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:
- x^2+6xy+9y^2
- x²+y²-z²=0
- 9x^2+4y=+8y^2+8x+14y+5=0
- 36x^2-16y^2=144
- Plot:
- (x^2)+(y-sqrt((|x|)))^2=1
- z=x^2+y^2
-
e^(y/x)=x*y+1
- x^(2)+(y-(x^(2))^(1/3))^(2)=1
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Identical expressions
- 3x^ two +8y^ two -12x+32y- one hundred = zero
- 3x squared plus 8y squared minus 12x plus 32y minus 100 equally 0
- 3x to the power of two plus 8y to the power of two minus 12x plus 32y minus one hundred equally zero
- 3x2+8y2-12x+32y-100=0
- 3x²+8y²-12x+32y-100=0
- 3x to the power of 2+8y to the power of 2-12x+32y-100=0
- 3x^2+8y^2-12x+32y-100=O
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Similar expressions
- 3x^2+8y^2-12x+32y+100=0
- 3x^2-8y^2-12x+32y-100=0
- 3x^2+8y^2-12x-32y-100=0
- 3x^2+8y^2+12x+32y-100=0
3x^2+8y^2-12x+32y-100=0 canonical form
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The solution
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2 2 -100 - 12*x + 3*x + 8*y + 32*y = 0
$$3 x^{2} - 12 x + 8 y^{2} + 32 y - 100 = 0$$
3*x^2 - 12*x + 8*y^2 + 32*y - 100 = 0
Detail solution
Given line equation of 2-order:
$$3 x^{2} - 12 x + 8 y^{2} + 32 y - 100 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = -6$$
$$a_{22} = 8$$
$$a_{23} = 16$$
$$a_{33} = -100$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}3 & 0\\0 & 8\end{matrix}\right|$$
$$\Delta = 24$$
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
$$3 x_{0} - 6 = 0$$
$$8 y_{0} + 16 = 0$$
then
$$x_{0} = 2$$
$$y_{0} = -2$$
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
$$a'_{33} = - 6 x_{0} + 16 y_{0} - 100$$
$$a'_{33} = -144$$
then equation turns into
$$3 x'^{2} + 8 y'^{2} - 144 = 0$$
Given equation is ellipse
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$3 x^{2} - 12 x + 8 y^{2} + 32 y - 100 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = -6$$
$$a_{22} = 8$$
$$a_{23} = 16$$
$$a_{33} = -100$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}3 & 0\\0 & 8\end{matrix}\right|$$
$$\Delta = 24$$
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
$$3 x_{0} - 6 = 0$$
$$8 y_{0} + 16 = 0$$
then
$$x_{0} = 2$$
$$y_{0} = -2$$
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
$$a'_{33} = - 6 x_{0} + 16 y_{0} - 100$$
$$a'_{33} = -144$$
then equation turns into
$$3 x'^{2} + 8 y'^{2} - 144 = 0$$
Given equation is ellipse
2 2
\tilde x \tilde y
---------- + ---------- = 1
2 2
// ___\\ // ___\\
||\/ 3 || ||\/ 2 ||
||-----|| ||-----||
|\ 3 /| |\ 4 /|
|-------| |-------|
\ 1/12 / \ 1/12 / - reduced to canonical form
The center of canonical coordinate system at point O
(2, -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:
$$3 x^{2} - 12 x + 8 y^{2} + 32 y - 100 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = -6$$
$$a_{22} = 8$$
$$a_{23} = 16$$
$$a_{33} = -100$$
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
$$I_{1} = 11$$
$$I_{3} = \left|\begin{matrix}3 & 0 & -6\\0 & 8 & 16\\-6 & 16 & -100\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & 0\\0 & 8 - \lambda\end{matrix}\right|$$
$$I_{1} = 11$$
$$I_{2} = 24$$
$$I_{3} = -3456$$
$$I{\left(\lambda \right)} = \lambda^{2} - 11 \lambda + 24$$
$$K_{2} = -1392$$
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
$$\lambda^{2} - 11 \lambda + 24 = 0$$
$$\lambda_{1} = 8$$
$$\lambda_{2} = 3$$
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
$$8 \tilde x^{2} + 3 \tilde y^{2} - 144 = 0$$
- reduced to canonical form
$$3 x^{2} - 12 x + 8 y^{2} + 32 y - 100 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = -6$$
$$a_{22} = 8$$
$$a_{23} = 16$$
$$a_{33} = -100$$
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
$$I_{1} = 11$$
|3 0|
I2 = | |
|0 8|$$I_{3} = \left|\begin{matrix}3 & 0 & -6\\0 & 8 & 16\\-6 & 16 & -100\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & 0\\0 & 8 - \lambda\end{matrix}\right|$$
|3 -6 | |8 16 |
K2 = | | + | |
|-6 -100| |16 -100|$$I_{1} = 11$$
$$I_{2} = 24$$
$$I_{3} = -3456$$
$$I{\left(\lambda \right)} = \lambda^{2} - 11 \lambda + 24$$
$$K_{2} = -1392$$
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
$$\lambda^{2} - 11 \lambda + 24 = 0$$
$$\lambda_{1} = 8$$
$$\lambda_{2} = 3$$
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
$$8 \tilde x^{2} + 3 \tilde y^{2} - 144 = 0$$
2 2
\tilde x \tilde y
---------- + ---------- = 1
2 2
// ___\\ // ___\\
||\/ 2 || ||\/ 3 ||
||-----|| ||-----||
|\ 4 /| |\ 3 /|
|-------| |-------|
\ 1/12 / \ 1/12 / - reduced to canonical form