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How to use it?
- Canonical form:
- x^2-10*x*y+7*y^2+2*x-10*y=a
- 4x^2-3y^2-2xy+z^2+6z+5=0
- 8x^2+6xy-5sqrt(10)x-3sqrt(10)y-22=0
- (y+3)^2=4x-2
- Plot:
- y^2-4*x+x^2=0
- z=/sqrt(3*x+2*y+2)
- x^4-y^4=x*y
-
sin(x+y)-1.2*x-0.1
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Identical expressions
- 8x^ two +6xy-5sqrt(ten)x-3sqrt(ten)y- twenty-two = zero
- 8x squared plus 6xy minus 5 square root of (10)x minus 3 square root of (10)y minus 22 equally 0
- 8x to the power of two plus 6xy minus 5 square root of (ten)x minus 3 square root of (ten)y minus twenty minus two equally zero
- 8x^2+6xy-5√(10)x-3√(10)y-22=0
- 8x2+6xy-5sqrt(10)x-3sqrt(10)y-22=0
- 8x2+6xy-5sqrt10x-3sqrt10y-22=0
- 8x²+6xy-5sqrt(10)x-3sqrt(10)y-22=0
- 8x to the power of 2+6xy-5sqrt(10)x-3sqrt(10)y-22=0
- 8x^2+6xy-5sqrt10x-3sqrt10y-22=0
- 8x^2+6xy-5sqrt(10)x-3sqrt(10)y-22=O
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Similar expressions
- 8x^2-6xy-5sqrt(10)x-3sqrt(10)y-22=0
- 8x^2+6xy+5sqrt(10)x-3sqrt(10)y-22=0
- 8x^2+6xy-5sqrt(10)x-3sqrt(10)y+22=0
- 8x^2+6xy-5sqrt(10)x+3sqrt(10)y-22=0
8x^2+6xy-5sqrt(10)x-3sqrt(10)y-22=0 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
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2 ____ ____ -22 + 8*x - 5*x*\/ 10 - 3*y*\/ 10 + 6*x*y = 0
$$8 x^{2} + 6 x y - 5 \sqrt{10} x - 3 \sqrt{10} y - 22 = 0$$
8*x^2 + 6*x*y - 5*sqrt(10)*x - 3*sqrt(10)*y - 22 = 0
Detail solution
Given line equation of 2-order:
$$8 x^{2} + 6 x y - 5 \sqrt{10} x - 3 \sqrt{10} y - 22 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = 3$$
$$a_{13} = - \frac{5 \sqrt{10}}{2}$$
$$a_{22} = 0$$
$$a_{23} = - \frac{3 \sqrt{10}}{2}$$
$$a_{33} = -22$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}8 & 3\\3 & 0\end{matrix}\right|$$
$$\Delta = -9$$
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
$$8 x_{0} + 3 y_{0} - \frac{5 \sqrt{10}}{2} = 0$$
$$3 x_{0} - \frac{3 \sqrt{10}}{2} = 0$$
then
$$x_{0} = \frac{\sqrt{10}}{2}$$
$$y_{0} = - \frac{\sqrt{10}}{2}$$
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
$$a'_{33} = - \frac{5 \sqrt{10} x_{0}}{2} - \frac{3 \sqrt{10} y_{0}}{2} - 22$$
$$a'_{33} = -27$$
then equation turns into
$$8 x'^{2} + 6 x' y' - 27 = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{4}{3}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{4}{3} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{3}{5}$$
$$\cos{\left(2 \phi \right)} = \frac{4}{5}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{3 \sqrt{10}}{10}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{10}}{10}$$
substitute coefficients
$$x' = \frac{3 \sqrt{10} \tilde x}{10} - \frac{\sqrt{10} \tilde y}{10}$$
$$y' = \frac{\sqrt{10} \tilde x}{10} + \frac{3 \sqrt{10} \tilde y}{10}$$
then the equation turns from
$$8 x'^{2} + 6 x' y' - 27 = 0$$
to
$$6 \left(\frac{\sqrt{10} \tilde x}{10} + \frac{3 \sqrt{10} \tilde y}{10}\right) \left(\frac{3 \sqrt{10} \tilde x}{10} - \frac{\sqrt{10} \tilde y}{10}\right) + 8 \left(\frac{3 \sqrt{10} \tilde x}{10} - \frac{\sqrt{10} \tilde y}{10}\right)^{2} - 27 = 0$$
simplify
$$9 \tilde x^{2} - \tilde y^{2} - 27 = 0$$
$$- 9 \tilde x^{2} + \tilde y^{2} + 27 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{3} - \frac{\tilde y^{2}}{27} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{3 \sqrt{10}}{10}, \ \frac{\sqrt{10}}{10}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{10}}{10}, \ \frac{3 \sqrt{10}}{10}\right)$$
$$8 x^{2} + 6 x y - 5 \sqrt{10} x - 3 \sqrt{10} y - 22 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = 3$$
$$a_{13} = - \frac{5 \sqrt{10}}{2}$$
$$a_{22} = 0$$
$$a_{23} = - \frac{3 \sqrt{10}}{2}$$
$$a_{33} = -22$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}8 & 3\\3 & 0\end{matrix}\right|$$
$$\Delta = -9$$
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
$$8 x_{0} + 3 y_{0} - \frac{5 \sqrt{10}}{2} = 0$$
$$3 x_{0} - \frac{3 \sqrt{10}}{2} = 0$$
then
$$x_{0} = \frac{\sqrt{10}}{2}$$
$$y_{0} = - \frac{\sqrt{10}}{2}$$
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
$$a'_{33} = - \frac{5 \sqrt{10} x_{0}}{2} - \frac{3 \sqrt{10} y_{0}}{2} - 22$$
$$a'_{33} = -27$$
then equation turns into
$$8 x'^{2} + 6 x' y' - 27 = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{4}{3}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{4}{3} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{3}{5}$$
$$\cos{\left(2 \phi \right)} = \frac{4}{5}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{3 \sqrt{10}}{10}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{10}}{10}$$
substitute coefficients
$$x' = \frac{3 \sqrt{10} \tilde x}{10} - \frac{\sqrt{10} \tilde y}{10}$$
$$y' = \frac{\sqrt{10} \tilde x}{10} + \frac{3 \sqrt{10} \tilde y}{10}$$
then the equation turns from
$$8 x'^{2} + 6 x' y' - 27 = 0$$
to
$$6 \left(\frac{\sqrt{10} \tilde x}{10} + \frac{3 \sqrt{10} \tilde y}{10}\right) \left(\frac{3 \sqrt{10} \tilde x}{10} - \frac{\sqrt{10} \tilde y}{10}\right) + 8 \left(\frac{3 \sqrt{10} \tilde x}{10} - \frac{\sqrt{10} \tilde y}{10}\right)^{2} - 27 = 0$$
simplify
$$9 \tilde x^{2} - \tilde y^{2} - 27 = 0$$
$$- 9 \tilde x^{2} + \tilde y^{2} + 27 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{3} - \frac{\tilde y^{2}}{27} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
____ ____ \/ 10 -\/ 10 (------, --------) 2 2
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{3 \sqrt{10}}{10}, \ \frac{\sqrt{10}}{10}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{10}}{10}, \ \frac{3 \sqrt{10}}{10}\right)$$
Invariants method
Given line equation of 2-order:
$$8 x^{2} + 6 x y - 5 \sqrt{10} x - 3 \sqrt{10} y - 22 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = 3$$
$$a_{13} = - \frac{5 \sqrt{10}}{2}$$
$$a_{22} = 0$$
$$a_{23} = - \frac{3 \sqrt{10}}{2}$$
$$a_{33} = -22$$
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
$$I_{1} = 8$$
$$I_{3} = \left|\begin{matrix}8 & 3 & - \frac{5 \sqrt{10}}{2}\\3 & 0 & - \frac{3 \sqrt{10}}{2}\\- \frac{5 \sqrt{10}}{2} & - \frac{3 \sqrt{10}}{2} & -22\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}8 - \lambda & 3\\3 & - \lambda\end{matrix}\right|$$
$$I_{1} = 8$$
$$I_{2} = -9$$
$$I_{3} = 243$$
$$I{\left(\lambda \right)} = \lambda^{2} - 8 \lambda - 9$$
$$K_{2} = -261$$
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
$$\lambda^{2} - 8 \lambda - 9 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = -1$$
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
$$9 \tilde x^{2} - \tilde y^{2} - 27 = 0$$
$$\frac{\tilde x^{2}}{3} - \frac{\tilde y^{2}}{27} = 1$$
- reduced to canonical form
$$8 x^{2} + 6 x y - 5 \sqrt{10} x - 3 \sqrt{10} y - 22 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = 3$$
$$a_{13} = - \frac{5 \sqrt{10}}{2}$$
$$a_{22} = 0$$
$$a_{23} = - \frac{3 \sqrt{10}}{2}$$
$$a_{33} = -22$$
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
$$I_{1} = 8$$
|8 3|
I2 = | |
|3 0|$$I_{3} = \left|\begin{matrix}8 & 3 & - \frac{5 \sqrt{10}}{2}\\3 & 0 & - \frac{3 \sqrt{10}}{2}\\- \frac{5 \sqrt{10}}{2} & - \frac{3 \sqrt{10}}{2} & -22\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}8 - \lambda & 3\\3 & - \lambda\end{matrix}\right|$$
| ____| | ____|
| -5*\/ 10 | | -3*\/ 10 |
| 8 ---------| | 0 ---------|
| 2 | | 2 |
K2 = | | + | |
| ____ | | ____ |
|-5*\/ 10 | |-3*\/ 10 |
|--------- -22 | |--------- -22 |
| 2 | | 2 |$$I_{1} = 8$$
$$I_{2} = -9$$
$$I_{3} = 243$$
$$I{\left(\lambda \right)} = \lambda^{2} - 8 \lambda - 9$$
$$K_{2} = -261$$
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
$$\lambda^{2} - 8 \lambda - 9 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = -1$$
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
$$9 \tilde x^{2} - \tilde y^{2} - 27 = 0$$
$$\frac{\tilde x^{2}}{3} - \frac{\tilde y^{2}}{27} = 1$$
- reduced to canonical form