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:
-
9x^2-25y^2-18x-100y-316=0
-
x1^2+4x2^2
- (x-x0)^2+(y-y0)^2=0
- 2x^2+10y^2+2z^2+6xy+2xz+6yz=36
- Plot:
-
sin(x^e)+cos(y^e)=1
- z=3/sqrt(x^2+y-9)
-
(-x^2)*y^3+(x^2+y^2-1)^3=0
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x^2+4y^2
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Identical expressions
- x1^ two +4x two ^2
- x1 squared plus 4x2 squared
- x1 to the power of two plus 4x two squared
- x12+4x22
- x1²+4x2²
- x1 to the power of 2+4x2 to the power of 2
-
Similar expressions
- 3x1^2+4x2^2+5x3^2-4x1x2+2x1x3=0
- x1^2-4x2^2
x1^2+4x2^2 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given line equation of 2-order:
$$x_{1}^{2} + 4 x_{2}^{2} = 0$$
This equation looks like:
$$a_{11} x_{2}^{2} + 2 a_{12} x_{1} x_{2} + a_{22} x_{1}^{2} + 2 a_{13} x_{2} + 2 a_{23} x_{1} + a_{33} = 0$$
where
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 1$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & 0\\0 & 1\end{matrix}\right|$$
$$\Delta = 4$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} x_{20} + a_{12} x_{10} + a_{13} = 0$$
$$a_{12} x_{20} + a_{22} x_{10} + a_{23} = 0$$
substitute coefficients
$$4 x_{20} = 0$$
$$x_{10} = 0$$
then
$$x_{20} = 0$$
$$x_{10} = 0$$
Thus, we have the equation in the coordinate system O'x'y'
$$a_{11} x2'^{2} + 2 a_{12} x1' x2' + a_{22} x1'^{2} + a'_{33} = 0$$
where
$$a'_{33} = a_{13} x_{20} + a_{23} x_{10} + a_{33}$$
or
$$a'_{33} = 0$$
$$a'_{33} = 0$$
then The equation is transformed to
$$x1'^{2} + 4 x2'^{2} = 0$$
Given equation is degenerate ellipse
$$\frac{\tilde x1^{2}}{1^{2}} + \frac{\tilde x2^{2}}{\left(\frac{1}{2}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( 1, \ 0\right)$$
$$\vec e_{2} = \left( 0, \ 1\right)$$
$$x_{1}^{2} + 4 x_{2}^{2} = 0$$
This equation looks like:
$$a_{11} x_{2}^{2} + 2 a_{12} x_{1} x_{2} + a_{22} x_{1}^{2} + 2 a_{13} x_{2} + 2 a_{23} x_{1} + a_{33} = 0$$
where
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 1$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & 0\\0 & 1\end{matrix}\right|$$
$$\Delta = 4$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} x_{20} + a_{12} x_{10} + a_{13} = 0$$
$$a_{12} x_{20} + a_{22} x_{10} + a_{23} = 0$$
substitute coefficients
$$4 x_{20} = 0$$
$$x_{10} = 0$$
then
$$x_{20} = 0$$
$$x_{10} = 0$$
Thus, we have the equation in the coordinate system O'x'y'
$$a_{11} x2'^{2} + 2 a_{12} x1' x2' + a_{22} x1'^{2} + a'_{33} = 0$$
where
$$a'_{33} = a_{13} x_{20} + a_{23} x_{10} + a_{33}$$
or
$$a'_{33} = 0$$
$$a'_{33} = 0$$
then The equation is transformed to
$$x1'^{2} + 4 x2'^{2} = 0$$
Given equation is degenerate ellipse
$$\frac{\tilde x1^{2}}{1^{2}} + \frac{\tilde x2^{2}}{\left(\frac{1}{2}\right)^{2}} = 0$$
- 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:
$$x_{1}^{2} + 4 x_{2}^{2} = 0$$
This equation looks like:
$$a_{11} x_{2}^{2} + 2 a_{12} x_{1} x_{2} + a_{22} x_{1}^{2} + 2 a_{13} x_{2} + 2 a_{23} x_{1} + a_{33} = 0$$
where
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 1$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = 5$$
$$I_{3} = \left|\begin{matrix}4 & 0 & 0\\0 & 1 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 4 & 0\\0 & - \lambda + 1\end{matrix}\right|$$
$$I_{1} = 5$$
$$I_{2} = 4$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 5 \lambda + 4$$
$$K_{2} = 0$$
Because
$$I_{3} = 0 \wedge I_{2} > 0$$
then by line type:
this equation is of type : degenerate ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + \lambda^{2} + I_{2} = 0$$
or
$$\lambda^{2} - 5 \lambda + 4 = 0$$
Solve this equation
$$\lambda_{1} = 4$$
$$\lambda_{2} = 1$$
then the canonical form of the equation will be
$$\tilde x1^{2} \lambda_{2} + \tilde x2^{2} \lambda_{1} + \frac{I_{3}}{I_{2}} = 0$$
or
$$\tilde x1^{2} + 4 \tilde x2^{2} = 0$$
$$\frac{\tilde x1^{2}}{1^{2}} + \frac{\tilde x2^{2}}{\left(\frac{1}{2}\right)^{2}} = 0$$
- reduced to canonical form
$$x_{1}^{2} + 4 x_{2}^{2} = 0$$
This equation looks like:
$$a_{11} x_{2}^{2} + 2 a_{12} x_{1} x_{2} + a_{22} x_{1}^{2} + 2 a_{13} x_{2} + 2 a_{23} x_{1} + a_{33} = 0$$
where
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 1$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = 5$$
|4 0|
I2 = | |
|0 1|$$I_{3} = \left|\begin{matrix}4 & 0 & 0\\0 & 1 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 4 & 0\\0 & - \lambda + 1\end{matrix}\right|$$
|4 0| |1 0|
K2 = | | + | |
|0 0| |0 0|$$I_{1} = 5$$
$$I_{2} = 4$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 5 \lambda + 4$$
$$K_{2} = 0$$
Because
$$I_{3} = 0 \wedge I_{2} > 0$$
then by line type:
this equation is of type : degenerate ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + \lambda^{2} + I_{2} = 0$$
or
$$\lambda^{2} - 5 \lambda + 4 = 0$$
Solve this equation
$$\lambda_{1} = 4$$
$$\lambda_{2} = 1$$
then the canonical form of the equation will be
$$\tilde x1^{2} \lambda_{2} + \tilde x2^{2} \lambda_{1} + \frac{I_{3}}{I_{2}} = 0$$
or
$$\tilde x1^{2} + 4 \tilde x2^{2} = 0$$
$$\frac{\tilde x1^{2}}{1^{2}} + \frac{\tilde x2^{2}}{\left(\frac{1}{2}\right)^{2}} = 0$$
- reduced to canonical form
The graph