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:
- 4*x^2−12*x*y+9*y^2−20*x+30*y+16=0.
- -x^2+2y^2-2xy-2x-2y+2=0
- -7x^2-6xy+1y^2-8=0
- x^2+4*y^2+8*z^2-16
- Plot:
- y-(|x|)=(1/2)
- z=3*x+6*y-x^2-x*y+y^2
- x^2-2=y
-
x^2=8y
-
Identical expressions
- -x^ two + two y^ two -2xy-2x-2y+2= zero
- minus x squared plus 2y squared minus 2xy minus 2x minus 2y plus 2 equally 0
- minus x to the power of two plus two y to the power of two minus 2xy minus 2x minus 2y plus 2 equally zero
- -x2+2y2-2xy-2x-2y+2=0
- -x²+2y²-2xy-2x-2y+2=0
- -x to the power of 2+2y to the power of 2-2xy-2x-2y+2=0
- -x^2+2y^2-2xy-2x-2y+2=O
-
Similar expressions
- -x^2-2y^2-2xy-2x-2y+2=0
- -x^2+2y^2+2xy-2x-2y+2=0
- -x^2+2y^2-2xy-2x-2y-2=0
- x^2+2y^2-2xy-2x-2y+2=0
- -x^2+2y^2-2xy-2x+2y+2=0
- -x^2+2y^2-2xy+2x-2y+2=0
-x^2+2y^2-2xy-2x-2y+2=0 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
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2 2 2 - x - 2*x - 2*y + 2*y - 2*x*y = 0
$$- x^{2} - 2 x y - 2 x + 2 y^{2} - 2 y + 2 = 0$$
-x^2 - 2*x*y - 2*x + 2*y^2 - 2*y + 2 = 0
Detail solution
Given line equation of 2-order:
$$- x^{2} - 2 x y - 2 x + 2 y^{2} - 2 y + 2 = 0$$
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
$$a_{11} = -1$$
$$a_{12} = -1$$
$$a_{13} = -1$$
$$a_{22} = 2$$
$$a_{23} = -1$$
$$a_{33} = 2$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-1 & -1\\-1 & 2\end{matrix}\right|$$
$$\Delta = -3$$
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
$$- x_{0} - y_{0} - 1 = 0$$
$$- x_{0} + 2 y_{0} - 1 = 0$$
then
$$x_{0} = -1$$
$$y_{0} = 0$$
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
$$a'_{33} = - x_{0} - y_{0} + 2$$
$$a'_{33} = 3$$
then equation turns into
$$- x'^{2} - 2 x' y' + 2 y'^{2} + 3 = 0$$
Rotate the resulting coordinate system by an angle φ
$$x' = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y' = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
φ - determined from the formula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
substitute coefficients
$$\cot{\left(2 \phi \right)} = \frac{3}{2}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{3}{2} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{2 \sqrt{13}}{13}$$
$$\cos{\left(2 \phi \right)} = \frac{3 \sqrt{13}}{13}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{3 \sqrt{13}}{26} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{\frac{1}{2} - \frac{3 \sqrt{13}}{26}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{3 \sqrt{13}}{26} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{13}}{26}}$$
$$y' = \tilde x \sqrt{\frac{1}{2} - \frac{3 \sqrt{13}}{26}} + \tilde y \sqrt{\frac{3 \sqrt{13}}{26} + \frac{1}{2}}$$
then the equation turns from
$$- x'^{2} - 2 x' y' + 2 y'^{2} + 3 = 0$$
to
$$2 \left(\tilde x \sqrt{\frac{1}{2} - \frac{3 \sqrt{13}}{26}} + \tilde y \sqrt{\frac{3 \sqrt{13}}{26} + \frac{1}{2}}\right)^{2} - 2 \left(\tilde x \sqrt{\frac{1}{2} - \frac{3 \sqrt{13}}{26}} + \tilde y \sqrt{\frac{3 \sqrt{13}}{26} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{3 \sqrt{13}}{26} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{13}}{26}}\right) - \left(\tilde x \sqrt{\frac{3 \sqrt{13}}{26} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{13}}{26}}\right)^{2} + 3 = 0$$
simplify
$$- \frac{9 \sqrt{13} \tilde x^{2}}{26} - 2 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{3 \sqrt{13}}{26}} \sqrt{\frac{3 \sqrt{13}}{26} + \frac{1}{2}} + \frac{\tilde x^{2}}{2} - \frac{6 \sqrt{13} \tilde x \tilde y}{13} + 6 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{13}}{26}} \sqrt{\frac{3 \sqrt{13}}{26} + \frac{1}{2}} + \frac{\tilde y^{2}}{2} + 2 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{3 \sqrt{13}}{26}} \sqrt{\frac{3 \sqrt{13}}{26} + \frac{1}{2}} + \frac{9 \sqrt{13} \tilde y^{2}}{26} + 3 = 0$$
$$- \frac{\sqrt{13} \tilde x^{2}}{2} + \frac{\tilde x^{2}}{2} + \frac{\tilde y^{2}}{2} + \frac{\sqrt{13} \tilde y^{2}}{2} + 3 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{3 \frac{1}{- \frac{1}{2} + \frac{\sqrt{13}}{2}}} - \frac{\tilde y^{2}}{3 \frac{1}{\frac{1}{2} + \frac{\sqrt{13}}{2}}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{3 \sqrt{13}}{26} + \frac{1}{2}}, \ \sqrt{\frac{1}{2} - \frac{3 \sqrt{13}}{26}}\right)$$
$$\vec e_2 = \left( - \sqrt{\frac{1}{2} - \frac{3 \sqrt{13}}{26}}, \ \sqrt{\frac{3 \sqrt{13}}{26} + \frac{1}{2}}\right)$$
$$- x^{2} - 2 x y - 2 x + 2 y^{2} - 2 y + 2 = 0$$
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
$$a_{11} = -1$$
$$a_{12} = -1$$
$$a_{13} = -1$$
$$a_{22} = 2$$
$$a_{23} = -1$$
$$a_{33} = 2$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-1 & -1\\-1 & 2\end{matrix}\right|$$
$$\Delta = -3$$
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
$$- x_{0} - y_{0} - 1 = 0$$
$$- x_{0} + 2 y_{0} - 1 = 0$$
then
$$x_{0} = -1$$
$$y_{0} = 0$$
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
$$a'_{33} = - x_{0} - y_{0} + 2$$
$$a'_{33} = 3$$
then equation turns into
$$- x'^{2} - 2 x' y' + 2 y'^{2} + 3 = 0$$
Rotate the resulting coordinate system by an angle φ
$$x' = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y' = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
φ - determined from the formula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
substitute coefficients
$$\cot{\left(2 \phi \right)} = \frac{3}{2}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{3}{2} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{2 \sqrt{13}}{13}$$
$$\cos{\left(2 \phi \right)} = \frac{3 \sqrt{13}}{13}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{3 \sqrt{13}}{26} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{\frac{1}{2} - \frac{3 \sqrt{13}}{26}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{3 \sqrt{13}}{26} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{13}}{26}}$$
$$y' = \tilde x \sqrt{\frac{1}{2} - \frac{3 \sqrt{13}}{26}} + \tilde y \sqrt{\frac{3 \sqrt{13}}{26} + \frac{1}{2}}$$
then the equation turns from
$$- x'^{2} - 2 x' y' + 2 y'^{2} + 3 = 0$$
to
$$2 \left(\tilde x \sqrt{\frac{1}{2} - \frac{3 \sqrt{13}}{26}} + \tilde y \sqrt{\frac{3 \sqrt{13}}{26} + \frac{1}{2}}\right)^{2} - 2 \left(\tilde x \sqrt{\frac{1}{2} - \frac{3 \sqrt{13}}{26}} + \tilde y \sqrt{\frac{3 \sqrt{13}}{26} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{3 \sqrt{13}}{26} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{13}}{26}}\right) - \left(\tilde x \sqrt{\frac{3 \sqrt{13}}{26} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{13}}{26}}\right)^{2} + 3 = 0$$
simplify
$$- \frac{9 \sqrt{13} \tilde x^{2}}{26} - 2 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{3 \sqrt{13}}{26}} \sqrt{\frac{3 \sqrt{13}}{26} + \frac{1}{2}} + \frac{\tilde x^{2}}{2} - \frac{6 \sqrt{13} \tilde x \tilde y}{13} + 6 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{13}}{26}} \sqrt{\frac{3 \sqrt{13}}{26} + \frac{1}{2}} + \frac{\tilde y^{2}}{2} + 2 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{3 \sqrt{13}}{26}} \sqrt{\frac{3 \sqrt{13}}{26} + \frac{1}{2}} + \frac{9 \sqrt{13} \tilde y^{2}}{26} + 3 = 0$$
$$- \frac{\sqrt{13} \tilde x^{2}}{2} + \frac{\tilde x^{2}}{2} + \frac{\tilde y^{2}}{2} + \frac{\sqrt{13} \tilde y^{2}}{2} + 3 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{3 \frac{1}{- \frac{1}{2} + \frac{\sqrt{13}}{2}}} - \frac{\tilde y^{2}}{3 \frac{1}{\frac{1}{2} + \frac{\sqrt{13}}{2}}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(-1, 0)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{3 \sqrt{13}}{26} + \frac{1}{2}}, \ \sqrt{\frac{1}{2} - \frac{3 \sqrt{13}}{26}}\right)$$
$$\vec e_2 = \left( - \sqrt{\frac{1}{2} - \frac{3 \sqrt{13}}{26}}, \ \sqrt{\frac{3 \sqrt{13}}{26} + \frac{1}{2}}\right)$$
Invariants method
Given line equation of 2-order:
$$- x^{2} - 2 x y - 2 x + 2 y^{2} - 2 y + 2 = 0$$
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
$$a_{11} = -1$$
$$a_{12} = -1$$
$$a_{13} = -1$$
$$a_{22} = 2$$
$$a_{23} = -1$$
$$a_{33} = 2$$
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
$$I_{1} = 1$$
$$I_{3} = \left|\begin{matrix}-1 & -1 & -1\\-1 & 2 & -1\\-1 & -1 & 2\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 1 & -1\\-1 & 2 - \lambda\end{matrix}\right|$$
$$I_{1} = 1$$
$$I_{2} = -3$$
$$I_{3} = -9$$
$$I{\left(\lambda \right)} = \lambda^{2} - \lambda - 3$$
$$K_{2} = 0$$
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
$$\lambda^{2} - \lambda - 3 = 0$$
$$\lambda_{1} = \frac{1}{2} - \frac{\sqrt{13}}{2}$$
$$\lambda_{2} = \frac{1}{2} + \frac{\sqrt{13}}{2}$$
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
$$\tilde x^{2} \left(\frac{1}{2} - \frac{\sqrt{13}}{2}\right) + \tilde y^{2} \left(\frac{1}{2} + \frac{\sqrt{13}}{2}\right) + 3 = 0$$
$$\frac{\tilde x^{2}}{3 \frac{1}{- \frac{1}{2} + \frac{\sqrt{13}}{2}}} - \frac{\tilde y^{2}}{3 \frac{1}{\frac{1}{2} + \frac{\sqrt{13}}{2}}} = 1$$
- reduced to canonical form
$$- x^{2} - 2 x y - 2 x + 2 y^{2} - 2 y + 2 = 0$$
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
$$a_{11} = -1$$
$$a_{12} = -1$$
$$a_{13} = -1$$
$$a_{22} = 2$$
$$a_{23} = -1$$
$$a_{33} = 2$$
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
$$I_{1} = 1$$
|-1 -1|
I2 = | |
|-1 2 |$$I_{3} = \left|\begin{matrix}-1 & -1 & -1\\-1 & 2 & -1\\-1 & -1 & 2\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 1 & -1\\-1 & 2 - \lambda\end{matrix}\right|$$
|-1 -1| |2 -1|
K2 = | | + | |
|-1 2 | |-1 2 |$$I_{1} = 1$$
$$I_{2} = -3$$
$$I_{3} = -9$$
$$I{\left(\lambda \right)} = \lambda^{2} - \lambda - 3$$
$$K_{2} = 0$$
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
$$\lambda^{2} - \lambda - 3 = 0$$
$$\lambda_{1} = \frac{1}{2} - \frac{\sqrt{13}}{2}$$
$$\lambda_{2} = \frac{1}{2} + \frac{\sqrt{13}}{2}$$
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
$$\tilde x^{2} \left(\frac{1}{2} - \frac{\sqrt{13}}{2}\right) + \tilde y^{2} \left(\frac{1}{2} + \frac{\sqrt{13}}{2}\right) + 3 = 0$$
$$\frac{\tilde x^{2}}{3 \frac{1}{- \frac{1}{2} + \frac{\sqrt{13}}{2}}} - \frac{\tilde y^{2}}{3 \frac{1}{\frac{1}{2} + \frac{\sqrt{13}}{2}}} = 1$$
- reduced to canonical form