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- Canonical form of a degenerate ellipse
- Canonical form of a parabola
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How to use it?
- Canonical form:
-
xy+x+y=0
- 9x^2-24xy+16y^2-8x+19y+4
- ax^2+by^2+c=0
- sin(x)-cos(x)
- Plot:
- (x-2)*(y-3)=0
-
y^2=x^2-x^4
- x^2/a^2-y^2/b^2=1
-
3*cos(x)*sin(y)-sin(x)*cos(y)-4*cos(y)=4
-
Identical expressions
- 9x^ two - two four xy+16y^2-8x+19y+4
- 9x squared minus 24xy plus 16y squared minus 8x plus 19y plus 4
- 9x to the power of two minus two four xy plus 16y squared minus 8x plus 19y plus 4
- 9x2-24xy+16y2-8x+19y+4
- 9x²-24xy+16y²-8x+19y+4
- 9x to the power of 2-24xy+16y to the power of 2-8x+19y+4
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Similar expressions
- 9x^2-24xy+16y^2+8x+19y+4
- 9x^2-24xy-16y^2-8x+19y+4
- 9x^2+24xy+16y^2-8x+19y+4
- 9x^2-24xy+16y^2-8x-19y+4
- 9x^2-24xy+16y^2-8x+19y-4
9x^2-24xy+16y^2-8x+19y+4 canonical form
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The solution
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2 2 4 - 8*x + 9*x + 16*y + 19*y - 24*x*y = 0
$$9 x^{2} - 24 x y - 8 x + 16 y^{2} + 19 y + 4 = 0$$
9*x^2 - 24*x*y - 8*x + 16*y^2 + 19*y + 4 = 0
Detail solution
Given line equation of 2-order:
$$9 x^{2} - 24 x y - 8 x + 16 y^{2} + 19 y + 4 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = -12$$
$$a_{13} = -4$$
$$a_{22} = 16$$
$$a_{23} = \frac{19}{2}$$
$$a_{33} = 4$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}9 & -12\\-12 & 16\end{matrix}\right|$$
$$\Delta = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{7}{24}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{7}{24} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{24}{25}$$
$$\cos{\left(2 \phi \right)} = \frac{7}{25}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{4}{5}$$
$$\sin{\left(\phi \right)} = \frac{3}{5}$$
substitute coefficients
$$x' = \frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}$$
$$y' = \frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}$$
then the equation turns from
$$9 x'^{2} - 24 x' y' - 8 x' + 16 y'^{2} + 19 y' + 4 = 0$$
to
$$16 \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right)^{2} - 24 \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right) + 19 \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) + 9 \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right)^{2} - 8 \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right) + 4 = 0$$
simplify
$$5 \tilde x + 25 \tilde y^{2} + 20 \tilde y + 4 = 0$$
$$\left(5 \tilde y + 2\right)^{2} = - 5 \tilde x$$
$$\left(\tilde y + \frac{2}{5}\right)^{2} = - \frac{\tilde x}{5}$$
$$\tilde y'^{2} = - \frac{\tilde x'}{5}$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = \frac{0 \cdot 4}{5} + \frac{0 \cdot 3}{5}$$
$$y_{0} = \frac{0 \cdot 3}{5} + \frac{0 \cdot 4}{5}$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{4}{5}, \ \frac{3}{5}\right)$$
$$\vec e_2 = \left( - \frac{3}{5}, \ \frac{4}{5}\right)$$
$$9 x^{2} - 24 x y - 8 x + 16 y^{2} + 19 y + 4 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = -12$$
$$a_{13} = -4$$
$$a_{22} = 16$$
$$a_{23} = \frac{19}{2}$$
$$a_{33} = 4$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}9 & -12\\-12 & 16\end{matrix}\right|$$
$$\Delta = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{7}{24}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{7}{24} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{24}{25}$$
$$\cos{\left(2 \phi \right)} = \frac{7}{25}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{4}{5}$$
$$\sin{\left(\phi \right)} = \frac{3}{5}$$
substitute coefficients
$$x' = \frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}$$
$$y' = \frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}$$
then the equation turns from
$$9 x'^{2} - 24 x' y' - 8 x' + 16 y'^{2} + 19 y' + 4 = 0$$
to
$$16 \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right)^{2} - 24 \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right) + 19 \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) + 9 \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right)^{2} - 8 \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right) + 4 = 0$$
simplify
$$5 \tilde x + 25 \tilde y^{2} + 20 \tilde y + 4 = 0$$
$$\left(5 \tilde y + 2\right)^{2} = - 5 \tilde x$$
$$\left(\tilde y + \frac{2}{5}\right)^{2} = - \frac{\tilde x}{5}$$
$$\tilde y'^{2} = - \frac{\tilde x'}{5}$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = \frac{0 \cdot 4}{5} + \frac{0 \cdot 3}{5}$$
$$y_{0} = \frac{0 \cdot 3}{5} + \frac{0 \cdot 4}{5}$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
(0, 0)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{4}{5}, \ \frac{3}{5}\right)$$
$$\vec e_2 = \left( - \frac{3}{5}, \ \frac{4}{5}\right)$$
Invariants method
Given line equation of 2-order:
$$9 x^{2} - 24 x y - 8 x + 16 y^{2} + 19 y + 4 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = -12$$
$$a_{13} = -4$$
$$a_{22} = 16$$
$$a_{23} = \frac{19}{2}$$
$$a_{33} = 4$$
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
$$I_{1} = 25$$
$$I_{3} = \left|\begin{matrix}9 & -12 & -4\\-12 & 16 & \frac{19}{2}\\-4 & \frac{19}{2} & 4\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & -12\\-12 & 16 - \lambda\end{matrix}\right|$$
$$I_{1} = 25$$
$$I_{2} = 0$$
$$I_{3} = - \frac{625}{4}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 25 \lambda$$
$$K_{2} = - \frac{25}{4}$$
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
$$5 \tilde x + 25 \tilde y^{2} = 0$$
$$\tilde y^{2} = \frac{\tilde x}{5}$$
- reduced to canonical form
$$9 x^{2} - 24 x y - 8 x + 16 y^{2} + 19 y + 4 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = -12$$
$$a_{13} = -4$$
$$a_{22} = 16$$
$$a_{23} = \frac{19}{2}$$
$$a_{33} = 4$$
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
$$I_{1} = 25$$
| 9 -12|
I2 = | |
|-12 16 |$$I_{3} = \left|\begin{matrix}9 & -12 & -4\\-12 & 16 & \frac{19}{2}\\-4 & \frac{19}{2} & 4\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & -12\\-12 & 16 - \lambda\end{matrix}\right|$$
|9 -4| | 16 19/2|
K2 = | | + | |
|-4 4 | |19/2 4 |$$I_{1} = 25$$
$$I_{2} = 0$$
$$I_{3} = - \frac{625}{4}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 25 \lambda$$
$$K_{2} = - \frac{25}{4}$$
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
$$5 \tilde x + 25 \tilde y^{2} = 0$$
$$\tilde y^{2} = \frac{\tilde x}{5}$$
- reduced to canonical form