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- 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:
- 4x^2+24xy+11y^2+64x+42y+51=0
-
9x^2-16y^2+18x+64y-199=0
- x^2+4x
- 17^2+12xy+8y^2-46x-28y+17=0
- Plot:
-
-1-(36*(x^2-y^2+5*x+6)-(36*x+132)*(2*x+5))/((x^2-y^2+5*x+6)^2+4*x^2*y+25*y^2+20*x*y^2)=0
-
x^2+(y-cbrt(x^2))^2=1
-
x^2+x*y^2=1
-
sqrt(x)+sqrt(y)=1
-
Identical expressions
- two x^ two +7y^2-28xy+2x+7y- one = zero
- 2x squared plus 7y squared minus 28xy plus 2x plus 7y minus 1 equally 0
- two x to the power of two plus 7y squared minus 28xy plus 2x plus 7y minus one equally zero
- 2x2+7y2-28xy+2x+7y-1=0
- 2x²+7y²-28xy+2x+7y-1=0
- 2x to the power of 2+7y to the power of 2-28xy+2x+7y-1=0
- 2x^2+7y^2-28xy+2x+7y-1=O
-
Similar expressions
- 2x^2+7y^2-28xy+2x+7y+1=0
- 2x^2+7y^2-28xy-2x+7y-1=0
- 2x^2+7y^2+28xy+2x+7y-1=0
- 2x^2-7y^2-28xy+2x+7y-1=0
- 2x^2+7y^2-28xy+2x-7y-1=0
2x^2+7y^2-28xy+2x+7y-1=0 canonical form
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The solution
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2 2 -1 + 2*x + 2*x + 7*y + 7*y - 28*x*y = 0
$$2 x^{2} - 28 x y + 2 x + 7 y^{2} + 7 y - 1 = 0$$
2*x^2 - 28*x*y + 2*x + 7*y^2 + 7*y - 1 = 0
Detail solution
Given line equation of 2-order:
$$2 x^{2} - 28 x y + 2 x + 7 y^{2} + 7 y - 1 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = -14$$
$$a_{13} = 1$$
$$a_{22} = 7$$
$$a_{23} = \frac{7}{2}$$
$$a_{33} = -1$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}2 & -14\\-14 & 7\end{matrix}\right|$$
$$\Delta = -182$$
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
$$2 x_{0} - 14 y_{0} + 1 = 0$$
$$- 14 x_{0} + 7 y_{0} + \frac{7}{2} = 0$$
then
$$x_{0} = \frac{4}{13}$$
$$y_{0} = \frac{3}{26}$$
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
$$a'_{33} = x_{0} + \frac{7 y_{0}}{2} - 1$$
$$a'_{33} = - \frac{15}{52}$$
then equation turns into
$$2 x'^{2} - 28 x' y' + 7 y'^{2} - \frac{15}{52} = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{5}{28}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{5}{28} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{28 \sqrt{809}}{809}$$
$$\cos{\left(2 \phi \right)} = \frac{5 \sqrt{809}}{809}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{5 \sqrt{809}}{1618} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{\frac{1}{2} - \frac{5 \sqrt{809}}{1618}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{5 \sqrt{809}}{1618} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{5 \sqrt{809}}{1618}}$$
$$y' = \tilde x \sqrt{\frac{1}{2} - \frac{5 \sqrt{809}}{1618}} + \tilde y \sqrt{\frac{5 \sqrt{809}}{1618} + \frac{1}{2}}$$
then the equation turns from
$$2 x'^{2} - 28 x' y' + 7 y'^{2} - \frac{15}{52} = 0$$
to
$$7 \left(\tilde x \sqrt{\frac{1}{2} - \frac{5 \sqrt{809}}{1618}} + \tilde y \sqrt{\frac{5 \sqrt{809}}{1618} + \frac{1}{2}}\right)^{2} - 28 \left(\tilde x \sqrt{\frac{1}{2} - \frac{5 \sqrt{809}}{1618}} + \tilde y \sqrt{\frac{5 \sqrt{809}}{1618} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{5 \sqrt{809}}{1618} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{5 \sqrt{809}}{1618}}\right) + 2 \left(\tilde x \sqrt{\frac{5 \sqrt{809}}{1618} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{5 \sqrt{809}}{1618}}\right)^{2} - \frac{15}{52} = 0$$
simplify
$$- 28 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{5 \sqrt{809}}{1618}} \sqrt{\frac{5 \sqrt{809}}{1618} + \frac{1}{2}} - \frac{25 \sqrt{809} \tilde x^{2}}{1618} + \frac{9 \tilde x^{2}}{2} - \frac{140 \sqrt{809} \tilde x \tilde y}{809} + 10 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{5 \sqrt{809}}{1618}} \sqrt{\frac{5 \sqrt{809}}{1618} + \frac{1}{2}} + \frac{25 \sqrt{809} \tilde y^{2}}{1618} + \frac{9 \tilde y^{2}}{2} + 28 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{5 \sqrt{809}}{1618}} \sqrt{\frac{5 \sqrt{809}}{1618} + \frac{1}{2}} - \frac{15}{52} = 0$$
$$- \frac{\sqrt{809} \tilde x^{2}}{2} + \frac{9 \tilde x^{2}}{2} + \frac{9 \tilde y^{2}}{2} + \frac{\sqrt{809} \tilde y^{2}}{2} - \frac{15}{52} = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{\frac{15}{52} \frac{1}{- \frac{9}{2} + \frac{\sqrt{809}}{2}}} - \frac{\tilde y^{2}}{\frac{15}{52} \frac{1}{\frac{9}{2} + \frac{\sqrt{809}}{2}}} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{5 \sqrt{809}}{1618} + \frac{1}{2}}, \ \sqrt{\frac{1}{2} - \frac{5 \sqrt{809}}{1618}}\right)$$
$$\vec e_2 = \left( - \sqrt{\frac{1}{2} - \frac{5 \sqrt{809}}{1618}}, \ \sqrt{\frac{5 \sqrt{809}}{1618} + \frac{1}{2}}\right)$$
$$2 x^{2} - 28 x y + 2 x + 7 y^{2} + 7 y - 1 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = -14$$
$$a_{13} = 1$$
$$a_{22} = 7$$
$$a_{23} = \frac{7}{2}$$
$$a_{33} = -1$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}2 & -14\\-14 & 7\end{matrix}\right|$$
$$\Delta = -182$$
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
$$2 x_{0} - 14 y_{0} + 1 = 0$$
$$- 14 x_{0} + 7 y_{0} + \frac{7}{2} = 0$$
then
$$x_{0} = \frac{4}{13}$$
$$y_{0} = \frac{3}{26}$$
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
$$a'_{33} = x_{0} + \frac{7 y_{0}}{2} - 1$$
$$a'_{33} = - \frac{15}{52}$$
then equation turns into
$$2 x'^{2} - 28 x' y' + 7 y'^{2} - \frac{15}{52} = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{5}{28}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{5}{28} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{28 \sqrt{809}}{809}$$
$$\cos{\left(2 \phi \right)} = \frac{5 \sqrt{809}}{809}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{5 \sqrt{809}}{1618} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{\frac{1}{2} - \frac{5 \sqrt{809}}{1618}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{5 \sqrt{809}}{1618} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{5 \sqrt{809}}{1618}}$$
$$y' = \tilde x \sqrt{\frac{1}{2} - \frac{5 \sqrt{809}}{1618}} + \tilde y \sqrt{\frac{5 \sqrt{809}}{1618} + \frac{1}{2}}$$
then the equation turns from
$$2 x'^{2} - 28 x' y' + 7 y'^{2} - \frac{15}{52} = 0$$
to
$$7 \left(\tilde x \sqrt{\frac{1}{2} - \frac{5 \sqrt{809}}{1618}} + \tilde y \sqrt{\frac{5 \sqrt{809}}{1618} + \frac{1}{2}}\right)^{2} - 28 \left(\tilde x \sqrt{\frac{1}{2} - \frac{5 \sqrt{809}}{1618}} + \tilde y \sqrt{\frac{5 \sqrt{809}}{1618} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{5 \sqrt{809}}{1618} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{5 \sqrt{809}}{1618}}\right) + 2 \left(\tilde x \sqrt{\frac{5 \sqrt{809}}{1618} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{5 \sqrt{809}}{1618}}\right)^{2} - \frac{15}{52} = 0$$
simplify
$$- 28 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{5 \sqrt{809}}{1618}} \sqrt{\frac{5 \sqrt{809}}{1618} + \frac{1}{2}} - \frac{25 \sqrt{809} \tilde x^{2}}{1618} + \frac{9 \tilde x^{2}}{2} - \frac{140 \sqrt{809} \tilde x \tilde y}{809} + 10 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{5 \sqrt{809}}{1618}} \sqrt{\frac{5 \sqrt{809}}{1618} + \frac{1}{2}} + \frac{25 \sqrt{809} \tilde y^{2}}{1618} + \frac{9 \tilde y^{2}}{2} + 28 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{5 \sqrt{809}}{1618}} \sqrt{\frac{5 \sqrt{809}}{1618} + \frac{1}{2}} - \frac{15}{52} = 0$$
$$- \frac{\sqrt{809} \tilde x^{2}}{2} + \frac{9 \tilde x^{2}}{2} + \frac{9 \tilde y^{2}}{2} + \frac{\sqrt{809} \tilde y^{2}}{2} - \frac{15}{52} = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{\frac{15}{52} \frac{1}{- \frac{9}{2} + \frac{\sqrt{809}}{2}}} - \frac{\tilde y^{2}}{\frac{15}{52} \frac{1}{\frac{9}{2} + \frac{\sqrt{809}}{2}}} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(4/13, 3/26)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{5 \sqrt{809}}{1618} + \frac{1}{2}}, \ \sqrt{\frac{1}{2} - \frac{5 \sqrt{809}}{1618}}\right)$$
$$\vec e_2 = \left( - \sqrt{\frac{1}{2} - \frac{5 \sqrt{809}}{1618}}, \ \sqrt{\frac{5 \sqrt{809}}{1618} + \frac{1}{2}}\right)$$
Invariants method
Given line equation of 2-order:
$$2 x^{2} - 28 x y + 2 x + 7 y^{2} + 7 y - 1 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = -14$$
$$a_{13} = 1$$
$$a_{22} = 7$$
$$a_{23} = \frac{7}{2}$$
$$a_{33} = -1$$
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
$$I_{1} = 9$$
$$I_{3} = \left|\begin{matrix}2 & -14 & 1\\-14 & 7 & \frac{7}{2}\\1 & \frac{7}{2} & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & -14\\-14 & 7 - \lambda\end{matrix}\right|$$
$$I_{1} = 9$$
$$I_{2} = -182$$
$$I_{3} = \frac{105}{2}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 9 \lambda - 182$$
$$K_{2} = - \frac{89}{4}$$
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
$$\lambda^{2} - 9 \lambda - 182 = 0$$
$$\lambda_{1} = \frac{9}{2} - \frac{\sqrt{809}}{2}$$
$$\lambda_{2} = \frac{9}{2} + \frac{\sqrt{809}}{2}$$
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
$$\tilde x^{2} \left(\frac{9}{2} - \frac{\sqrt{809}}{2}\right) + \tilde y^{2} \left(\frac{9}{2} + \frac{\sqrt{809}}{2}\right) - \frac{15}{52} = 0$$
$$\frac{\tilde x^{2}}{\frac{15}{52} \frac{1}{- \frac{9}{2} + \frac{\sqrt{809}}{2}}} - \frac{\tilde y^{2}}{\frac{15}{52} \frac{1}{\frac{9}{2} + \frac{\sqrt{809}}{2}}} = -1$$
- reduced to canonical form
$$2 x^{2} - 28 x y + 2 x + 7 y^{2} + 7 y - 1 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = -14$$
$$a_{13} = 1$$
$$a_{22} = 7$$
$$a_{23} = \frac{7}{2}$$
$$a_{33} = -1$$
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
$$I_{1} = 9$$
| 2 -14|
I2 = | |
|-14 7 |$$I_{3} = \left|\begin{matrix}2 & -14 & 1\\-14 & 7 & \frac{7}{2}\\1 & \frac{7}{2} & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & -14\\-14 & 7 - \lambda\end{matrix}\right|$$
|2 1 | | 7 7/2|
K2 = | | + | |
|1 -1| |7/2 -1 |$$I_{1} = 9$$
$$I_{2} = -182$$
$$I_{3} = \frac{105}{2}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 9 \lambda - 182$$
$$K_{2} = - \frac{89}{4}$$
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
$$\lambda^{2} - 9 \lambda - 182 = 0$$
$$\lambda_{1} = \frac{9}{2} - \frac{\sqrt{809}}{2}$$
$$\lambda_{2} = \frac{9}{2} + \frac{\sqrt{809}}{2}$$
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
$$\tilde x^{2} \left(\frac{9}{2} - \frac{\sqrt{809}}{2}\right) + \tilde y^{2} \left(\frac{9}{2} + \frac{\sqrt{809}}{2}\right) - \frac{15}{52} = 0$$
$$\frac{\tilde x^{2}}{\frac{15}{52} \frac{1}{- \frac{9}{2} + \frac{\sqrt{809}}{2}}} - \frac{\tilde y^{2}}{\frac{15}{52} \frac{1}{\frac{9}{2} + \frac{\sqrt{809}}{2}}} = -1$$
- reduced to canonical form