Mister Exam
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How to use it?
- Canonical form:
- 4*x^2−12*x*y+9*y^2−20*x+30*y+16=0.
- -x^2+2y^2-2xy-2x-2y+2=0
- -7x^2-6xy+1y^2-8=0
- x^2+4*y^2+8*z^2-16
- Plot:
- y-(|x|)=(1/2)
- z=3*x+6*y-x^2-x*y+y^2
- x^2-2=y
-
x^2=8y
-
Identical expressions
- four *x^ two − twelve *x*y+ nine *y^ two − twenty *x+ thirty *y+ sixteen = zero .
- 4 multiply by x squared −12 multiply by x multiply by y plus 9 multiply by y squared −20 multiply by x plus 30 multiply by y plus 16 equally 0.
- four multiply by x to the power of two − twelve multiply by x multiply by y plus nine multiply by y to the power of two − twenty multiply by x plus thirty multiply by y plus sixteen equally zero .
- 4*x2−12*x*y+9*y2−20*x+30*y+16=0.
- 4*x²−12*x*y+9*y²−20*x+30*y+16=0.
- 4*x to the power of 2−12*x*y+9*y to the power of 2−20*x+30*y+16=0.
- 4x^2−12xy+9y^2−20x+30y+16=0.
- 4x2−12xy+9y2−20x+30y+16=0.
- 4*x^2−12*x*y+9*y^2−20*x+30*y+16=O.
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Similar expressions
- 4*x^2−12*x*y-9*y^2−20*x+30*y+16=0.
- 4*x^2−12*x*y+9*y^2−20*x+30*y-16=0.
- 4*x^2−12*x*y+9*y^2−20*x-30*y+16=0.
4*x^2−12*x*y+9*y^2−20*x+30*y+16=0. canonical form
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The solution
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2 2 16 - 20*x + 4*x + 9*y + 30*y - 12*x*y = 0
$$4 x^{2} - 12 x y - 20 x + 9 y^{2} + 30 y + 16 = 0$$
4*x^2 - 12*x*y - 20*x + 9*y^2 + 30*y + 16 = 0
Detail solution
Given line equation of 2-order:
$$4 x^{2} - 12 x y - 20 x + 9 y^{2} + 30 y + 16 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = -6$$
$$a_{13} = -10$$
$$a_{22} = 9$$
$$a_{23} = 15$$
$$a_{33} = 16$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & -6\\-6 & 9\end{matrix}\right|$$
$$\Delta = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{5}{12}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{5}{12} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{12}{13}$$
$$\cos{\left(2 \phi \right)} = \frac{5}{13}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{3 \sqrt{13}}{13}$$
$$\sin{\left(\phi \right)} = \frac{2 \sqrt{13}}{13}$$
substitute coefficients
$$x' = \frac{3 \sqrt{13} \tilde x}{13} - \frac{2 \sqrt{13} \tilde y}{13}$$
$$y' = \frac{2 \sqrt{13} \tilde x}{13} + \frac{3 \sqrt{13} \tilde y}{13}$$
then the equation turns from
$$4 x'^{2} - 12 x' y' - 20 x' + 9 y'^{2} + 30 y' + 16 = 0$$
to
$$9 \left(\frac{2 \sqrt{13} \tilde x}{13} + \frac{3 \sqrt{13} \tilde y}{13}\right)^{2} - 12 \left(\frac{2 \sqrt{13} \tilde x}{13} + \frac{3 \sqrt{13} \tilde y}{13}\right) \left(\frac{3 \sqrt{13} \tilde x}{13} - \frac{2 \sqrt{13} \tilde y}{13}\right) + 30 \left(\frac{2 \sqrt{13} \tilde x}{13} + \frac{3 \sqrt{13} \tilde y}{13}\right) + 4 \left(\frac{3 \sqrt{13} \tilde x}{13} - \frac{2 \sqrt{13} \tilde y}{13}\right)^{2} - 20 \left(\frac{3 \sqrt{13} \tilde x}{13} - \frac{2 \sqrt{13} \tilde y}{13}\right) + 16 = 0$$
simplify
$$13 \tilde y^{2} + 10 \sqrt{13} \tilde y + 16 = 0$$
$$13 \tilde y^{2} + 10 \sqrt{13} \tilde y = -16$$
$$\left(\sqrt{13} \tilde y + 5\right)^{2} = 25$$
$$\left(\tilde y + \frac{5 \sqrt{13}}{13}\right)^{2} = \frac{25}{13}$$
$$\tilde y'^{2} = \frac{25}{13}$$
Given equation is two parallel straight lines
- reduced to canonical form
where replacement made
$$\tilde y' = \tilde y + \frac{5 \sqrt{13}}{13}$$
$$\tilde x' = \tilde x$$
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \frac{3 \sqrt{13}}{13} - \frac{\left(-5\right) \sqrt{13}}{13} \frac{2 \sqrt{13}}{13}$$
$$y_{0} = \frac{\left(-5\right) \sqrt{13}}{13} \frac{3 \sqrt{13}}{13} + 0 \frac{2 \sqrt{13}}{13}$$
$$x_{0} = \frac{10}{13}$$
$$y_{0} = - \frac{15}{13}$$
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{3 \sqrt{13}}{13}, \ \frac{2 \sqrt{13}}{13}\right)$$
$$\vec e_2 = \left( - \frac{2 \sqrt{13}}{13}, \ \frac{3 \sqrt{13}}{13}\right)$$
$$4 x^{2} - 12 x y - 20 x + 9 y^{2} + 30 y + 16 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = -6$$
$$a_{13} = -10$$
$$a_{22} = 9$$
$$a_{23} = 15$$
$$a_{33} = 16$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & -6\\-6 & 9\end{matrix}\right|$$
$$\Delta = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{5}{12}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{5}{12} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{12}{13}$$
$$\cos{\left(2 \phi \right)} = \frac{5}{13}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{3 \sqrt{13}}{13}$$
$$\sin{\left(\phi \right)} = \frac{2 \sqrt{13}}{13}$$
substitute coefficients
$$x' = \frac{3 \sqrt{13} \tilde x}{13} - \frac{2 \sqrt{13} \tilde y}{13}$$
$$y' = \frac{2 \sqrt{13} \tilde x}{13} + \frac{3 \sqrt{13} \tilde y}{13}$$
then the equation turns from
$$4 x'^{2} - 12 x' y' - 20 x' + 9 y'^{2} + 30 y' + 16 = 0$$
to
$$9 \left(\frac{2 \sqrt{13} \tilde x}{13} + \frac{3 \sqrt{13} \tilde y}{13}\right)^{2} - 12 \left(\frac{2 \sqrt{13} \tilde x}{13} + \frac{3 \sqrt{13} \tilde y}{13}\right) \left(\frac{3 \sqrt{13} \tilde x}{13} - \frac{2 \sqrt{13} \tilde y}{13}\right) + 30 \left(\frac{2 \sqrt{13} \tilde x}{13} + \frac{3 \sqrt{13} \tilde y}{13}\right) + 4 \left(\frac{3 \sqrt{13} \tilde x}{13} - \frac{2 \sqrt{13} \tilde y}{13}\right)^{2} - 20 \left(\frac{3 \sqrt{13} \tilde x}{13} - \frac{2 \sqrt{13} \tilde y}{13}\right) + 16 = 0$$
simplify
$$13 \tilde y^{2} + 10 \sqrt{13} \tilde y + 16 = 0$$
$$13 \tilde y^{2} + 10 \sqrt{13} \tilde y = -16$$
$$\left(\sqrt{13} \tilde y + 5\right)^{2} = 25$$
$$\left(\tilde y + \frac{5 \sqrt{13}}{13}\right)^{2} = \frac{25}{13}$$
$$\tilde y'^{2} = \frac{25}{13}$$
Given equation is two parallel straight lines
- reduced to canonical form
where replacement made
$$\tilde y' = \tilde y + \frac{5 \sqrt{13}}{13}$$
$$\tilde x' = \tilde x$$
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \frac{3 \sqrt{13}}{13} - \frac{\left(-5\right) \sqrt{13}}{13} \frac{2 \sqrt{13}}{13}$$
$$y_{0} = \frac{\left(-5\right) \sqrt{13}}{13} \frac{3 \sqrt{13}}{13} + 0 \frac{2 \sqrt{13}}{13}$$
$$x_{0} = \frac{10}{13}$$
$$y_{0} = - \frac{15}{13}$$
The center of canonical coordinate system at point O
10 -15 (--, ----) 13 13
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{3 \sqrt{13}}{13}, \ \frac{2 \sqrt{13}}{13}\right)$$
$$\vec e_2 = \left( - \frac{2 \sqrt{13}}{13}, \ \frac{3 \sqrt{13}}{13}\right)$$
Invariants method
Given line equation of 2-order:
$$4 x^{2} - 12 x y - 20 x + 9 y^{2} + 30 y + 16 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = -6$$
$$a_{13} = -10$$
$$a_{22} = 9$$
$$a_{23} = 15$$
$$a_{33} = 16$$
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
$$I_{1} = 13$$
$$I_{3} = \left|\begin{matrix}4 & -6 & -10\\-6 & 9 & 15\\-10 & 15 & 16\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & -6\\-6 & 9 - \lambda\end{matrix}\right|$$
$$I_{1} = 13$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 13 \lambda$$
$$K_{2} = -117$$
Because
$$I_{2} = 0 \wedge I_{3} = 0 \wedge K_{2} < 0 \wedge I_{1} \neq 0$$
then by line type:
this equation is of type : two parallel lines
$$I_{1} \tilde y^{2} + \frac{K_{2}}{I_{1}} = 0$$
or
$$13 \tilde y^{2} - 9 = 0$$
- reduced to canonical form
$$4 x^{2} - 12 x y - 20 x + 9 y^{2} + 30 y + 16 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = -6$$
$$a_{13} = -10$$
$$a_{22} = 9$$
$$a_{23} = 15$$
$$a_{33} = 16$$
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
$$I_{1} = 13$$
|4 -6|
I2 = | |
|-6 9 |$$I_{3} = \left|\begin{matrix}4 & -6 & -10\\-6 & 9 & 15\\-10 & 15 & 16\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & -6\\-6 & 9 - \lambda\end{matrix}\right|$$
| 4 -10| |9 15|
K2 = | | + | |
|-10 16 | |15 16|$$I_{1} = 13$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 13 \lambda$$
$$K_{2} = -117$$
Because
$$I_{2} = 0 \wedge I_{3} = 0 \wedge K_{2} < 0 \wedge I_{1} \neq 0$$
then by line type:
this equation is of type : two parallel lines
$$I_{1} \tilde y^{2} + \frac{K_{2}}{I_{1}} = 0$$
or
$$13 \tilde y^{2} - 9 = 0$$
None
- reduced to canonical form