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:
- 4*x^2+4*y^2-4x+8y+59=0
- 4x^2-12xy+13y^2-36x+62y+69=0
- 3(x+2)^2+7(y-1)=36
- 0x^2+5y^2-3z^2+14yz−2y+10z-3=0.
- Plot:
-
x^2+(y-|x|^(1/2))^2=1
- x^2/36+y^2/36=1
- x^4-y^4=x*y
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((x+5/2)^2-(y-11^(1/2)/2)^2+2.5)*((x+5/2)^2-(y+11^(1/2)/2)^2+2.5)=444-36*x^2
- Sum of series:
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36
- The double integral of:
- 36
- Limit of the function:
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36
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Identical expressions
- three (x+ two)^ two + seven (y- one)= thirty-six
- 3(x plus 2) squared plus 7(y minus 1) equally 36
- three (x plus two) to the power of two plus seven (y minus one) equally thirty minus six
- 3(x+2)2+7(y-1)=36
- 3x+22+7y-1=36
- 3(x+2)²+7(y-1)=36
- 3(x+2) to the power of 2+7(y-1)=36
- 3x+2^2+7y-1=36
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Similar expressions
- 3(x+2)^2-7(y-1)=36
- 3(x-2)^2+7(y-1)=36
- 3(x+2)^2+7(y+1)=36
3(x+2)^2+7(y-1)=36 canonical form
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The solution
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2 -43 + 3*(2 + x) + 7*y = 0
$$7 y + 3 \left(x + 2\right)^{2} - 43 = 0$$
7*y + 3*(x + 2)^2 - 43 = 0
Detail solution
Given line equation of 2-order:
$$7 y + 3 \left(x + 2\right)^{2} - 43 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = 6$$
$$a_{22} = 0$$
$$a_{23} = \frac{7}{2}$$
$$a_{33} = -31$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}3 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
Given equation is straight line
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 0$$
$$y_{0} = 0 \cdot 0$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$7 y + 3 \left(x + 2\right)^{2} - 43 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = 6$$
$$a_{22} = 0$$
$$a_{23} = \frac{7}{2}$$
$$a_{33} = -31$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}3 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
Given equation is straight line
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 0$$
$$y_{0} = 0 \cdot 0$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
(0, 0)
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
Invariants method
Given line equation of 2-order:
$$7 y + 3 \left(x + 2\right)^{2} - 43 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = 6$$
$$a_{22} = 0$$
$$a_{23} = \frac{7}{2}$$
$$a_{33} = -31$$
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
$$I_{1} = 3$$
$$I_{3} = \left|\begin{matrix}3 & 0 & 6\\0 & 0 & \frac{7}{2}\\6 & \frac{7}{2} & -31\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & 0\\0 & - \lambda\end{matrix}\right|$$
$$I_{1} = 3$$
$$I_{2} = 0$$
$$I_{3} = - \frac{147}{4}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 3 \lambda$$
$$K_{2} = - \frac{565}{4}$$
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
$$7 \tilde x + 3 \tilde y^{2} = 0$$
$$\tilde y^{2} = \frac{7 \tilde x}{3}$$
- reduced to canonical form
$$7 y + 3 \left(x + 2\right)^{2} - 43 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = 6$$
$$a_{22} = 0$$
$$a_{23} = \frac{7}{2}$$
$$a_{33} = -31$$
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
$$I_{1} = 3$$
|3 0|
I2 = | |
|0 0|$$I_{3} = \left|\begin{matrix}3 & 0 & 6\\0 & 0 & \frac{7}{2}\\6 & \frac{7}{2} & -31\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & 0\\0 & - \lambda\end{matrix}\right|$$
|3 6 | | 0 7/2|
K2 = | | + | |
|6 -31| |7/2 -31|$$I_{1} = 3$$
$$I_{2} = 0$$
$$I_{3} = - \frac{147}{4}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 3 \lambda$$
$$K_{2} = - \frac{565}{4}$$
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
$$7 \tilde x + 3 \tilde y^{2} = 0$$
$$\tilde y^{2} = \frac{7 \tilde x}{3}$$
- reduced to canonical form