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:
- 25x^2+9y^2-9z^2=225
- z^2+2y^2-x=0
- 5x^2+y^2+10x-6y-10z+14=0
- x^2-4x+2+y^2+2y
- Plot:
-
-0.47y-1.02y^3+0.5ln((1+y)/(1-y))=x(1-exp(2.2*x/2))
- y=1/3*x^3-2*x^2
- x^2+y^2=16
- x^2*(y-x^2/3)^2=1
-
Identical expressions
- 4x^ two -1 two xy-13y^2-36x+62y+ sixty-nine = zero
- 4x squared minus 12xy minus 13y squared minus 36x plus 62y plus 69 equally 0
- 4x to the power of two minus 1 two xy minus 13y squared minus 36x plus 62y plus sixty minus nine equally zero
- 4x2-12xy-13y2-36x+62y+69=0
- 4x²-12xy-13y²-36x+62y+69=0
- 4x to the power of 2-12xy-13y to the power of 2-36x+62y+69=0
- 4x^2-12xy-13y^2-36x+62y+69=O
-
Similar expressions
- 4x^2+12xy-13y^2-36x+62y+69=0
- 4x^2-12xy-13y^2-36x-62y+69=0
- 4x^2-12xy-13y^2-36x+62y-69=0
- 4x^2-12xy+13y^2-36x+62y+69=0
- 4x^2-12xy-13y^2+36x+62y+69=0
4x^2-12xy-13y^2-36x+62y+69=0 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
2 2 69 - 36*x - 13*y + 4*x + 62*y - 12*x*y = 0
$$4 x^{2} - 12 x y - 13 y^{2} - 36 x + 62 y + 69 = 0$$
4*x^2 - 12*x*y - 36*x - 13*y^2 + 62*y + 69 = 0
Detail solution
Given line equation of 2-order:
$$4 x^{2} - 12 x y - 13 y^{2} - 36 x + 62 y + 69 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 4$$
$$a_{12} = -6$$
$$a_{13} = -18$$
$$a_{22} = -13$$
$$a_{23} = 31$$
$$a_{33} = 69$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & -6\\-6 & -13\end{matrix}\right|$$
$$\Delta = -88$$
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
$$4 x_{0} - 6 y_{0} - 18 = 0$$
$$- 6 x_{0} - 13 y_{0} + 31 = 0$$
then
$$x_{0} = \frac{105}{22}$$
$$y_{0} = \frac{2}{11}$$
Thus, we have the equation in the coordinate system O'x'y'
$$a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} + a'_{33} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = - 18 x_{0} + 31 y_{0} + 69$$
$$a'_{33} = - \frac{124}{11}$$
then The equation is transformed to
$$4 x'^{2} - 12 x' y' - 13 y'^{2} - \frac{124}{11} = 0$$
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
$$\cot{\left(2 \phi \right)} = - \frac{17}{12}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{17}{12} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{12 \sqrt{433}}{433}$$
$$\cos{\left(2 \phi \right)} = \frac{17 \sqrt{433}}{433}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{- \cos^{2}{\left(\phi \right)} + 1}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{17 \sqrt{433}}{866} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = - \sqrt{- \frac{17 \sqrt{433}}{866} + \frac{1}{2}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{17 \sqrt{433}}{866} + \frac{1}{2}} + \tilde y \sqrt{- \frac{17 \sqrt{433}}{866} + \frac{1}{2}}$$
$$y' = - \tilde x \sqrt{- \frac{17 \sqrt{433}}{866} + \frac{1}{2}} + \tilde y \sqrt{\frac{17 \sqrt{433}}{866} + \frac{1}{2}}$$
then the equation turns from
$$4 x'^{2} - 12 x' y' - 13 y'^{2} - \frac{124}{11} = 0$$
to
$$- 13 \left(- \tilde x \sqrt{- \frac{17 \sqrt{433}}{866} + \frac{1}{2}} + \tilde y \sqrt{\frac{17 \sqrt{433}}{866} + \frac{1}{2}}\right)^{2} - 12 \left(- \tilde x \sqrt{- \frac{17 \sqrt{433}}{866} + \frac{1}{2}} + \tilde y \sqrt{\frac{17 \sqrt{433}}{866} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{17 \sqrt{433}}{866} + \frac{1}{2}} + \tilde y \sqrt{- \frac{17 \sqrt{433}}{866} + \frac{1}{2}}\right) + 4 \left(\tilde x \sqrt{\frac{17 \sqrt{433}}{866} + \frac{1}{2}} + \tilde y \sqrt{- \frac{17 \sqrt{433}}{866} + \frac{1}{2}}\right)^{2} - \frac{124}{11} = 0$$
simplify
$$- \frac{9 \tilde x^{2}}{2} + 12 \tilde x^{2} \sqrt{- \frac{17 \sqrt{433}}{866} + \frac{1}{2}} \sqrt{\frac{17 \sqrt{433}}{866} + \frac{1}{2}} + \frac{289 \sqrt{433} \tilde x^{2}}{866} - \frac{204 \sqrt{433} \tilde x \tilde y}{433} + 34 \tilde x \tilde y \sqrt{- \frac{17 \sqrt{433}}{866} + \frac{1}{2}} \sqrt{\frac{17 \sqrt{433}}{866} + \frac{1}{2}} - \frac{289 \sqrt{433} \tilde y^{2}}{866} - \frac{9 \tilde y^{2}}{2} - 12 \tilde y^{2} \sqrt{- \frac{17 \sqrt{433}}{866} + \frac{1}{2}} \sqrt{\frac{17 \sqrt{433}}{866} + \frac{1}{2}} - \frac{124}{11} = 0$$
$$- \frac{\sqrt{433} \tilde x^{2}}{2} + \frac{9 \tilde x^{2}}{2} + \frac{9 \tilde y^{2}}{2} + \frac{\sqrt{433} \tilde y^{2}}{2} + \frac{124}{11} = 0$$
Given equation is hyperbole
$$\tilde x^{2} \left(- \frac{99}{248} + \frac{11 \sqrt{433}}{248}\right) - \tilde y^{2} \cdot \left(\frac{99}{248} + \frac{11 \sqrt{433}}{248}\right) = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( \sqrt{\frac{17 \sqrt{433}}{866} + \frac{1}{2}}, \ - \sqrt{- \frac{17 \sqrt{433}}{866} + \frac{1}{2}}\right)$$
$$\vec e_{2} = \left( \sqrt{- \frac{17 \sqrt{433}}{866} + \frac{1}{2}}, \ \sqrt{\frac{17 \sqrt{433}}{866} + \frac{1}{2}}\right)$$
$$4 x^{2} - 12 x y - 13 y^{2} - 36 x + 62 y + 69 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 4$$
$$a_{12} = -6$$
$$a_{13} = -18$$
$$a_{22} = -13$$
$$a_{23} = 31$$
$$a_{33} = 69$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & -6\\-6 & -13\end{matrix}\right|$$
$$\Delta = -88$$
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
$$4 x_{0} - 6 y_{0} - 18 = 0$$
$$- 6 x_{0} - 13 y_{0} + 31 = 0$$
then
$$x_{0} = \frac{105}{22}$$
$$y_{0} = \frac{2}{11}$$
Thus, we have the equation in the coordinate system O'x'y'
$$a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} + a'_{33} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = - 18 x_{0} + 31 y_{0} + 69$$
$$a'_{33} = - \frac{124}{11}$$
then The equation is transformed to
$$4 x'^{2} - 12 x' y' - 13 y'^{2} - \frac{124}{11} = 0$$
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
$$\cot{\left(2 \phi \right)} = - \frac{17}{12}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{17}{12} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{12 \sqrt{433}}{433}$$
$$\cos{\left(2 \phi \right)} = \frac{17 \sqrt{433}}{433}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{- \cos^{2}{\left(\phi \right)} + 1}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{17 \sqrt{433}}{866} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = - \sqrt{- \frac{17 \sqrt{433}}{866} + \frac{1}{2}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{17 \sqrt{433}}{866} + \frac{1}{2}} + \tilde y \sqrt{- \frac{17 \sqrt{433}}{866} + \frac{1}{2}}$$
$$y' = - \tilde x \sqrt{- \frac{17 \sqrt{433}}{866} + \frac{1}{2}} + \tilde y \sqrt{\frac{17 \sqrt{433}}{866} + \frac{1}{2}}$$
then the equation turns from
$$4 x'^{2} - 12 x' y' - 13 y'^{2} - \frac{124}{11} = 0$$
to
$$- 13 \left(- \tilde x \sqrt{- \frac{17 \sqrt{433}}{866} + \frac{1}{2}} + \tilde y \sqrt{\frac{17 \sqrt{433}}{866} + \frac{1}{2}}\right)^{2} - 12 \left(- \tilde x \sqrt{- \frac{17 \sqrt{433}}{866} + \frac{1}{2}} + \tilde y \sqrt{\frac{17 \sqrt{433}}{866} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{17 \sqrt{433}}{866} + \frac{1}{2}} + \tilde y \sqrt{- \frac{17 \sqrt{433}}{866} + \frac{1}{2}}\right) + 4 \left(\tilde x \sqrt{\frac{17 \sqrt{433}}{866} + \frac{1}{2}} + \tilde y \sqrt{- \frac{17 \sqrt{433}}{866} + \frac{1}{2}}\right)^{2} - \frac{124}{11} = 0$$
simplify
$$- \frac{9 \tilde x^{2}}{2} + 12 \tilde x^{2} \sqrt{- \frac{17 \sqrt{433}}{866} + \frac{1}{2}} \sqrt{\frac{17 \sqrt{433}}{866} + \frac{1}{2}} + \frac{289 \sqrt{433} \tilde x^{2}}{866} - \frac{204 \sqrt{433} \tilde x \tilde y}{433} + 34 \tilde x \tilde y \sqrt{- \frac{17 \sqrt{433}}{866} + \frac{1}{2}} \sqrt{\frac{17 \sqrt{433}}{866} + \frac{1}{2}} - \frac{289 \sqrt{433} \tilde y^{2}}{866} - \frac{9 \tilde y^{2}}{2} - 12 \tilde y^{2} \sqrt{- \frac{17 \sqrt{433}}{866} + \frac{1}{2}} \sqrt{\frac{17 \sqrt{433}}{866} + \frac{1}{2}} - \frac{124}{11} = 0$$
$$- \frac{\sqrt{433} \tilde x^{2}}{2} + \frac{9 \tilde x^{2}}{2} + \frac{9 \tilde y^{2}}{2} + \frac{\sqrt{433} \tilde y^{2}}{2} + \frac{124}{11} = 0$$
Given equation is hyperbole
$$\tilde x^{2} \left(- \frac{99}{248} + \frac{11 \sqrt{433}}{248}\right) - \tilde y^{2} \cdot \left(\frac{99}{248} + \frac{11 \sqrt{433}}{248}\right) = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
105 (---, 2/11) 22
Basis of the canonical coordinate system
$$\vec e_{1} = \left( \sqrt{\frac{17 \sqrt{433}}{866} + \frac{1}{2}}, \ - \sqrt{- \frac{17 \sqrt{433}}{866} + \frac{1}{2}}\right)$$
$$\vec e_{2} = \left( \sqrt{- \frac{17 \sqrt{433}}{866} + \frac{1}{2}}, \ \sqrt{\frac{17 \sqrt{433}}{866} + \frac{1}{2}}\right)$$
Invariants method
Given line equation of 2-order:
$$4 x^{2} - 12 x y - 13 y^{2} - 36 x + 62 y + 69 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 4$$
$$a_{12} = -6$$
$$a_{13} = -18$$
$$a_{22} = -13$$
$$a_{23} = 31$$
$$a_{33} = 69$$
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
$$I_{1} = -9$$
$$I_{3} = \left|\begin{matrix}4 & -6 & -18\\-6 & -13 & 31\\-18 & 31 & 69\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 4 & -6\\-6 & - \lambda - 13\end{matrix}\right|$$
$$I_{1} = -9$$
$$I_{2} = -88$$
$$I_{3} = 992$$
$$I{\left(\lambda \right)} = \lambda^{2} + 9 \lambda - 88$$
$$K_{2} = -1906$$
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:
$$- I_{1} \lambda + \lambda^{2} + I_{2} = 0$$
or
$$\lambda^{2} + 9 \lambda - 88 = 0$$
Solve this equation
$$\lambda_{1} = - \frac{\sqrt{433}}{2} - \frac{9}{2}$$
$$\lambda_{2} = - \frac{9}{2} + \frac{\sqrt{433}}{2}$$
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
$$\tilde x^{2} \left(- \frac{\sqrt{433}}{2} - \frac{9}{2}\right) + \tilde y^{2} \left(- \frac{9}{2} + \frac{\sqrt{433}}{2}\right) - \frac{124}{11} = 0$$
$$\tilde x^{2} \cdot \left(\frac{99}{248} + \frac{11 \sqrt{433}}{248}\right) - \tilde y^{2} \left(- \frac{99}{248} + \frac{11 \sqrt{433}}{248}\right) = -1$$
- reduced to canonical form
$$4 x^{2} - 12 x y - 13 y^{2} - 36 x + 62 y + 69 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 4$$
$$a_{12} = -6$$
$$a_{13} = -18$$
$$a_{22} = -13$$
$$a_{23} = 31$$
$$a_{33} = 69$$
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
$$I_{1} = -9$$
|4 -6 |
I2 = | |
|-6 -13|$$I_{3} = \left|\begin{matrix}4 & -6 & -18\\-6 & -13 & 31\\-18 & 31 & 69\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 4 & -6\\-6 & - \lambda - 13\end{matrix}\right|$$
| 4 -18| |-13 31|
K2 = | | + | |
|-18 69 | |31 69|$$I_{1} = -9$$
$$I_{2} = -88$$
$$I_{3} = 992$$
$$I{\left(\lambda \right)} = \lambda^{2} + 9 \lambda - 88$$
$$K_{2} = -1906$$
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:
$$- I_{1} \lambda + \lambda^{2} + I_{2} = 0$$
or
$$\lambda^{2} + 9 \lambda - 88 = 0$$
Solve this equation
$$\lambda_{1} = - \frac{\sqrt{433}}{2} - \frac{9}{2}$$
$$\lambda_{2} = - \frac{9}{2} + \frac{\sqrt{433}}{2}$$
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
$$\tilde x^{2} \left(- \frac{\sqrt{433}}{2} - \frac{9}{2}\right) + \tilde y^{2} \left(- \frac{9}{2} + \frac{\sqrt{433}}{2}\right) - \frac{124}{11} = 0$$
$$\tilde x^{2} \cdot \left(\frac{99}{248} + \frac{11 \sqrt{433}}{248}\right) - \tilde y^{2} \left(- \frac{99}{248} + \frac{11 \sqrt{433}}{248}\right) = -1$$
- reduced to canonical form
The graph