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How to use it?
- Canonical form:
-
9x^2-24xy+16y^2+18x+226y+209=0
- x+y
- x^2+y^2+6x-4y+13=0
- x^2+6x+y+11=0
- Plot:
-
(-x^2)*y^3+(x^2+y^2-1)^3=0
-
x^2+(y+cbrt(x^2))^2=1
-
sin(x^e)+cos(y^e)=1
- f*(x)=100-x*y-x*y*z
-
Identical expressions
- 9x^ two - two 4xy+16y^2+18x+226y+ two hundred and nine = zero
- 9x squared minus 24xy plus 16y squared plus 18x plus 226y plus 209 equally 0
- 9x to the power of two minus two 4xy plus 16y squared plus 18x plus 226y plus two hundred and nine equally zero
- 9x2-24xy+16y2+18x+226y+209=0
- 9x²-24xy+16y²+18x+226y+209=0
- 9x to the power of 2-24xy+16y to the power of 2+18x+226y+209=0
- 9x^2-24xy+16y^2+18x+226y+209=O
-
Similar expressions
- 9x^2-24xy-16y^2+18x+226y+209=0
- 9x^2-24xy+16y^2+18x-226y+209=0
- 9x^2-24xy+16y^2-18x+226y+209=0
- 9x^2-24xy+16y^2+18x+226y-209=0
- 9x^2+24xy+16y^2+18x+226y+209=0
9x^2-24xy+16y^2+18x+226y+209=0 canonical form
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The solution
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2 2 209 + 9*x + 16*y + 18*x + 226*y - 24*x*y = 0
$$9 x^{2} - 24 x y + 18 x + 16 y^{2} + 226 y + 209 = 0$$
9*x^2 - 24*x*y + 18*x + 16*y^2 + 226*y + 209 = 0
Detail solution
Given line equation of 2-order:
$$9 x^{2} - 24 x y + 18 x + 16 y^{2} + 226 y + 209 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = -12$$
$$a_{13} = 9$$
$$a_{22} = 16$$
$$a_{23} = 113$$
$$a_{33} = 209$$
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
$$9 x'^{2} - 24 x' y' + 18 x' + 16 y'^{2} + 226 y' + 209 = 0$$
to
$$16 \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right)^{2} - 24 \cdot \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right) + 226 \cdot \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) + 9 \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right)^{2} + 18 \cdot \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right) + 209 = 0$$
simplify
$$150 \tilde x + 25 \tilde y^{2} + 170 \tilde y + 209 = 0$$
$$\left(5 \tilde y + 17\right)^{2} = 80 - 150 \tilde x$$
$$\left(\tilde y + \frac{17}{5}\right)^{2} = \frac{16}{5} - 6 \tilde x$$
$$\tilde y'^{2} = \frac{16}{5} - 6 \tilde x$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot \frac{4}{5} + 0 \cdot \frac{3}{5}$$
$$y_{0} = 0 \cdot \frac{3}{5} + 0 \cdot \frac{4}{5}$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( \frac{4}{5}, \ \frac{3}{5}\right)$$
$$\vec e_{2} = \left( - \frac{3}{5}, \ \frac{4}{5}\right)$$
$$9 x^{2} - 24 x y + 18 x + 16 y^{2} + 226 y + 209 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = -12$$
$$a_{13} = 9$$
$$a_{22} = 16$$
$$a_{23} = 113$$
$$a_{33} = 209$$
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
$$9 x'^{2} - 24 x' y' + 18 x' + 16 y'^{2} + 226 y' + 209 = 0$$
to
$$16 \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right)^{2} - 24 \cdot \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right) + 226 \cdot \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) + 9 \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right)^{2} + 18 \cdot \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right) + 209 = 0$$
simplify
$$150 \tilde x + 25 \tilde y^{2} + 170 \tilde y + 209 = 0$$
$$\left(5 \tilde y + 17\right)^{2} = 80 - 150 \tilde x$$
$$\left(\tilde y + \frac{17}{5}\right)^{2} = \frac{16}{5} - 6 \tilde x$$
$$\tilde y'^{2} = \frac{16}{5} - 6 \tilde x$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot \frac{4}{5} + 0 \cdot \frac{3}{5}$$
$$y_{0} = 0 \cdot \frac{3}{5} + 0 \cdot \frac{4}{5}$$
$$x_{0} = 0$$
$$y_{0} = 0$$
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:
$$9 x^{2} - 24 x y + 18 x + 16 y^{2} + 226 y + 209 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = -12$$
$$a_{13} = 9$$
$$a_{22} = 16$$
$$a_{23} = 113$$
$$a_{33} = 209$$
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
$$I_{1} = 25$$
$$I_{3} = \left|\begin{matrix}9 & -12 & 9\\-12 & 16 & 113\\9 & 113 & 209\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & -12\\-12 & 16 - \lambda\end{matrix}\right|$$
$$I_{1} = 25$$
$$I_{2} = 0$$
$$I_{3} = -140625$$
$$I{\left(\lambda \right)} = \lambda^{2} - 25 \lambda$$
$$K_{2} = -7625$$
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
$$150 \tilde x + 25 \tilde y^{2} = 0$$
$$\tilde y^{2} = 6 \tilde x$$
- reduced to canonical form
$$9 x^{2} - 24 x y + 18 x + 16 y^{2} + 226 y + 209 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = -12$$
$$a_{13} = 9$$
$$a_{22} = 16$$
$$a_{23} = 113$$
$$a_{33} = 209$$
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 |$$I_{3} = \left|\begin{matrix}9 & -12 & 9\\-12 & 16 & 113\\9 & 113 & 209\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & -12\\-12 & 16 - \lambda\end{matrix}\right|$$
|9 9 | |16 113|
K2 = | | + | |
|9 209| |113 209|$$I_{1} = 25$$
$$I_{2} = 0$$
$$I_{3} = -140625$$
$$I{\left(\lambda \right)} = \lambda^{2} - 25 \lambda$$
$$K_{2} = -7625$$
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
$$150 \tilde x + 25 \tilde y^{2} = 0$$
$$\tilde y^{2} = 6 \tilde x$$
- reduced to canonical form
The graph