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^2+y^2+z^2=2x+2y-1
- (-x+1)^2+(3-y)^2/2^2=1
- 852
- 3x^2-8xy+6y^2-8x+4y-3=0
- Plot:
-
y^2=x(1-x^2)^3
- y^2-(8/5)*x+x^2=0
-
y^2=x*y-2
- y=x3+4*x^2+4*x
-
Identical expressions
- three x^ two -8xy+6y^ two -8x+4y-3= zero
- 3x squared minus 8xy plus 6y squared minus 8x plus 4y minus 3 equally 0
- three x to the power of two minus 8xy plus 6y to the power of two minus 8x plus 4y minus 3 equally zero
- 3x2-8xy+6y2-8x+4y-3=0
- 3x²-8xy+6y²-8x+4y-3=0
- 3x to the power of 2-8xy+6y to the power of 2-8x+4y-3=0
- 3x^2-8xy+6y^2-8x+4y-3=O
-
Similar expressions
- 3x^2+8xy+6y^2-8x+4y-3=0
- 3x^2-8xy-6y^2-8x+4y-3=0
- 3x^2-8xy+6y^2+8x+4y-3=0
- 3x^2-8xy+6y^2-8x-4y-3=0
- 3x^2-8xy+6y^2-8x+4y+3=0
3x^2-8xy+6y^2-8x+4y-3=0 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
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2 2 -3 - 8*x + 3*x + 4*y + 6*y - 8*x*y = 0
$$3 x^{2} - 8 x y - 8 x + 6 y^{2} + 4 y - 3 = 0$$
3*x^2 - 8*x*y - 8*x + 6*y^2 + 4*y - 3 = 0
Detail solution
Given line equation of 2-order:
$$3 x^{2} - 8 x y - 8 x + 6 y^{2} + 4 y - 3 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = -4$$
$$a_{13} = -4$$
$$a_{22} = 6$$
$$a_{23} = 2$$
$$a_{33} = -3$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}3 & -4\\-4 & 6\end{matrix}\right|$$
$$\Delta = 2$$
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
$$3 x_{0} - 4 y_{0} - 4 = 0$$
$$- 4 x_{0} + 6 y_{0} + 2 = 0$$
then
$$x_{0} = 8$$
$$y_{0} = 5$$
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
$$a'_{33} = - 4 x_{0} + 2 y_{0} - 3$$
$$a'_{33} = -25$$
then equation turns into
$$3 x'^{2} - 8 x' y' + 6 y'^{2} - 25 = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{3}{8}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{3}{8} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{8 \sqrt{73}}{73}$$
$$\cos{\left(2 \phi \right)} = \frac{3 \sqrt{73}}{73}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{3 \sqrt{73}}{146} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{\frac{1}{2} - \frac{3 \sqrt{73}}{146}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{3 \sqrt{73}}{146} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{73}}{146}}$$
$$y' = \tilde x \sqrt{\frac{1}{2} - \frac{3 \sqrt{73}}{146}} + \tilde y \sqrt{\frac{3 \sqrt{73}}{146} + \frac{1}{2}}$$
then the equation turns from
$$3 x'^{2} - 8 x' y' + 6 y'^{2} - 25 = 0$$
to
$$6 \left(\tilde x \sqrt{\frac{1}{2} - \frac{3 \sqrt{73}}{146}} + \tilde y \sqrt{\frac{3 \sqrt{73}}{146} + \frac{1}{2}}\right)^{2} - 8 \left(\tilde x \sqrt{\frac{1}{2} - \frac{3 \sqrt{73}}{146}} + \tilde y \sqrt{\frac{3 \sqrt{73}}{146} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{3 \sqrt{73}}{146} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{73}}{146}}\right) + 3 \left(\tilde x \sqrt{\frac{3 \sqrt{73}}{146} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{73}}{146}}\right)^{2} - 25 = 0$$
simplify
$$- 8 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{3 \sqrt{73}}{146}} \sqrt{\frac{3 \sqrt{73}}{146} + \frac{1}{2}} - \frac{9 \sqrt{73} \tilde x^{2}}{146} + \frac{9 \tilde x^{2}}{2} - \frac{24 \sqrt{73} \tilde x \tilde y}{73} + 6 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{73}}{146}} \sqrt{\frac{3 \sqrt{73}}{146} + \frac{1}{2}} + \frac{9 \sqrt{73} \tilde y^{2}}{146} + 8 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{3 \sqrt{73}}{146}} \sqrt{\frac{3 \sqrt{73}}{146} + \frac{1}{2}} + \frac{9 \tilde y^{2}}{2} - 25 = 0$$
$$- \frac{\sqrt{73} \tilde x^{2}}{2} + \frac{9 \tilde x^{2}}{2} + \frac{\sqrt{73} \tilde y^{2}}{2} + \frac{9 \tilde y^{2}}{2} - 25 = 0$$
Given equation is ellipse
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{3 \sqrt{73}}{146} + \frac{1}{2}}, \ \sqrt{\frac{1}{2} - \frac{3 \sqrt{73}}{146}}\right)$$
$$\vec e_2 = \left( - \sqrt{\frac{1}{2} - \frac{3 \sqrt{73}}{146}}, \ \sqrt{\frac{3 \sqrt{73}}{146} + \frac{1}{2}}\right)$$
$$3 x^{2} - 8 x y - 8 x + 6 y^{2} + 4 y - 3 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = -4$$
$$a_{13} = -4$$
$$a_{22} = 6$$
$$a_{23} = 2$$
$$a_{33} = -3$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}3 & -4\\-4 & 6\end{matrix}\right|$$
$$\Delta = 2$$
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
$$3 x_{0} - 4 y_{0} - 4 = 0$$
$$- 4 x_{0} + 6 y_{0} + 2 = 0$$
then
$$x_{0} = 8$$
$$y_{0} = 5$$
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
$$a'_{33} = - 4 x_{0} + 2 y_{0} - 3$$
$$a'_{33} = -25$$
then equation turns into
$$3 x'^{2} - 8 x' y' + 6 y'^{2} - 25 = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{3}{8}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{3}{8} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{8 \sqrt{73}}{73}$$
$$\cos{\left(2 \phi \right)} = \frac{3 \sqrt{73}}{73}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{3 \sqrt{73}}{146} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{\frac{1}{2} - \frac{3 \sqrt{73}}{146}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{3 \sqrt{73}}{146} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{73}}{146}}$$
$$y' = \tilde x \sqrt{\frac{1}{2} - \frac{3 \sqrt{73}}{146}} + \tilde y \sqrt{\frac{3 \sqrt{73}}{146} + \frac{1}{2}}$$
then the equation turns from
$$3 x'^{2} - 8 x' y' + 6 y'^{2} - 25 = 0$$
to
$$6 \left(\tilde x \sqrt{\frac{1}{2} - \frac{3 \sqrt{73}}{146}} + \tilde y \sqrt{\frac{3 \sqrt{73}}{146} + \frac{1}{2}}\right)^{2} - 8 \left(\tilde x \sqrt{\frac{1}{2} - \frac{3 \sqrt{73}}{146}} + \tilde y \sqrt{\frac{3 \sqrt{73}}{146} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{3 \sqrt{73}}{146} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{73}}{146}}\right) + 3 \left(\tilde x \sqrt{\frac{3 \sqrt{73}}{146} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{73}}{146}}\right)^{2} - 25 = 0$$
simplify
$$- 8 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{3 \sqrt{73}}{146}} \sqrt{\frac{3 \sqrt{73}}{146} + \frac{1}{2}} - \frac{9 \sqrt{73} \tilde x^{2}}{146} + \frac{9 \tilde x^{2}}{2} - \frac{24 \sqrt{73} \tilde x \tilde y}{73} + 6 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{73}}{146}} \sqrt{\frac{3 \sqrt{73}}{146} + \frac{1}{2}} + \frac{9 \sqrt{73} \tilde y^{2}}{146} + 8 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{3 \sqrt{73}}{146}} \sqrt{\frac{3 \sqrt{73}}{146} + \frac{1}{2}} + \frac{9 \tilde y^{2}}{2} - 25 = 0$$
$$- \frac{\sqrt{73} \tilde x^{2}}{2} + \frac{9 \tilde x^{2}}{2} + \frac{\sqrt{73} \tilde y^{2}}{2} + \frac{9 \tilde y^{2}}{2} - 25 = 0$$
Given equation is ellipse
2 2
\tilde x \tilde y
------------------------ + ------------------------ = 1
2 2
/ 1 \ / 1 \
|---------------------| |---------------------|
| ____________ | | ____________ |
| / ____ | | / ____ |
| / 9 \/ 73 | | / 9 \/ 73 |
| / - - ------ *1/5| | / - + ------ *1/5|
\\/ 2 2 / \\/ 2 2 / - reduced to canonical form
The center of canonical coordinate system at point O
(8, 5)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{3 \sqrt{73}}{146} + \frac{1}{2}}, \ \sqrt{\frac{1}{2} - \frac{3 \sqrt{73}}{146}}\right)$$
$$\vec e_2 = \left( - \sqrt{\frac{1}{2} - \frac{3 \sqrt{73}}{146}}, \ \sqrt{\frac{3 \sqrt{73}}{146} + \frac{1}{2}}\right)$$
Invariants method
Given line equation of 2-order:
$$3 x^{2} - 8 x y - 8 x + 6 y^{2} + 4 y - 3 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = -4$$
$$a_{13} = -4$$
$$a_{22} = 6$$
$$a_{23} = 2$$
$$a_{33} = -3$$
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
$$I_{1} = 9$$
$$I_{3} = \left|\begin{matrix}3 & -4 & -4\\-4 & 6 & 2\\-4 & 2 & -3\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & -4\\-4 & 6 - \lambda\end{matrix}\right|$$
$$I_{1} = 9$$
$$I_{2} = 2$$
$$I_{3} = -50$$
$$I{\left(\lambda \right)} = \lambda^{2} - 9 \lambda + 2$$
$$K_{2} = -47$$
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
$$\lambda^{2} - 9 \lambda + 2 = 0$$
$$\lambda_{1} = \frac{9}{2} - \frac{\sqrt{73}}{2}$$
$$\lambda_{2} = \frac{\sqrt{73}}{2} + \frac{9}{2}$$
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
$$\tilde x^{2} \left(\frac{9}{2} - \frac{\sqrt{73}}{2}\right) + \tilde y^{2} \left(\frac{\sqrt{73}}{2} + \frac{9}{2}\right) - 25 = 0$$
- reduced to canonical form
$$3 x^{2} - 8 x y - 8 x + 6 y^{2} + 4 y - 3 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = -4$$
$$a_{13} = -4$$
$$a_{22} = 6$$
$$a_{23} = 2$$
$$a_{33} = -3$$
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
$$I_{1} = 9$$
|3 -4|
I2 = | |
|-4 6 |$$I_{3} = \left|\begin{matrix}3 & -4 & -4\\-4 & 6 & 2\\-4 & 2 & -3\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & -4\\-4 & 6 - \lambda\end{matrix}\right|$$
|3 -4| |6 2 |
K2 = | | + | |
|-4 -3| |2 -3|$$I_{1} = 9$$
$$I_{2} = 2$$
$$I_{3} = -50$$
$$I{\left(\lambda \right)} = \lambda^{2} - 9 \lambda + 2$$
$$K_{2} = -47$$
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
$$\lambda^{2} - 9 \lambda + 2 = 0$$
$$\lambda_{1} = \frac{9}{2} - \frac{\sqrt{73}}{2}$$
$$\lambda_{2} = \frac{\sqrt{73}}{2} + \frac{9}{2}$$
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
$$\tilde x^{2} \left(\frac{9}{2} - \frac{\sqrt{73}}{2}\right) + \tilde y^{2} \left(\frac{\sqrt{73}}{2} + \frac{9}{2}\right) - 25 = 0$$
2 2
\tilde x \tilde y
------------------------ + ------------------------ = 1
2 2
/ 1 \ / 1 \
|---------------------| |---------------------|
| ____________ | | ____________ |
| / ____ | | / ____ |
| / 9 \/ 73 | | / 9 \/ 73 |
| / - - ------ *1/5| | / - + ------ *1/5|
\\/ 2 2 / \\/ 2 2 / - reduced to canonical form