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How to use it?
- Canonical form:
-
25x^2+25y^2-14xy+64x-64y-224=0
- 319
- 15x^2+6y^2-10z^2+90x-72y+60z+231=0
- factorial(x+1)
- Plot:
-
4x^2+4y^2=25
-
xy=4
-
(x^2+y^2-1)^3-(x^2)*y^3=0
- xyz
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Identical expressions
- two 5x^ two +25y^2-14xy+64x-64y- two hundred and twenty-four = zero
- 25x squared plus 25y squared minus 14xy plus 64x minus 64y minus 224 equally 0
- two 5x to the power of two plus 25y squared minus 14xy plus 64x minus 64y minus two hundred and twenty minus four equally zero
- 25x2+25y2-14xy+64x-64y-224=0
- 25x²+25y²-14xy+64x-64y-224=0
- 25x to the power of 2+25y to the power of 2-14xy+64x-64y-224=0
- 25x^2+25y^2-14xy+64x-64y-224=O
-
Similar expressions
- 25x^2+25y^2-14xy+64x+64y-224=0
- 25x^2+25y^2-14xy+64x-64y+224=0
- 25x^2+25y^2-14xy-64x-64y-224=0
- 25x^2+25y^2+14xy+64x-64y-224=0
- 25x^2-25y^2-14xy+64x-64y-224=0
25x^2+25y^2-14xy+64x-64y-224=0 canonical form
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The solution
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2 2 -224 - 64*y + 25*x + 25*y + 64*x - 14*x*y = 0
$$25 x^{2} - 14 x y + 25 y^{2} + 64 x - 64 y - 224 = 0$$
25*x^2 - 14*x*y + 64*x + 25*y^2 - 64*y - 224 = 0
Detail solution
Given line equation of 2-order:
$$25 x^{2} - 14 x y + 25 y^{2} + 64 x - 64 y - 224 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 25$$
$$a_{12} = -7$$
$$a_{13} = 32$$
$$a_{22} = 25$$
$$a_{23} = -32$$
$$a_{33} = -224$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}25 & -7\\-7 & 25\end{matrix}\right|$$
$$\Delta = 576$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} x_{0} + a_{12} y_{0} + a_{13} = 0$$
$$a_{12} x_{0} + a_{22} y_{0} + a_{23} = 0$$
substitute coefficients
$$25 x_{0} - 7 y_{0} + 32 = 0$$
$$- 7 x_{0} + 25 y_{0} - 32 = 0$$
then
$$x_{0} = -1$$
$$y_{0} = 1$$
Thus, we have the equation in the coordinate system O'x'y'
$$a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} + a'_{33} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = 32 x_{0} - 32 y_{0} - 224$$
$$a'_{33} = -288$$
then The equation is transformed to
$$25 x'^{2} - 14 x' y' + 25 y'^{2} - 288 = 0$$
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
$$\cot{\left(2 \phi \right)} = 0$$
then
$$\phi = \frac{\pi}{4}$$
$$\sin{\left(2 \phi \right)} = 1$$
$$\cos{\left(2 \phi \right)} = 0$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{- \cos^{2}{\left(\phi \right)} + 1}$$
$$\cos{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
substitute coefficients
$$x' = \frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}$$
$$y' = \frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}$$
then the equation turns from
$$25 x'^{2} - 14 x' y' + 25 y'^{2} - 288 = 0$$
to
$$25 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right)^{2} - 14 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right) \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right) + 25 \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right)^{2} - 288 = 0$$
simplify
$$18 \tilde x^{2} + 32 \tilde y^{2} - 288 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{4^{2}} + \frac{\tilde y^{2}}{3^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
$$\vec e_{2} = \left( - \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
$$25 x^{2} - 14 x y + 25 y^{2} + 64 x - 64 y - 224 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 25$$
$$a_{12} = -7$$
$$a_{13} = 32$$
$$a_{22} = 25$$
$$a_{23} = -32$$
$$a_{33} = -224$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}25 & -7\\-7 & 25\end{matrix}\right|$$
$$\Delta = 576$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} x_{0} + a_{12} y_{0} + a_{13} = 0$$
$$a_{12} x_{0} + a_{22} y_{0} + a_{23} = 0$$
substitute coefficients
$$25 x_{0} - 7 y_{0} + 32 = 0$$
$$- 7 x_{0} + 25 y_{0} - 32 = 0$$
then
$$x_{0} = -1$$
$$y_{0} = 1$$
Thus, we have the equation in the coordinate system O'x'y'
$$a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} + a'_{33} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = 32 x_{0} - 32 y_{0} - 224$$
$$a'_{33} = -288$$
then The equation is transformed to
$$25 x'^{2} - 14 x' y' + 25 y'^{2} - 288 = 0$$
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
$$\cot{\left(2 \phi \right)} = 0$$
then
$$\phi = \frac{\pi}{4}$$
$$\sin{\left(2 \phi \right)} = 1$$
$$\cos{\left(2 \phi \right)} = 0$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{- \cos^{2}{\left(\phi \right)} + 1}$$
$$\cos{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
substitute coefficients
$$x' = \frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}$$
$$y' = \frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}$$
then the equation turns from
$$25 x'^{2} - 14 x' y' + 25 y'^{2} - 288 = 0$$
to
$$25 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right)^{2} - 14 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right) \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right) + 25 \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right)^{2} - 288 = 0$$
simplify
$$18 \tilde x^{2} + 32 \tilde y^{2} - 288 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{4^{2}} + \frac{\tilde y^{2}}{3^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(-1, 1)
Basis of the canonical coordinate system
$$\vec e_{1} = \left( \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
$$\vec e_{2} = \left( - \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
Invariants method
Given line equation of 2-order:
$$25 x^{2} - 14 x y + 25 y^{2} + 64 x - 64 y - 224 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 25$$
$$a_{12} = -7$$
$$a_{13} = 32$$
$$a_{22} = 25$$
$$a_{23} = -32$$
$$a_{33} = -224$$
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
$$I_{1} = 50$$
$$I_{3} = \left|\begin{matrix}25 & -7 & 32\\-7 & 25 & -32\\32 & -32 & -224\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 25 & -7\\-7 & - \lambda + 25\end{matrix}\right|$$
$$I_{1} = 50$$
$$I_{2} = 576$$
$$I_{3} = -165888$$
$$I{\left(\lambda \right)} = \lambda^{2} - 50 \lambda + 576$$
$$K_{2} = -13248$$
Because
$$I_{2} > 0 \wedge I_{1} I_{3} < 0$$
then by line type:
this equation is of type : ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + \lambda^{2} + I_{2} = 0$$
or
$$\lambda^{2} - 50 \lambda + 576 = 0$$
Solve this equation
$$\lambda_{1} = 32$$
$$\lambda_{2} = 18$$
then the canonical form of the equation will be
$$\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2} + \frac{I_{3}}{I_{2}} = 0$$
or
$$32 \tilde x^{2} + 18 \tilde y^{2} - 288 = 0$$
$$\frac{\tilde x^{2}}{3^{2}} + \frac{\tilde y^{2}}{4^{2}} = 1$$
- reduced to canonical form
$$25 x^{2} - 14 x y + 25 y^{2} + 64 x - 64 y - 224 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 25$$
$$a_{12} = -7$$
$$a_{13} = 32$$
$$a_{22} = 25$$
$$a_{23} = -32$$
$$a_{33} = -224$$
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
$$I_{1} = 50$$
|25 -7|
I2 = | |
|-7 25|$$I_{3} = \left|\begin{matrix}25 & -7 & 32\\-7 & 25 & -32\\32 & -32 & -224\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 25 & -7\\-7 & - \lambda + 25\end{matrix}\right|$$
|25 32 | |25 -32 |
K2 = | | + | |
|32 -224| |-32 -224|$$I_{1} = 50$$
$$I_{2} = 576$$
$$I_{3} = -165888$$
$$I{\left(\lambda \right)} = \lambda^{2} - 50 \lambda + 576$$
$$K_{2} = -13248$$
Because
$$I_{2} > 0 \wedge I_{1} I_{3} < 0$$
then by line type:
this equation is of type : ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + \lambda^{2} + I_{2} = 0$$
or
$$\lambda^{2} - 50 \lambda + 576 = 0$$
Solve this equation
$$\lambda_{1} = 32$$
$$\lambda_{2} = 18$$
then the canonical form of the equation will be
$$\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2} + \frac{I_{3}}{I_{2}} = 0$$
or
$$32 \tilde x^{2} + 18 \tilde y^{2} - 288 = 0$$
$$\frac{\tilde x^{2}}{3^{2}} + \frac{\tilde y^{2}}{4^{2}} = 1$$
- reduced to canonical form
The graph