Mister Exam
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- 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:
- (2)x^2+(-2)xy+(2)y^2+(-2)x+(-2)y+1=0
- 25x^2-120xy+144y^2-242x-298y+491=0
- 11x^2+24xy+4y^2+42x+64y+51=0
- 3x^2+3y^2-18x-18y+51=0
- Plot:
-
-0.47y-1.02y^3+0.5ln((1+y)/(1-y))=x(1-exp(2.2*x/2))
- z=x^2+y^2/4
- y*(x)=x+4/x^2
- (7/10)*x^2+(y-sqrt(|x|))^2=10
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Identical expressions
- two 5x^ two -12 zero xy+144y^2-242x-298y+ four hundred and ninety-one =0
- 25x squared minus 120xy plus 144y squared minus 242x minus 298y plus 491 equally 0
- two 5x to the power of two minus 12 zero xy plus 144y squared minus 242x minus 298y plus four hundred and ninety minus one equally 0
- 25x2-120xy+144y2-242x-298y+491=0
- 25x²-120xy+144y²-242x-298y+491=0
- 25x to the power of 2-120xy+144y to the power of 2-242x-298y+491=0
- 25x^2-120xy+144y^2-242x-298y+491=O
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Similar expressions
- 25x^2-120xy-144y^2-242x-298y+491=0
- 25x^2-120xy+144y^2+242x-298y+491=0
- 25x^2-120xy+144y^2-242x+298y+491=0
- 25x^2-120xy+144y^2-242x-298y-491=0
- 25x^2+120xy+144y^2-242x-298y+491=0
25x^2-120xy+144y^2-242x-298y+491=0 canonical form
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The solution
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2 2 491 - 298*y - 242*x + 25*x + 144*y - 120*x*y = 0
$$25 x^{2} - 120 x y - 242 x + 144 y^{2} - 298 y + 491 = 0$$
25*x^2 - 120*x*y - 242*x + 144*y^2 - 298*y + 491 = 0
Detail solution
Given line equation of 2-order:
$$25 x^{2} - 120 x y - 242 x + 144 y^{2} - 298 y + 491 = 0$$
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
$$a_{11} = 25$$
$$a_{12} = -60$$
$$a_{13} = -121$$
$$a_{22} = 144$$
$$a_{23} = -149$$
$$a_{33} = 491$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}25 & -60\\-60 & 144\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
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
$$\cot{\left(2 \phi \right)} = \frac{119}{120}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{119}{120} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{120}{169}$$
$$\cos{\left(2 \phi \right)} = \frac{119}{169}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{12}{13}$$
$$\sin{\left(\phi \right)} = \frac{5}{13}$$
substitute coefficients
$$x' = \frac{12 \tilde x}{13} - \frac{5 \tilde y}{13}$$
$$y' = \frac{5 \tilde x}{13} + \frac{12 \tilde y}{13}$$
then the equation turns from
$$25 x'^{2} - 120 x' y' - 242 x' + 144 y'^{2} - 298 y' + 491 = 0$$
to
$$144 \left(\frac{5 \tilde x}{13} + \frac{12 \tilde y}{13}\right)^{2} - 120 \left(\frac{5 \tilde x}{13} + \frac{12 \tilde y}{13}\right) \left(\frac{12 \tilde x}{13} - \frac{5 \tilde y}{13}\right) - 298 \left(\frac{5 \tilde x}{13} + \frac{12 \tilde y}{13}\right) + 25 \left(\frac{12 \tilde x}{13} - \frac{5 \tilde y}{13}\right)^{2} - 242 \left(\frac{12 \tilde x}{13} - \frac{5 \tilde y}{13}\right) + 491 = 0$$
simplify
$$- 338 \tilde x + 169 \tilde y^{2} - 182 \tilde y + 491 = 0$$
$$338 \tilde x - 169 \tilde y^{2} + 182 \tilde y - 491 = 0$$
$$\left(13 \tilde y + 7\right)^{2} = 338 \tilde x - 442$$
$$\left(\tilde y + \frac{7}{13}\right)^{2} = 2 \tilde x - \frac{34}{13}$$
$$\tilde y'^{2} = 2 \tilde x - \frac{34}{13}$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = \frac{0 \cdot 12}{13} + \frac{0 \cdot 5}{13}$$
$$y_{0} = \frac{0 \cdot 5}{13} + \frac{0 \cdot 12}{13}$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{12}{13}, \ \frac{5}{13}\right)$$
$$\vec e_2 = \left( - \frac{5}{13}, \ \frac{12}{13}\right)$$
$$25 x^{2} - 120 x y - 242 x + 144 y^{2} - 298 y + 491 = 0$$
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
$$a_{11} = 25$$
$$a_{12} = -60$$
$$a_{13} = -121$$
$$a_{22} = 144$$
$$a_{23} = -149$$
$$a_{33} = 491$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}25 & -60\\-60 & 144\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
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
$$\cot{\left(2 \phi \right)} = \frac{119}{120}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{119}{120} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{120}{169}$$
$$\cos{\left(2 \phi \right)} = \frac{119}{169}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{12}{13}$$
$$\sin{\left(\phi \right)} = \frac{5}{13}$$
substitute coefficients
$$x' = \frac{12 \tilde x}{13} - \frac{5 \tilde y}{13}$$
$$y' = \frac{5 \tilde x}{13} + \frac{12 \tilde y}{13}$$
then the equation turns from
$$25 x'^{2} - 120 x' y' - 242 x' + 144 y'^{2} - 298 y' + 491 = 0$$
to
$$144 \left(\frac{5 \tilde x}{13} + \frac{12 \tilde y}{13}\right)^{2} - 120 \left(\frac{5 \tilde x}{13} + \frac{12 \tilde y}{13}\right) \left(\frac{12 \tilde x}{13} - \frac{5 \tilde y}{13}\right) - 298 \left(\frac{5 \tilde x}{13} + \frac{12 \tilde y}{13}\right) + 25 \left(\frac{12 \tilde x}{13} - \frac{5 \tilde y}{13}\right)^{2} - 242 \left(\frac{12 \tilde x}{13} - \frac{5 \tilde y}{13}\right) + 491 = 0$$
simplify
$$- 338 \tilde x + 169 \tilde y^{2} - 182 \tilde y + 491 = 0$$
$$338 \tilde x - 169 \tilde y^{2} + 182 \tilde y - 491 = 0$$
$$\left(13 \tilde y + 7\right)^{2} = 338 \tilde x - 442$$
$$\left(\tilde y + \frac{7}{13}\right)^{2} = 2 \tilde x - \frac{34}{13}$$
$$\tilde y'^{2} = 2 \tilde x - \frac{34}{13}$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = \frac{0 \cdot 12}{13} + \frac{0 \cdot 5}{13}$$
$$y_{0} = \frac{0 \cdot 5}{13} + \frac{0 \cdot 12}{13}$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
(0, 0)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{12}{13}, \ \frac{5}{13}\right)$$
$$\vec e_2 = \left( - \frac{5}{13}, \ \frac{12}{13}\right)$$
Invariants method
Given line equation of 2-order:
$$25 x^{2} - 120 x y - 242 x + 144 y^{2} - 298 y + 491 = 0$$
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
$$a_{11} = 25$$
$$a_{12} = -60$$
$$a_{13} = -121$$
$$a_{22} = 144$$
$$a_{23} = -149$$
$$a_{33} = 491$$
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
$$I_{1} = 169$$
$$I_{3} = \left|\begin{matrix}25 & -60 & -121\\-60 & 144 & -149\\-121 & -149 & 491\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}25 - \lambda & -60\\-60 & 144 - \lambda\end{matrix}\right|$$
$$I_{1} = 169$$
$$I_{2} = 0$$
$$I_{3} = -4826809$$
$$I{\left(\lambda \right)} = \lambda^{2} - 169 \lambda$$
$$K_{2} = 46137$$
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
$$338 \tilde x + 169 \tilde y^{2} = 0$$
$$\tilde y^{2} = 2 \tilde x$$
- reduced to canonical form
$$25 x^{2} - 120 x y - 242 x + 144 y^{2} - 298 y + 491 = 0$$
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
$$a_{11} = 25$$
$$a_{12} = -60$$
$$a_{13} = -121$$
$$a_{22} = 144$$
$$a_{23} = -149$$
$$a_{33} = 491$$
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
$$I_{1} = 169$$
|25 -60|
I2 = | |
|-60 144|$$I_{3} = \left|\begin{matrix}25 & -60 & -121\\-60 & 144 & -149\\-121 & -149 & 491\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}25 - \lambda & -60\\-60 & 144 - \lambda\end{matrix}\right|$$
| 25 -121| |144 -149|
K2 = | | + | |
|-121 491 | |-149 491 |$$I_{1} = 169$$
$$I_{2} = 0$$
$$I_{3} = -4826809$$
$$I{\left(\lambda \right)} = \lambda^{2} - 169 \lambda$$
$$K_{2} = 46137$$
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
$$338 \tilde x + 169 \tilde y^{2} = 0$$
$$\tilde y^{2} = 2 \tilde x$$
- reduced to canonical form