{3}{4}} = 1$$
- 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:
$$9 x^{2} - 16 y^{2} - 12 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -16$$
$$a_{23} = 0$$
$$a_{33} = -12$$
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
$$I_{1} = -7$$
$$I_{3} = \left|\begin{matrix}9 & 0 & 0\\0 & -16 & 0\\0 & 0 & -12\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & 0\\0 & - \lambda - 16\end{matrix}\right|$$
$$I_{1} = -7$$
$$I_{2} = -144$$
$$I_{3} = 1728$$
$$I{\left(\lambda \right)} = \lambda^{2} + 7 \lambda - 144$$
$$K_{2} = 84$$
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
$$\lambda^{2} + 7 \lambda - 144 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = -16$$
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
$$9 \tilde x^{2} - 16 \tilde y^{2} - 12 = 0$$
$$\frac{\tilde x^{2}}{\frac{4}{3}} - \frac{\tilde y^{2}}{\frac{3}{4}} = 1$$
- reduced to canonical form
$$9 x^{2} - 16 y^{2} - 12 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -16$$
$$a_{23} = 0$$
$$a_{33} = -12$$
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
$$I_{1} = -7$$
|9 0 |
I2 = | |
|0 -16|$$I_{3} = \left|\begin{matrix}9 & 0 & 0\\0 & -16 & 0\\0 & 0 & -12\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & 0\\0 & - \lambda - 16\end{matrix}\right|$$
|9 0 | |-16 0 |
K2 = | | + | |
|0 -12| | 0 -12|$$I_{1} = -7$$
$$I_{2} = -144$$
$$I_{3} = 1728$$
$$I{\left(\lambda \right)} = \lambda^{2} + 7 \lambda - 144$$
$$K_{2} = 84$$
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
$$\lambda^{2} + 7 \lambda - 144 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = -16$$
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
$$9 \tilde x^{2} - 16 \tilde y^{2} - 12 = 0$$
$$\frac{\tilde x^{2}}{\frac{4}{3}} - \frac{\tilde y^{2}}{\frac{3}{4}} = 1$$
- reduced to canonical form