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- Canonical form of a double hyperboloid
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- Canonical form of a degenerate ellipse
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How to use it?
- Canonical form:
- 4y^2+z^2-x-16y-4z+20=0
- x2a2+y2b2+z2c2=1
- x^2+y^2+2x+6y+6=0
- x-exp(-x)
- Plot:
- x^2+(y-(sqrt((|x|))))^2=1
- x^2+(y-(sqrt(x))^3*(x^2))^2=1
- sqrt(x^2-0,36)/x-2
- tan(x*y)=1
- Sum of series:
-
64
-
Identical expressions
- 4x^ two +16y^ two = sixty-four
- 4x squared plus 16y squared equally 64
- 4x to the power of two plus 16y to the power of two equally sixty minus four
- 4x2+16y2=64
- 4x²+16y²=64
- 4x to the power of 2+16y to the power of 2=64
-
Similar expressions
- 4x^2-16y^2=64
4x^2+16y^2=64 canonical form
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The solution
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2 2 -64 + 4*x + 16*y = 0
$$4 x^{2} + 16 y^{2} - 64 = 0$$
4*x^2 + 16*y^2 - 64 = 0
Detail solution
Given line equation of 2-order:
$$4 x^{2} + 16 y^{2} - 64 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 16$$
$$a_{23} = 0$$
$$a_{33} = -64$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & 0\\0 & 16\end{matrix}\right|$$
$$\Delta = 64$$
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
$$4 x_{0} = 0$$
$$16 y_{0} = 0$$
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
$$a'_{33} = -64$$
$$a'_{33} = -64$$
then equation turns into
$$4 x'^{2} + 16 y'^{2} - 64 = 0$$
Given equation is ellipse
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$4 x^{2} + 16 y^{2} - 64 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 16$$
$$a_{23} = 0$$
$$a_{33} = -64$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & 0\\0 & 16\end{matrix}\right|$$
$$\Delta = 64$$
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
$$4 x_{0} = 0$$
$$16 y_{0} = 0$$
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
$$a'_{33} = -64$$
$$a'_{33} = -64$$
then equation turns into
$$4 x'^{2} + 16 y'^{2} - 64 = 0$$
Given equation is ellipse
2 2
\tilde x \tilde y
--------- + --------- = 1
2 2
/ 1 \ / 1 \
|-----| |-----|
\2*1/8/ \4*1/8/ - 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:
$$4 x^{2} + 16 y^{2} - 64 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 16$$
$$a_{23} = 0$$
$$a_{33} = -64$$
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
$$I_{1} = 20$$
$$I_{3} = \left|\begin{matrix}4 & 0 & 0\\0 & 16 & 0\\0 & 0 & -64\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & 0\\0 & 16 - \lambda\end{matrix}\right|$$
$$I_{1} = 20$$
$$I_{2} = 64$$
$$I_{3} = -4096$$
$$I{\left(\lambda \right)} = \lambda^{2} - 20 \lambda + 64$$
$$K_{2} = -1280$$
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
$$\lambda^{2} - 20 \lambda + 64 = 0$$
$$\lambda_{1} = 16$$
$$\lambda_{2} = 4$$
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
$$16 \tilde x^{2} + 4 \tilde y^{2} - 64 = 0$$
- reduced to canonical form
$$4 x^{2} + 16 y^{2} - 64 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 16$$
$$a_{23} = 0$$
$$a_{33} = -64$$
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
$$I_{1} = 20$$
|4 0 |
I2 = | |
|0 16|$$I_{3} = \left|\begin{matrix}4 & 0 & 0\\0 & 16 & 0\\0 & 0 & -64\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & 0\\0 & 16 - \lambda\end{matrix}\right|$$
|4 0 | |16 0 |
K2 = | | + | |
|0 -64| |0 -64|$$I_{1} = 20$$
$$I_{2} = 64$$
$$I_{3} = -4096$$
$$I{\left(\lambda \right)} = \lambda^{2} - 20 \lambda + 64$$
$$K_{2} = -1280$$
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
$$\lambda^{2} - 20 \lambda + 64 = 0$$
$$\lambda_{1} = 16$$
$$\lambda_{2} = 4$$
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
$$16 \tilde x^{2} + 4 \tilde y^{2} - 64 = 0$$
2 2
\tilde x \tilde y
--------- + --------- = 1
2 2
/ 1 \ / 1 \
|-----| |-----|
\4*1/8/ \2*1/8/ - reduced to canonical form