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:
- 8x^2-12xy+17y^2+16x-12y+3=0
- 3x^2+0xy-6y^2-12x-108y-492=0
- sqrt(16-y^2)=sqrt(16-a^2*x^2)
- x^2+5y^2+z^2-2xy+6xz-2yz-2x-6y+2z=0
- Plot:
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xy=4
-
xy=6
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4x^2+4y^2=25
- xyz
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Identical expressions
- 8x^ two -1 two xy+17y^2+16x-12y+ three = zero
- 8x squared minus 12xy plus 17y squared plus 16x minus 12y plus 3 equally 0
- 8x to the power of two minus 1 two xy plus 17y squared plus 16x minus 12y plus three equally zero
- 8x2-12xy+17y2+16x-12y+3=0
- 8x²-12xy+17y²+16x-12y+3=0
- 8x to the power of 2-12xy+17y to the power of 2+16x-12y+3=0
- 8x^2-12xy+17y^2+16x-12y+3=O
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Similar expressions
- 8x^2+12xy+17y^2+16x-12y+3=0
- 8x^2-12xy+17y^2+16x+12y+3=0
- 8x^2-12xy-17y^2+16x-12y+3=0
- 8x^2-12xy+17y^2+16x-12y-3=0
- 8x^2-12xy+17y^2-16x-12y+3=0
8x^2-12xy+17y^2+16x-12y+3=0 canonical form
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The solution
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2 2 3 - 12*y + 8*x + 16*x + 17*y - 12*x*y = 0
$$8 x^{2} - 12 x y + 16 x + 17 y^{2} - 12 y + 3 = 0$$
8*x^2 - 12*x*y + 16*x + 17*y^2 - 12*y + 3 = 0
Detail solution
Given line equation of 2-order:
$$8 x^{2} - 12 x y + 16 x + 17 y^{2} - 12 y + 3 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = -6$$
$$a_{13} = 8$$
$$a_{22} = 17$$
$$a_{23} = -6$$
$$a_{33} = 3$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}8 & -6\\-6 & 17\end{matrix}\right|$$
$$\Delta = 100$$
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
$$8 x_{0} - 6 y_{0} + 8 = 0$$
$$- 6 x_{0} + 17 y_{0} - 6 = 0$$
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
$$a'_{33} = 8 x_{0} - 6 y_{0} + 3$$
$$a'_{33} = -5$$
then equation turns into
$$8 x'^{2} - 12 x' y' + 17 y'^{2} - 5 = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{3}{4}$$
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
$$8 x'^{2} - 12 x' y' + 17 y'^{2} - 5 = 0$$
to
$$17 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right)^{2} - 12 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right) + 8 \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right)^{2} - 5 = 0$$
simplify
$$5 \tilde x^{2} + 20 \tilde y^{2} - 5 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{5} \sqrt{5}}{\frac{1}{5} \sqrt{5}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{10} \sqrt{5}}{\frac{1}{5} \sqrt{5}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{2 \sqrt{5}}{5}, \ \frac{\sqrt{5}}{5}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{5}}{5}, \ \frac{2 \sqrt{5}}{5}\right)$$
$$8 x^{2} - 12 x y + 16 x + 17 y^{2} - 12 y + 3 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = -6$$
$$a_{13} = 8$$
$$a_{22} = 17$$
$$a_{23} = -6$$
$$a_{33} = 3$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}8 & -6\\-6 & 17\end{matrix}\right|$$
$$\Delta = 100$$
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
$$8 x_{0} - 6 y_{0} + 8 = 0$$
$$- 6 x_{0} + 17 y_{0} - 6 = 0$$
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
$$a'_{33} = 8 x_{0} - 6 y_{0} + 3$$
$$a'_{33} = -5$$
then equation turns into
$$8 x'^{2} - 12 x' y' + 17 y'^{2} - 5 = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{3}{4}$$
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
$$8 x'^{2} - 12 x' y' + 17 y'^{2} - 5 = 0$$
to
$$17 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right)^{2} - 12 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right) + 8 \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right)^{2} - 5 = 0$$
simplify
$$5 \tilde x^{2} + 20 \tilde y^{2} - 5 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{5} \sqrt{5}}{\frac{1}{5} \sqrt{5}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{10} \sqrt{5}}{\frac{1}{5} \sqrt{5}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(-1, 0)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{2 \sqrt{5}}{5}, \ \frac{\sqrt{5}}{5}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{5}}{5}, \ \frac{2 \sqrt{5}}{5}\right)$$
Invariants method
Given line equation of 2-order:
$$8 x^{2} - 12 x y + 16 x + 17 y^{2} - 12 y + 3 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = -6$$
$$a_{13} = 8$$
$$a_{22} = 17$$
$$a_{23} = -6$$
$$a_{33} = 3$$
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
$$I_{1} = 25$$
$$I_{3} = \left|\begin{matrix}8 & -6 & 8\\-6 & 17 & -6\\8 & -6 & 3\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}8 - \lambda & -6\\-6 & 17 - \lambda\end{matrix}\right|$$
$$I_{1} = 25$$
$$I_{2} = 100$$
$$I_{3} = -500$$
$$I{\left(\lambda \right)} = \lambda^{2} - 25 \lambda + 100$$
$$K_{2} = -25$$
Because
$$I_{2} > 0 \wedge I_{1} I_{3} < 0$$
then by line type:
this equation is of type : ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 25 \lambda + 100 = 0$$
$$\lambda_{1} = 20$$
$$\lambda_{2} = 5$$
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
$$20 \tilde x^{2} + 5 \tilde y^{2} - 5 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{10} \sqrt{5}}{\frac{1}{5} \sqrt{5}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{5} \sqrt{5}}{\frac{1}{5} \sqrt{5}}\right)^{2}} = 1$$
- reduced to canonical form
$$8 x^{2} - 12 x y + 16 x + 17 y^{2} - 12 y + 3 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = -6$$
$$a_{13} = 8$$
$$a_{22} = 17$$
$$a_{23} = -6$$
$$a_{33} = 3$$
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
$$I_{1} = 25$$
|8 -6|
I2 = | |
|-6 17|$$I_{3} = \left|\begin{matrix}8 & -6 & 8\\-6 & 17 & -6\\8 & -6 & 3\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}8 - \lambda & -6\\-6 & 17 - \lambda\end{matrix}\right|$$
|8 8| |17 -6|
K2 = | | + | |
|8 3| |-6 3 |$$I_{1} = 25$$
$$I_{2} = 100$$
$$I_{3} = -500$$
$$I{\left(\lambda \right)} = \lambda^{2} - 25 \lambda + 100$$
$$K_{2} = -25$$
Because
$$I_{2} > 0 \wedge I_{1} I_{3} < 0$$
then by line type:
this equation is of type : ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 25 \lambda + 100 = 0$$
$$\lambda_{1} = 20$$
$$\lambda_{2} = 5$$
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
$$20 \tilde x^{2} + 5 \tilde y^{2} - 5 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{10} \sqrt{5}}{\frac{1}{5} \sqrt{5}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{5} \sqrt{5}}{\frac{1}{5} \sqrt{5}}\right)^{2}} = 1$$
- reduced to canonical form