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:
- 2+x^2-y=0
- -1/4y^2+y-x=0
- -4x1^2+4x1x2-7x2^2
- -6-(x-2)^4
- Plot:
- (x-y^2)*((|z|)-1)=0
-
(x²+y²-1)((x-2)²+y²-1)((x+2)²+y²-1)((x-1)²+(y-sqrt(3))²-1)((x+1)²+(y-sqrt(3))²-1)((x-1)²+(y+sqrt(3))²-1)((x+1)²+(y+sqrt(3))²-1)(x²+y²)((x-2)²+y²)((x+2)²+y²)((x-1)²+(y-sqrt(3))²)((x+1)²+(y-sqrt(3))²)((x-1)²+(y+sqrt(3))²)((x+1)²+(y+sqrt(3))²)=0
- z=x^2+y^2-x*y+9*x-6*y+20
- y^2=4*(x-1)/x^2
-
Identical expressions
- -4x1^ two +4x1x two -7x2^2
- minus 4x1 squared plus 4x1x2 minus 7x2 squared
- minus 4x1 to the power of two plus 4x1x two minus 7x2 squared
- -4x12+4x1x2-7x22
- -4x1²+4x1x2-7x2²
- -4x1 to the power of 2+4x1x2-7x2 to the power of 2
-
Similar expressions
- -4x1^2-4x1x2-7x2^2
- 4x1^2+4x1x2-7x2^2
- -4x1^2+4x1x2+7x2^2
-4x1^2+4x1x2-7x2^2 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
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2 2 - 7*x2 - 4*x1 + 4*x1*x2 = 0
$$- 4 x_{1}^{2} + 4 x_{1} x_{2} - 7 x_{2}^{2} = 0$$
-4*x1^2 + 4*x1*x2 - 7*x2^2 = 0
Detail solution
Given line equation of 2-order:
$$- 4 x_{1}^{2} + 4 x_{1} x_{2} - 7 x_{2}^{2} = 0$$
This equation looks like:
$$a_{11} x_{2}^{2} + 2 a_{12} x_{1} x_{2} + 2 a_{13} x_{2} + a_{22} x_{1}^{2} + 2 a_{23} x_{1} + a_{33} = 0$$
where
$$a_{11} = -7$$
$$a_{12} = 2$$
$$a_{13} = 0$$
$$a_{22} = -4$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-7 & 2\\2 & -4\end{matrix}\right|$$
$$\Delta = 24$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} x_{20} + a_{12} x_{10} + a_{13} = 0$$
$$a_{12} x_{20} + a_{22} x_{10} + a_{23} = 0$$
substitute coefficients
$$2 x_{10} - 7 x_{20} = 0$$
$$- 4 x_{10} + 2 x_{20} = 0$$
then
$$x_{20} = 0$$
$$x_{10} = 0$$
Thus, we have the equation in the coordinate system O'x'y'
$$a'_{33} + a_{11} x2'^{2} + 2 a_{12} x1' x2' + a_{22} x1'^{2} = 0$$
where
$$a'_{33} = a_{13} x_{20} + a_{23} x_{10} + a_{33}$$
or
$$a'_{33} = 0$$
$$a'_{33} = 0$$
then equation turns into
$$- 4 x1'^{2} + 4 x1' x2' - 7 x2'^{2} = 0$$
Rotate the resulting coordinate system by an angle φ
$$x2' = - \tilde x1 \sin{\left(\phi \right)} + \tilde x2 \cos{\left(\phi \right)}$$
$$x1' = \tilde x1 \cos{\left(\phi \right)} + \tilde x2 \sin{\left(\phi \right)}$$
φ - determined from the formula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
substitute coefficients
$$\cot{\left(2 \phi \right)} = - \frac{3}{4}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{3}{4} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{4}{5}$$
$$\cos{\left(2 \phi \right)} = \frac{3}{5}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{2 \sqrt{5}}{5}$$
$$\sin{\left(\phi \right)} = - \frac{\sqrt{5}}{5}$$
substitute coefficients
$$x2' = \frac{\sqrt{5} \tilde x1}{5} + \frac{2 \sqrt{5} \tilde x2}{5}$$
$$x1' = \frac{2 \sqrt{5} \tilde x1}{5} - \frac{\sqrt{5} \tilde x2}{5}$$
then the equation turns from
$$- 4 x1'^{2} + 4 x1' x2' - 7 x2'^{2} = 0$$
to
$$- 7 \left(\frac{\sqrt{5} \tilde x1}{5} + \frac{2 \sqrt{5} \tilde x2}{5}\right)^{2} + 4 \left(\frac{\sqrt{5} \tilde x1}{5} + \frac{2 \sqrt{5} \tilde x2}{5}\right) \left(\frac{2 \sqrt{5} \tilde x1}{5} - \frac{\sqrt{5} \tilde x2}{5}\right) - 4 \left(\frac{2 \sqrt{5} \tilde x1}{5} - \frac{\sqrt{5} \tilde x2}{5}\right)^{2} = 0$$
simplify
$$- 3 \tilde x1^{2} - 8 \tilde x2^{2} = 0$$
$$3 \tilde x1^{2} + 8 \tilde x2^{2} = 0$$
Given equation is degenerate ellipse
$$\frac{\tilde x1^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}} + \frac{\tilde x2^{2}}{\left(\frac{\sqrt{2}}{4}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{2 \sqrt{5}}{5}, \ - \frac{\sqrt{5}}{5}\right)$$
$$\vec e_2 = \left( \frac{\sqrt{5}}{5}, \ \frac{2 \sqrt{5}}{5}\right)$$
$$- 4 x_{1}^{2} + 4 x_{1} x_{2} - 7 x_{2}^{2} = 0$$
This equation looks like:
$$a_{11} x_{2}^{2} + 2 a_{12} x_{1} x_{2} + 2 a_{13} x_{2} + a_{22} x_{1}^{2} + 2 a_{23} x_{1} + a_{33} = 0$$
where
$$a_{11} = -7$$
$$a_{12} = 2$$
$$a_{13} = 0$$
$$a_{22} = -4$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-7 & 2\\2 & -4\end{matrix}\right|$$
$$\Delta = 24$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} x_{20} + a_{12} x_{10} + a_{13} = 0$$
$$a_{12} x_{20} + a_{22} x_{10} + a_{23} = 0$$
substitute coefficients
$$2 x_{10} - 7 x_{20} = 0$$
$$- 4 x_{10} + 2 x_{20} = 0$$
then
$$x_{20} = 0$$
$$x_{10} = 0$$
Thus, we have the equation in the coordinate system O'x'y'
$$a'_{33} + a_{11} x2'^{2} + 2 a_{12} x1' x2' + a_{22} x1'^{2} = 0$$
where
$$a'_{33} = a_{13} x_{20} + a_{23} x_{10} + a_{33}$$
or
$$a'_{33} = 0$$
$$a'_{33} = 0$$
then equation turns into
$$- 4 x1'^{2} + 4 x1' x2' - 7 x2'^{2} = 0$$
Rotate the resulting coordinate system by an angle φ
$$x2' = - \tilde x1 \sin{\left(\phi \right)} + \tilde x2 \cos{\left(\phi \right)}$$
$$x1' = \tilde x1 \cos{\left(\phi \right)} + \tilde x2 \sin{\left(\phi \right)}$$
φ - determined from the formula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
substitute coefficients
$$\cot{\left(2 \phi \right)} = - \frac{3}{4}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{3}{4} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{4}{5}$$
$$\cos{\left(2 \phi \right)} = \frac{3}{5}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{2 \sqrt{5}}{5}$$
$$\sin{\left(\phi \right)} = - \frac{\sqrt{5}}{5}$$
substitute coefficients
$$x2' = \frac{\sqrt{5} \tilde x1}{5} + \frac{2 \sqrt{5} \tilde x2}{5}$$
$$x1' = \frac{2 \sqrt{5} \tilde x1}{5} - \frac{\sqrt{5} \tilde x2}{5}$$
then the equation turns from
$$- 4 x1'^{2} + 4 x1' x2' - 7 x2'^{2} = 0$$
to
$$- 7 \left(\frac{\sqrt{5} \tilde x1}{5} + \frac{2 \sqrt{5} \tilde x2}{5}\right)^{2} + 4 \left(\frac{\sqrt{5} \tilde x1}{5} + \frac{2 \sqrt{5} \tilde x2}{5}\right) \left(\frac{2 \sqrt{5} \tilde x1}{5} - \frac{\sqrt{5} \tilde x2}{5}\right) - 4 \left(\frac{2 \sqrt{5} \tilde x1}{5} - \frac{\sqrt{5} \tilde x2}{5}\right)^{2} = 0$$
simplify
$$- 3 \tilde x1^{2} - 8 \tilde x2^{2} = 0$$
$$3 \tilde x1^{2} + 8 \tilde x2^{2} = 0$$
Given equation is degenerate ellipse
$$\frac{\tilde x1^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}} + \frac{\tilde x2^{2}}{\left(\frac{\sqrt{2}}{4}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
(0, 0)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{2 \sqrt{5}}{5}, \ - \frac{\sqrt{5}}{5}\right)$$
$$\vec e_2 = \left( \frac{\sqrt{5}}{5}, \ \frac{2 \sqrt{5}}{5}\right)$$
Invariants method
Given line equation of 2-order:
$$- 4 x_{1}^{2} + 4 x_{1} x_{2} - 7 x_{2}^{2} = 0$$
This equation looks like:
$$a_{11} x_{2}^{2} + 2 a_{12} x_{1} x_{2} + 2 a_{13} x_{2} + a_{22} x_{1}^{2} + 2 a_{23} x_{1} + a_{33} = 0$$
where
$$a_{11} = -7$$
$$a_{12} = 2$$
$$a_{13} = 0$$
$$a_{22} = -4$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = -11$$
$$I_{3} = \left|\begin{matrix}-7 & 2 & 0\\2 & -4 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 7 & 2\\2 & - \lambda - 4\end{matrix}\right|$$
$$I_{1} = -11$$
$$I_{2} = 24$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} + 11 \lambda + 24$$
$$K_{2} = 0$$
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:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} + 11 \lambda + 24 = 0$$
$$\lambda_{1} = -3$$
$$\lambda_{2} = -8$$
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
$$- 8 \tilde x1^{2} - 3 \tilde x2^{2} = 0$$
$$\frac{\tilde x1^{2}}{\left(\frac{\sqrt{2}}{4}\right)^{2}} + \frac{\tilde x2^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}} = 0$$
- reduced to canonical form
$$- 4 x_{1}^{2} + 4 x_{1} x_{2} - 7 x_{2}^{2} = 0$$
This equation looks like:
$$a_{11} x_{2}^{2} + 2 a_{12} x_{1} x_{2} + 2 a_{13} x_{2} + a_{22} x_{1}^{2} + 2 a_{23} x_{1} + a_{33} = 0$$
where
$$a_{11} = -7$$
$$a_{12} = 2$$
$$a_{13} = 0$$
$$a_{22} = -4$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = -11$$
|-7 2 |
I2 = | |
|2 -4|$$I_{3} = \left|\begin{matrix}-7 & 2 & 0\\2 & -4 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 7 & 2\\2 & - \lambda - 4\end{matrix}\right|$$
|-7 0| |-4 0|
K2 = | | + | |
|0 0| |0 0|$$I_{1} = -11$$
$$I_{2} = 24$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} + 11 \lambda + 24$$
$$K_{2} = 0$$
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:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} + 11 \lambda + 24 = 0$$
$$\lambda_{1} = -3$$
$$\lambda_{2} = -8$$
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
$$- 8 \tilde x1^{2} - 3 \tilde x2^{2} = 0$$
$$\frac{\tilde x1^{2}}{\left(\frac{\sqrt{2}}{4}\right)^{2}} + \frac{\tilde x2^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}} = 0$$
- reduced to canonical form