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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:
- 9x^2+16y^2-24xy-20x+110y-50=0
- 32y-2(1+8y)^2
- 9x2+12xy+4y2-24x-16y+3=0
- tga+ctgb/ctga+tgb=tga+ctgb
- Plot:
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xy=6
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4x^2+4y^2=25
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(x^2+y^2-1)^3-(x^2)*y^3=0
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xy=4
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Identical expressions
- 9x^ two +16y^ two -24xy-2 zero x+110y- fifty =0
- 9x squared plus 16y squared minus 24xy minus 20x plus 110y minus 50 equally 0
- 9x to the power of two plus 16y to the power of two minus 24xy minus 2 zero x plus 110y minus fifty equally 0
- 9x2+16y2-24xy-20x+110y-50=0
- 9x²+16y²-24xy-20x+110y-50=0
- 9x to the power of 2+16y to the power of 2-24xy-20x+110y-50=0
- 9x^2+16y^2-24xy-20x+110y-50=O
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Similar expressions
- 9x^2+16y^2-24xy-20x-110y-50=0
- 9x^2-16y^2-24xy-20x+110y-50=0
- 9x^2+16y^2-24xy+20x+110y-50=0
- 9x^2+16y^2+24xy-20x+110y-50=0
- 9x^2+16y^2-24xy-20x+110y+50=0
9x^2+16y^2-24xy-20x+110y-50=0 canonical form
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The solution
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2 2 -50 - 20*x + 9*x + 16*y + 110*y - 24*x*y = 0
$$9 x^{2} - 24 x y - 20 x + 16 y^{2} + 110 y - 50 = 0$$
9*x^2 - 24*x*y - 20*x + 16*y^2 + 110*y - 50 = 0
Detail solution
Given line equation of 2-order:
$$9 x^{2} - 24 x y - 20 x + 16 y^{2} + 110 y - 50 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = -12$$
$$a_{13} = -10$$
$$a_{22} = 16$$
$$a_{23} = 55$$
$$a_{33} = -50$$
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
$$9 x'^{2} - 24 x' y' - 20 x' + 16 y'^{2} + 110 y' - 50 = 0$$
to
$$16 \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right)^{2} - 24 \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right) + 110 \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) + 9 \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right)^{2} - 20 \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right) - 50 = 0$$
simplify
$$50 \tilde x + 25 \tilde y^{2} + 100 \tilde y - 50 = 0$$
$$\left(5 \tilde y + 10\right)^{2} = 150 - 50 \tilde x$$
$$\left(\tilde y + 2\right)^{2} = 6 - 2 \tilde x$$
$$\tilde y'^{2} = 6 - 2 \tilde x$$
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
$$\vec e_1 = \left( \frac{4}{5}, \ \frac{3}{5}\right)$$
$$\vec e_2 = \left( - \frac{3}{5}, \ \frac{4}{5}\right)$$
$$9 x^{2} - 24 x y - 20 x + 16 y^{2} + 110 y - 50 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = -12$$
$$a_{13} = -10$$
$$a_{22} = 16$$
$$a_{23} = 55$$
$$a_{33} = -50$$
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
$$9 x'^{2} - 24 x' y' - 20 x' + 16 y'^{2} + 110 y' - 50 = 0$$
to
$$16 \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right)^{2} - 24 \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right) + 110 \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) + 9 \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right)^{2} - 20 \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right) - 50 = 0$$
simplify
$$50 \tilde x + 25 \tilde y^{2} + 100 \tilde y - 50 = 0$$
$$\left(5 \tilde y + 10\right)^{2} = 150 - 50 \tilde x$$
$$\left(\tilde y + 2\right)^{2} = 6 - 2 \tilde x$$
$$\tilde y'^{2} = 6 - 2 \tilde x$$
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:
$$9 x^{2} - 24 x y - 20 x + 16 y^{2} + 110 y - 50 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = -12$$
$$a_{13} = -10$$
$$a_{22} = 16$$
$$a_{23} = 55$$
$$a_{33} = -50$$
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
$$I_{1} = 25$$
$$I_{3} = \left|\begin{matrix}9 & -12 & -10\\-12 & 16 & 55\\-10 & 55 & -50\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & -12\\-12 & 16 - \lambda\end{matrix}\right|$$
$$I_{1} = 25$$
$$I_{2} = 0$$
$$I_{3} = -15625$$
$$I{\left(\lambda \right)} = \lambda^{2} - 25 \lambda$$
$$K_{2} = -4375$$
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
$$50 \tilde x + 25 \tilde y^{2} = 0$$
$$\tilde y^{2} = 2 \tilde x$$
- reduced to canonical form
$$9 x^{2} - 24 x y - 20 x + 16 y^{2} + 110 y - 50 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = -12$$
$$a_{13} = -10$$
$$a_{22} = 16$$
$$a_{23} = 55$$
$$a_{33} = -50$$
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 |$$I_{3} = \left|\begin{matrix}9 & -12 & -10\\-12 & 16 & 55\\-10 & 55 & -50\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & -12\\-12 & 16 - \lambda\end{matrix}\right|$$
| 9 -10| |16 55 |
K2 = | | + | |
|-10 -50| |55 -50|$$I_{1} = 25$$
$$I_{2} = 0$$
$$I_{3} = -15625$$
$$I{\left(\lambda \right)} = \lambda^{2} - 25 \lambda$$
$$K_{2} = -4375$$
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
$$50 \tilde x + 25 \tilde y^{2} = 0$$
$$\tilde y^{2} = 2 \tilde x$$
- reduced to canonical form