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:
- x=-3-2/3*sqrt(8+2y-y^2)
- 11x^2-20xy-4y^2-8y+1=0
- 4x^2-4xy+y^2-14x-8y-2=0
- 2x^2+2(sqrt8)xy-5y^2
- Plot:
-
x^2/3+y^2/4=1
-
x^2/2+y^2/8=1
- 4*(x^2)+9*(y^2)=36
- 2*x^3-9*x^2-3=y
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Identical expressions
- 4x^ two -4xy+y^ two -14x-8y- two = zero
- 4x squared minus 4xy plus y squared minus 14x minus 8y minus 2 equally 0
- 4x to the power of two minus 4xy plus y to the power of two minus 14x minus 8y minus two equally zero
- 4x2-4xy+y2-14x-8y-2=0
- 4x²-4xy+y²-14x-8y-2=0
- 4x to the power of 2-4xy+y to the power of 2-14x-8y-2=0
- 4x^2-4xy+y^2-14x-8y-2=O
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Similar expressions
- 4x^2-4xy+y^2-14x+8y-2=0
- 4x^2-4xy+y^2-14x-8y+2=0
- 4x^2-4xy-y^2-14x-8y-2=0
- 4x^2-4xy+y^2+14x-8y-2=0
- 4x^2+4xy+y^2-14x-8y-2=0
4x^2-4xy+y^2-14x-8y-2=0 canonical form
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The solution
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2 2 -2 + y - 14*x - 8*y + 4*x - 4*x*y = 0
$$4 x^{2} - 4 x y - 14 x + y^{2} - 8 y - 2 = 0$$
4*x^2 - 4*x*y - 14*x + y^2 - 8*y - 2 = 0
Detail solution
Given line equation of 2-order:
$$4 x^{2} - 4 x y - 14 x + y^{2} - 8 y - 2 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = -2$$
$$a_{13} = -7$$
$$a_{22} = 1$$
$$a_{23} = -4$$
$$a_{33} = -2$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & -2\\-2 & 1\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
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
$$\cot{\left(2 \phi \right)} = - \frac{3}{4}$$
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
$$x' = \frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}$$
$$y' = - \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}$$
then the equation turns from
$$4 x'^{2} - 4 x' y' - 14 x' + y'^{2} - 8 y' - 2 = 0$$
to
$$\left(- \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right)^{2} - 4 \left(- \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) \left(\frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}\right) - 8 \left(- \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) + 4 \left(\frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}\right)^{2} - 14 \left(\frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}\right) - 2 = 0$$
simplify
$$5 \tilde x^{2} - 4 \sqrt{5} \tilde x - 6 \sqrt{5} \tilde y - 2 = 0$$
$$- 5 \tilde x^{2} + 4 \sqrt{5} \tilde x + 6 \sqrt{5} \tilde y + 2 = 0$$
$$\left(\sqrt{5} \tilde x + 2\right)^{2} = 6 \sqrt{5} \tilde y + 6$$
$$\left(\tilde x + \frac{2 \sqrt{5}}{5}\right)^{2} = \frac{6 \sqrt{5} \left(\tilde y + \frac{\sqrt{5}}{5}\right)}{5}$$
$$\tilde x'^{2} = \frac{6 \sqrt{5} \tilde y'}{5}$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \frac{2 \sqrt{5}}{5} + 0 \left(- \frac{\sqrt{5}}{5}\right)$$
$$y_{0} = 0 \left(- \frac{\sqrt{5}}{5}\right) + 0 \frac{2 \sqrt{5}}{5}$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{2 \sqrt{5}}{5}, \ - \frac{\sqrt{5}}{5}\right)$$
$$\vec e_2 = \left( \frac{\sqrt{5}}{5}, \ \frac{2 \sqrt{5}}{5}\right)$$
$$4 x^{2} - 4 x y - 14 x + y^{2} - 8 y - 2 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = -2$$
$$a_{13} = -7$$
$$a_{22} = 1$$
$$a_{23} = -4$$
$$a_{33} = -2$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & -2\\-2 & 1\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
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
$$\cot{\left(2 \phi \right)} = - \frac{3}{4}$$
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
$$x' = \frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}$$
$$y' = - \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}$$
then the equation turns from
$$4 x'^{2} - 4 x' y' - 14 x' + y'^{2} - 8 y' - 2 = 0$$
to
$$\left(- \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right)^{2} - 4 \left(- \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) \left(\frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}\right) - 8 \left(- \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) + 4 \left(\frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}\right)^{2} - 14 \left(\frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}\right) - 2 = 0$$
simplify
$$5 \tilde x^{2} - 4 \sqrt{5} \tilde x - 6 \sqrt{5} \tilde y - 2 = 0$$
$$- 5 \tilde x^{2} + 4 \sqrt{5} \tilde x + 6 \sqrt{5} \tilde y + 2 = 0$$
$$\left(\sqrt{5} \tilde x + 2\right)^{2} = 6 \sqrt{5} \tilde y + 6$$
$$\left(\tilde x + \frac{2 \sqrt{5}}{5}\right)^{2} = \frac{6 \sqrt{5} \left(\tilde y + \frac{\sqrt{5}}{5}\right)}{5}$$
$$\tilde x'^{2} = \frac{6 \sqrt{5} \tilde y'}{5}$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \frac{2 \sqrt{5}}{5} + 0 \left(- \frac{\sqrt{5}}{5}\right)$$
$$y_{0} = 0 \left(- \frac{\sqrt{5}}{5}\right) + 0 \frac{2 \sqrt{5}}{5}$$
$$x_{0} = 0$$
$$y_{0} = 0$$
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:
$$4 x^{2} - 4 x y - 14 x + y^{2} - 8 y - 2 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = -2$$
$$a_{13} = -7$$
$$a_{22} = 1$$
$$a_{23} = -4$$
$$a_{33} = -2$$
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
$$I_{1} = 5$$
$$I_{3} = \left|\begin{matrix}4 & -2 & -7\\-2 & 1 & -4\\-7 & -4 & -2\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & -2\\-2 & 1 - \lambda\end{matrix}\right|$$
$$I_{1} = 5$$
$$I_{2} = 0$$
$$I_{3} = -225$$
$$I{\left(\lambda \right)} = \lambda^{2} - 5 \lambda$$
$$K_{2} = -75$$
Because
$$I_{2} = 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : parabola
$$I_{1} \tilde y^{2} + 2 \tilde x \sqrt{- \frac{I_{3}}{I_{1}}} = 0$$
or
$$6 \sqrt{5} \tilde x + 5 \tilde y^{2} = 0$$
$$\tilde y^{2} = \frac{6 \sqrt{5}}{5} \tilde x$$
- reduced to canonical form
$$4 x^{2} - 4 x y - 14 x + y^{2} - 8 y - 2 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = -2$$
$$a_{13} = -7$$
$$a_{22} = 1$$
$$a_{23} = -4$$
$$a_{33} = -2$$
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
$$I_{1} = 5$$
|4 -2|
I2 = | |
|-2 1 |$$I_{3} = \left|\begin{matrix}4 & -2 & -7\\-2 & 1 & -4\\-7 & -4 & -2\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & -2\\-2 & 1 - \lambda\end{matrix}\right|$$
|4 -7| |1 -4|
K2 = | | + | |
|-7 -2| |-4 -2|$$I_{1} = 5$$
$$I_{2} = 0$$
$$I_{3} = -225$$
$$I{\left(\lambda \right)} = \lambda^{2} - 5 \lambda$$
$$K_{2} = -75$$
Because
$$I_{2} = 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : parabola
$$I_{1} \tilde y^{2} + 2 \tilde x \sqrt{- \frac{I_{3}}{I_{1}}} = 0$$
or
$$6 \sqrt{5} \tilde x + 5 \tilde y^{2} = 0$$
$$\tilde y^{2} = \frac{6 \sqrt{5}}{5} \tilde x$$
- reduced to canonical form