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How to use it?
- Canonical form:
- sqrt(x+1)-11/(10*x)=0
- 24xy+24y+8xy2 +8y2
- cos(x)*cos(x)/sin(x)
- 6x1^2+5x2^2+7x3^2-4x1x2+4x1x3=0
- Plot:
- 27*x=(3*y+x)^3
- z=/sqrt(3*x+2*y+2)
- y^2=4*x
-
x^4*y-x*y^3=-2
-
Identical expressions
- eight *x^ two - four *x*y- five *y^ two - fifty-two *x+ twenty-two *y+ fifty-three = zero
- 8 multiply by x squared minus 4 multiply by x multiply by y minus 5 multiply by y squared minus 52 multiply by x plus 22 multiply by y plus 53 equally 0
- eight multiply by x to the power of two minus four multiply by x multiply by y minus five multiply by y to the power of two minus fifty minus two multiply by x plus twenty minus two multiply by y plus fifty minus three equally zero
- 8*x2-4*x*y-5*y2-52*x+22*y+53=0
- 8*x²-4*x*y-5*y²-52*x+22*y+53=0
- 8*x to the power of 2-4*x*y-5*y to the power of 2-52*x+22*y+53=0
- 8x^2-4xy-5y^2-52x+22y+53=0
- 8x2-4xy-5y2-52x+22y+53=0
- 8*x^2-4*x*y-5*y^2-52*x+22*y+53=O
-
Similar expressions
- 8*x^2-4*x*y+5*y^2-52*x+22*y+53=0
- 8*x^2+4*x*y-5*y^2-52*x+22*y+53=0
- 8*x^2-4*x*y-5*y^2-52*x+22*y-53=0
- 8*x^2-4*x*y-5*y^2-52*x-22*y+53=0
- 8*x^2-4*x*y-5*y^2+52*x+22*y+53=0
8*x^2-4*x*y-5*y^2-52*x+22*y+53=0 canonical form
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The solution
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2 2 53 - 52*x - 5*y + 8*x + 22*y - 4*x*y = 0
$$8 x^{2} - 4 x y - 52 x - 5 y^{2} + 22 y + 53 = 0$$
8*x^2 - 4*x*y - 52*x - 5*y^2 + 22*y + 53 = 0
Detail solution
Given line equation of 2-order:
$$8 x^{2} - 4 x y - 52 x - 5 y^{2} + 22 y + 53 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = -2$$
$$a_{13} = -26$$
$$a_{22} = -5$$
$$a_{23} = 11$$
$$a_{33} = 53$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}8 & -2\\-2 & -5\end{matrix}\right|$$
$$\Delta = -44$$
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
$$8 x_{0} - 2 y_{0} - 26 = 0$$
$$- 2 x_{0} - 5 y_{0} + 11 = 0$$
then
$$x_{0} = \frac{38}{11}$$
$$y_{0} = \frac{9}{11}$$
Thus, we have the equation in the coordinate system O'x'y'
$$a'_{33} + a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = - 26 x_{0} + 11 y_{0} + 53$$
$$a'_{33} = - \frac{306}{11}$$
then equation turns into
$$8 x'^{2} - 4 x' y' - 5 y'^{2} - \frac{306}{11} = 0$$
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
$$\cot{\left(2 \phi \right)} = - \frac{13}{4}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{13}{4} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{4 \sqrt{185}}{185}$$
$$\cos{\left(2 \phi \right)} = \frac{13 \sqrt{185}}{185}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{13 \sqrt{185}}{370} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = - \sqrt{\frac{1}{2} - \frac{13 \sqrt{185}}{370}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{13 \sqrt{185}}{370} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{13 \sqrt{185}}{370}}$$
$$y' = - \tilde x \sqrt{\frac{1}{2} - \frac{13 \sqrt{185}}{370}} + \tilde y \sqrt{\frac{13 \sqrt{185}}{370} + \frac{1}{2}}$$
then the equation turns from
$$8 x'^{2} - 4 x' y' - 5 y'^{2} - \frac{306}{11} = 0$$
to
$$- 5 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{13 \sqrt{185}}{370}} + \tilde y \sqrt{\frac{13 \sqrt{185}}{370} + \frac{1}{2}}\right)^{2} - 4 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{13 \sqrt{185}}{370}} + \tilde y \sqrt{\frac{13 \sqrt{185}}{370} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{13 \sqrt{185}}{370} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{13 \sqrt{185}}{370}}\right) + 8 \left(\tilde x \sqrt{\frac{13 \sqrt{185}}{370} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{13 \sqrt{185}}{370}}\right)^{2} - \frac{306}{11} = 0$$
simplify
$$4 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{13 \sqrt{185}}{370}} \sqrt{\frac{13 \sqrt{185}}{370} + \frac{1}{2}} + \frac{3 \tilde x^{2}}{2} + \frac{169 \sqrt{185} \tilde x^{2}}{370} - \frac{52 \sqrt{185} \tilde x \tilde y}{185} + 26 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{13 \sqrt{185}}{370}} \sqrt{\frac{13 \sqrt{185}}{370} + \frac{1}{2}} - \frac{169 \sqrt{185} \tilde y^{2}}{370} - 4 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{13 \sqrt{185}}{370}} \sqrt{\frac{13 \sqrt{185}}{370} + \frac{1}{2}} + \frac{3 \tilde y^{2}}{2} - \frac{306}{11} = 0$$
$$\frac{3 \tilde x^{2}}{2} + \frac{\sqrt{185} \tilde x^{2}}{2} - \frac{\sqrt{185} \tilde y^{2}}{2} + \frac{3 \tilde y^{2}}{2} - \frac{306}{11} = 0$$
Given equation is hyperbole
$$\tilde x^{2} \cdot \left(\frac{11}{204} + \frac{11 \sqrt{185}}{612}\right) - \tilde y^{2} \left(- \frac{11}{204} + \frac{11 \sqrt{185}}{612}\right) = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( \sqrt{\frac{13 \sqrt{185}}{370} + \frac{1}{2}}, \ - \sqrt{\frac{1}{2} - \frac{13 \sqrt{185}}{370}}\right)$$
$$\vec e_{2} = \left( \sqrt{\frac{1}{2} - \frac{13 \sqrt{185}}{370}}, \ \sqrt{\frac{13 \sqrt{185}}{370} + \frac{1}{2}}\right)$$
$$8 x^{2} - 4 x y - 52 x - 5 y^{2} + 22 y + 53 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = -2$$
$$a_{13} = -26$$
$$a_{22} = -5$$
$$a_{23} = 11$$
$$a_{33} = 53$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}8 & -2\\-2 & -5\end{matrix}\right|$$
$$\Delta = -44$$
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
$$8 x_{0} - 2 y_{0} - 26 = 0$$
$$- 2 x_{0} - 5 y_{0} + 11 = 0$$
then
$$x_{0} = \frac{38}{11}$$
$$y_{0} = \frac{9}{11}$$
Thus, we have the equation in the coordinate system O'x'y'
$$a'_{33} + a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = - 26 x_{0} + 11 y_{0} + 53$$
$$a'_{33} = - \frac{306}{11}$$
then equation turns into
$$8 x'^{2} - 4 x' y' - 5 y'^{2} - \frac{306}{11} = 0$$
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
$$\cot{\left(2 \phi \right)} = - \frac{13}{4}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{13}{4} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{4 \sqrt{185}}{185}$$
$$\cos{\left(2 \phi \right)} = \frac{13 \sqrt{185}}{185}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{13 \sqrt{185}}{370} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = - \sqrt{\frac{1}{2} - \frac{13 \sqrt{185}}{370}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{13 \sqrt{185}}{370} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{13 \sqrt{185}}{370}}$$
$$y' = - \tilde x \sqrt{\frac{1}{2} - \frac{13 \sqrt{185}}{370}} + \tilde y \sqrt{\frac{13 \sqrt{185}}{370} + \frac{1}{2}}$$
then the equation turns from
$$8 x'^{2} - 4 x' y' - 5 y'^{2} - \frac{306}{11} = 0$$
to
$$- 5 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{13 \sqrt{185}}{370}} + \tilde y \sqrt{\frac{13 \sqrt{185}}{370} + \frac{1}{2}}\right)^{2} - 4 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{13 \sqrt{185}}{370}} + \tilde y \sqrt{\frac{13 \sqrt{185}}{370} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{13 \sqrt{185}}{370} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{13 \sqrt{185}}{370}}\right) + 8 \left(\tilde x \sqrt{\frac{13 \sqrt{185}}{370} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{13 \sqrt{185}}{370}}\right)^{2} - \frac{306}{11} = 0$$
simplify
$$4 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{13 \sqrt{185}}{370}} \sqrt{\frac{13 \sqrt{185}}{370} + \frac{1}{2}} + \frac{3 \tilde x^{2}}{2} + \frac{169 \sqrt{185} \tilde x^{2}}{370} - \frac{52 \sqrt{185} \tilde x \tilde y}{185} + 26 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{13 \sqrt{185}}{370}} \sqrt{\frac{13 \sqrt{185}}{370} + \frac{1}{2}} - \frac{169 \sqrt{185} \tilde y^{2}}{370} - 4 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{13 \sqrt{185}}{370}} \sqrt{\frac{13 \sqrt{185}}{370} + \frac{1}{2}} + \frac{3 \tilde y^{2}}{2} - \frac{306}{11} = 0$$
$$\frac{3 \tilde x^{2}}{2} + \frac{\sqrt{185} \tilde x^{2}}{2} - \frac{\sqrt{185} \tilde y^{2}}{2} + \frac{3 \tilde y^{2}}{2} - \frac{306}{11} = 0$$
Given equation is hyperbole
$$\tilde x^{2} \cdot \left(\frac{11}{204} + \frac{11 \sqrt{185}}{612}\right) - \tilde y^{2} \left(- \frac{11}{204} + \frac{11 \sqrt{185}}{612}\right) = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
38 (--, 9/11) 11
Basis of the canonical coordinate system
$$\vec e_{1} = \left( \sqrt{\frac{13 \sqrt{185}}{370} + \frac{1}{2}}, \ - \sqrt{\frac{1}{2} - \frac{13 \sqrt{185}}{370}}\right)$$
$$\vec e_{2} = \left( \sqrt{\frac{1}{2} - \frac{13 \sqrt{185}}{370}}, \ \sqrt{\frac{13 \sqrt{185}}{370} + \frac{1}{2}}\right)$$
Invariants method
Given line equation of 2-order:
$$8 x^{2} - 4 x y - 52 x - 5 y^{2} + 22 y + 53 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = -2$$
$$a_{13} = -26$$
$$a_{22} = -5$$
$$a_{23} = 11$$
$$a_{33} = 53$$
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
$$I_{1} = 3$$
$$I_{3} = \left|\begin{matrix}8 & -2 & -26\\-2 & -5 & 11\\-26 & 11 & 53\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}8 - \lambda & -2\\-2 & - \lambda - 5\end{matrix}\right|$$
$$I_{1} = 3$$
$$I_{2} = -44$$
$$I_{3} = 1224$$
$$I{\left(\lambda \right)} = \lambda^{2} - 3 \lambda - 44$$
$$K_{2} = -638$$
Because
$$I_{2} < 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 3 \lambda - 44 = 0$$
Solve this equation
$$\lambda_{1} = \frac{3}{2} - \frac{\sqrt{185}}{2}$$
$$\lambda_{2} = \frac{3}{2} + \frac{\sqrt{185}}{2}$$
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
$$\tilde x^{2} \cdot \left(\frac{3}{2} - \frac{\sqrt{185}}{2}\right) + \tilde y^{2} \cdot \left(\frac{3}{2} + \frac{\sqrt{185}}{2}\right) - \frac{306}{11} = 0$$
$$\tilde x^{2} \left(- \frac{11}{204} + \frac{11 \sqrt{185}}{612}\right) - \tilde y^{2} \cdot \left(\frac{11}{204} + \frac{11 \sqrt{185}}{612}\right) = -1$$
- reduced to canonical form
$$8 x^{2} - 4 x y - 52 x - 5 y^{2} + 22 y + 53 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = -2$$
$$a_{13} = -26$$
$$a_{22} = -5$$
$$a_{23} = 11$$
$$a_{33} = 53$$
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
$$I_{1} = 3$$
|8 -2|
I2 = | |
|-2 -5|$$I_{3} = \left|\begin{matrix}8 & -2 & -26\\-2 & -5 & 11\\-26 & 11 & 53\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}8 - \lambda & -2\\-2 & - \lambda - 5\end{matrix}\right|$$
| 8 -26| |-5 11|
K2 = | | + | |
|-26 53 | |11 53|$$I_{1} = 3$$
$$I_{2} = -44$$
$$I_{3} = 1224$$
$$I{\left(\lambda \right)} = \lambda^{2} - 3 \lambda - 44$$
$$K_{2} = -638$$
Because
$$I_{2} < 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 3 \lambda - 44 = 0$$
Solve this equation
$$\lambda_{1} = \frac{3}{2} - \frac{\sqrt{185}}{2}$$
$$\lambda_{2} = \frac{3}{2} + \frac{\sqrt{185}}{2}$$
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
$$\tilde x^{2} \cdot \left(\frac{3}{2} - \frac{\sqrt{185}}{2}\right) + \tilde y^{2} \cdot \left(\frac{3}{2} + \frac{\sqrt{185}}{2}\right) - \frac{306}{11} = 0$$
$$\tilde x^{2} \left(- \frac{11}{204} + \frac{11 \sqrt{185}}{612}\right) - \tilde y^{2} \cdot \left(\frac{11}{204} + \frac{11 \sqrt{185}}{612}\right) = -1$$
- reduced to canonical form
The graph