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:
- (2)x^2+(-2)xy+(2)y^2+(-2)x+(-2)y+1=0
- 25x^2-120xy+144y^2-242x-298y+491=0
- x^2/6-y^2/5+z^2-1=0
- 3x^2+3y^2-18x-18y+51=0
- Plot:
- 3*x-84*y=54
- y=1/3*x^3-2*x^2
- x^2*(y-x^2/3)^2=1
-
0=-x^6-2(1-4y^2)xy-2(1-y^2)y^4
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Identical expressions
- - nine *x^ two + four *y^ two + ninety *x-16y- two hundred and forty-five = zero
- minus 9 multiply by x squared plus 4 multiply by y squared plus 90 multiply by x minus 16y minus 245 equally 0
- minus nine multiply by x to the power of two plus four multiply by y to the power of two plus ninety multiply by x minus 16y minus two hundred and forty minus five equally zero
- -9*x2+4*y2+90*x-16y-245=0
- -9*x²+4*y²+90*x-16y-245=0
- -9*x to the power of 2+4*y to the power of 2+90*x-16y-245=0
- -9x^2+4y^2+90x-16y-245=0
- -9x2+4y2+90x-16y-245=0
- -9*x^2+4*y^2+90*x-16y-245=O
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Similar expressions
- -9*x^2+4*y^2+90*x-16y+245=0
- -9*x^2+4*y^2+90*x-16*y-245=0
- -9*x^2+4*y^2-90*x-16y-245=0
- -9*x^2+4*y^2+90*x+16y-245=0
- 9*x^2+4*y^2+90*x-16y-245=0
- -9*x^2-4*y^2+90*x-16y-245=0
-9*x^2+4*y^2+90*x-16y-245=0 canonical form
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The solution
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2 2 -245 - 16*y - 9*x + 4*y + 90*x = 0
$$- 9 x^{2} + 4 y^{2} + 90 x - 16 y - 245 = 0$$
-9*x^2 + 90*x + 4*y^2 - 16*y - 245 = 0
Detail solution
Given line equation of 2-order:
$$- 9 x^{2} + 4 y^{2} + 90 x - 16 y - 245 = 0$$
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
$$a_{11} = -9$$
$$a_{12} = 0$$
$$a_{13} = 45$$
$$a_{22} = 4$$
$$a_{23} = -8$$
$$a_{33} = -245$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-9 & 0\\0 & 4\end{matrix}\right|$$
$$\Delta = -36$$
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
$$- 9 x_{0} + 45 = 0$$
$$4 y_{0} - 8 = 0$$
then
$$x_{0} = 5$$
$$y_{0} = 2$$
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
$$a'_{33} = 45 x_{0} - 8 y_{0} - 245$$
$$a'_{33} = -36$$
then The equation is transformed to
$$- 9 x'^{2} + 4 y'^{2} - 36 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{4} - \frac{\tilde y^{2}}{9} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( 1, \ 0\right)$$
$$\vec e_{2} = \left( 0, \ 1\right)$$
$$- 9 x^{2} + 4 y^{2} + 90 x - 16 y - 245 = 0$$
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
$$a_{11} = -9$$
$$a_{12} = 0$$
$$a_{13} = 45$$
$$a_{22} = 4$$
$$a_{23} = -8$$
$$a_{33} = -245$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-9 & 0\\0 & 4\end{matrix}\right|$$
$$\Delta = -36$$
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
$$- 9 x_{0} + 45 = 0$$
$$4 y_{0} - 8 = 0$$
then
$$x_{0} = 5$$
$$y_{0} = 2$$
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
$$a'_{33} = 45 x_{0} - 8 y_{0} - 245$$
$$a'_{33} = -36$$
then The equation is transformed to
$$- 9 x'^{2} + 4 y'^{2} - 36 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{4} - \frac{\tilde y^{2}}{9} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(5, 2)
Basis of the canonical coordinate system
$$\vec e_{1} = \left( 1, \ 0\right)$$
$$\vec e_{2} = \left( 0, \ 1\right)$$
Invariants method
Given line equation of 2-order:
$$- 9 x^{2} + 4 y^{2} + 90 x - 16 y - 245 = 0$$
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
$$a_{11} = -9$$
$$a_{12} = 0$$
$$a_{13} = 45$$
$$a_{22} = 4$$
$$a_{23} = -8$$
$$a_{33} = -245$$
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
$$I_{1} = -5$$
$$I_{3} = \left|\begin{matrix}-9 & 0 & 45\\0 & 4 & -8\\45 & -8 & -245\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 9 & 0\\0 & - \lambda + 4\end{matrix}\right|$$
$$I_{1} = -5$$
$$I_{2} = -36$$
$$I_{3} = 1296$$
$$I{\left(\lambda \right)} = \lambda^{2} + 5 \lambda - 36$$
$$K_{2} = -864$$
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
$$\lambda^{2} + 5 \lambda - 36 = 0$$
Solve this equation
$$\lambda_{1} = 4$$
$$\lambda_{2} = -9$$
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
$$4 \tilde x^{2} - 9 \tilde y^{2} - 36 = 0$$
$$\frac{\tilde x^{2}}{9} - \frac{\tilde y^{2}}{4} = 1$$
- reduced to canonical form
$$- 9 x^{2} + 4 y^{2} + 90 x - 16 y - 245 = 0$$
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
$$a_{11} = -9$$
$$a_{12} = 0$$
$$a_{13} = 45$$
$$a_{22} = 4$$
$$a_{23} = -8$$
$$a_{33} = -245$$
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
$$I_{1} = -5$$
|-9 0|
I2 = | |
|0 4|$$I_{3} = \left|\begin{matrix}-9 & 0 & 45\\0 & 4 & -8\\45 & -8 & -245\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 9 & 0\\0 & - \lambda + 4\end{matrix}\right|$$
|-9 45 | |4 -8 |
K2 = | | + | |
|45 -245| |-8 -245|$$I_{1} = -5$$
$$I_{2} = -36$$
$$I_{3} = 1296$$
$$I{\left(\lambda \right)} = \lambda^{2} + 5 \lambda - 36$$
$$K_{2} = -864$$
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
$$\lambda^{2} + 5 \lambda - 36 = 0$$
Solve this equation
$$\lambda_{1} = 4$$
$$\lambda_{2} = -9$$
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
$$4 \tilde x^{2} - 9 \tilde y^{2} - 36 = 0$$
$$\frac{\tilde x^{2}}{9} - \frac{\tilde y^{2}}{4} = 1$$
- reduced to canonical form
The graph