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How to use it?
- Canonical form:
- 4x^2+24xy+11y^2+64x+42y+51=0
- y=-2x^2+8x-6
- x+2y+z,y-z-x+y
- u(xx)-4u(xy)+5u(yy)-3u(x)+u(y)+u=0
- Plot:
-
x^2+(y-cbrt(x^2))^2=1
- x^2+(y-(x^2)^(1/3))^2=1
-
3x^2+5y^2=25
-
sqrt(x)+sqrt(y)=1
-
Identical expressions
- 4x^ two + two 4xy+11y^2+64x+42y+ fifty-one = zero
- 4x squared plus 24xy plus 11y squared plus 64x plus 42y plus 51 equally 0
- 4x to the power of two plus two 4xy plus 11y squared plus 64x plus 42y plus fifty minus one equally zero
- 4x2+24xy+11y2+64x+42y+51=0
- 4x²+24xy+11y²+64x+42y+51=0
- 4x to the power of 2+24xy+11y to the power of 2+64x+42y+51=0
- 4x^2+24xy+11y^2+64x+42y+51=O
-
Similar expressions
- 4x^2+24xy-11y^2+64x+42y+51=0
- 4x^2-24xy+11y^2+64x+42y+51=0
- 4x^2+24xy+11y^2+64x-42y+51=0
- 4x^2+24xy+11y^2-64x+42y+51=0
- 4x^2+24xy+11y^2+64x+42y-51=0
4x^2+24xy+11y^2+64x+42y+51=0 canonical form
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The solution
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2 2 51 + 4*x + 11*y + 42*y + 64*x + 24*x*y = 0
$$4 x^{2} + 24 x y + 64 x + 11 y^{2} + 42 y + 51 = 0$$
4*x^2 + 24*x*y + 64*x + 11*y^2 + 42*y + 51 = 0
Detail solution
Given line equation of 2-order:
$$4 x^{2} + 24 x y + 64 x + 11 y^{2} + 42 y + 51 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 12$$
$$a_{13} = 32$$
$$a_{22} = 11$$
$$a_{23} = 21$$
$$a_{33} = 51$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & 12\\12 & 11\end{matrix}\right|$$
$$\Delta = -100$$
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
$$4 x_{0} + 12 y_{0} + 32 = 0$$
$$12 x_{0} + 11 y_{0} + 21 = 0$$
then
$$x_{0} = 1$$
$$y_{0} = -3$$
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
$$a'_{33} = 32 x_{0} + 21 y_{0} + 51$$
$$a'_{33} = 20$$
then equation turns into
$$4 x'^{2} + 24 x' y' + 11 y'^{2} + 20 = 0$$
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
$$\cot{\left(2 \phi \right)} = - \frac{7}{24}$$
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
$$4 x'^{2} + 24 x' y' + 11 y'^{2} + 20 = 0$$
to
$$11 \left(- \frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right)^{2} + 24 \left(- \frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) \left(\frac{4 \tilde x}{5} + \frac{3 \tilde y}{5}\right) + 4 \left(\frac{4 \tilde x}{5} + \frac{3 \tilde y}{5}\right)^{2} + 20 = 0$$
simplify
$$- 5 \tilde x^{2} + 20 \tilde y^{2} + 20 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{4} - \frac{\tilde y^{2}}{1} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{4}{5}, \ - \frac{3}{5}\right)$$
$$\vec e_2 = \left( \frac{3}{5}, \ \frac{4}{5}\right)$$
$$4 x^{2} + 24 x y + 64 x + 11 y^{2} + 42 y + 51 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 12$$
$$a_{13} = 32$$
$$a_{22} = 11$$
$$a_{23} = 21$$
$$a_{33} = 51$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & 12\\12 & 11\end{matrix}\right|$$
$$\Delta = -100$$
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
$$4 x_{0} + 12 y_{0} + 32 = 0$$
$$12 x_{0} + 11 y_{0} + 21 = 0$$
then
$$x_{0} = 1$$
$$y_{0} = -3$$
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
$$a'_{33} = 32 x_{0} + 21 y_{0} + 51$$
$$a'_{33} = 20$$
then equation turns into
$$4 x'^{2} + 24 x' y' + 11 y'^{2} + 20 = 0$$
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
$$\cot{\left(2 \phi \right)} = - \frac{7}{24}$$
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
$$4 x'^{2} + 24 x' y' + 11 y'^{2} + 20 = 0$$
to
$$11 \left(- \frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right)^{2} + 24 \left(- \frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) \left(\frac{4 \tilde x}{5} + \frac{3 \tilde y}{5}\right) + 4 \left(\frac{4 \tilde x}{5} + \frac{3 \tilde y}{5}\right)^{2} + 20 = 0$$
simplify
$$- 5 \tilde x^{2} + 20 \tilde y^{2} + 20 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{4} - \frac{\tilde y^{2}}{1} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(1, -3)
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:
$$4 x^{2} + 24 x y + 64 x + 11 y^{2} + 42 y + 51 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 12$$
$$a_{13} = 32$$
$$a_{22} = 11$$
$$a_{23} = 21$$
$$a_{33} = 51$$
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
$$I_{1} = 15$$
$$I_{3} = \left|\begin{matrix}4 & 12 & 32\\12 & 11 & 21\\32 & 21 & 51\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & 12\\12 & 11 - \lambda\end{matrix}\right|$$
$$I_{1} = 15$$
$$I_{2} = -100$$
$$I_{3} = -2000$$
$$I{\left(\lambda \right)} = \lambda^{2} - 15 \lambda - 100$$
$$K_{2} = -700$$
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
$$\lambda^{2} - 15 \lambda - 100 = 0$$
$$\lambda_{1} = 20$$
$$\lambda_{2} = -5$$
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
$$20 \tilde x^{2} - 5 \tilde y^{2} + 20 = 0$$
$$\frac{\tilde x^{2}}{1} - \frac{\tilde y^{2}}{4} = -1$$
- reduced to canonical form
$$4 x^{2} + 24 x y + 64 x + 11 y^{2} + 42 y + 51 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 12$$
$$a_{13} = 32$$
$$a_{22} = 11$$
$$a_{23} = 21$$
$$a_{33} = 51$$
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
$$I_{1} = 15$$
|4 12|
I2 = | |
|12 11|$$I_{3} = \left|\begin{matrix}4 & 12 & 32\\12 & 11 & 21\\32 & 21 & 51\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & 12\\12 & 11 - \lambda\end{matrix}\right|$$
|4 32| |11 21|
K2 = | | + | |
|32 51| |21 51|$$I_{1} = 15$$
$$I_{2} = -100$$
$$I_{3} = -2000$$
$$I{\left(\lambda \right)} = \lambda^{2} - 15 \lambda - 100$$
$$K_{2} = -700$$
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
$$\lambda^{2} - 15 \lambda - 100 = 0$$
$$\lambda_{1} = 20$$
$$\lambda_{2} = -5$$
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
$$20 \tilde x^{2} - 5 \tilde y^{2} + 20 = 0$$
$$\frac{\tilde x^{2}}{1} - \frac{\tilde y^{2}}{4} = -1$$
- reduced to canonical form