Mister Exam
Other calculators
- Canonical form of a elliptical paraboloid
- Canonical form of a double hyperboloid
- Canonical form of a imaginary ellipsoid
- Canonical form of a degenerate ellipse
- Canonical form of a parabola
-
How to use it?
- Canonical form:
- x=-3-2/3*sqrt(8+2y-y^2)
- 11x^2-20xy-4y^2-8y+1=0
- 4x^2-4xy+y^2-14x-8y-2=0
- 2x^2+2(sqrt8)xy-5y^2
- Plot:
-
x^2/3+y^2/4=1
-
x^2/2+y^2/8=1
- 4*(x^2)+9*(y^2)=36
- 2*x^3-9*x^2-3=y
-
Identical expressions
- two x^ two + two (sqrt8)xy-5y^2
- 2x squared plus 2( square root of 8)xy minus 5y squared
- two x to the power of two plus two ( square root of 8)xy minus 5y squared
- 2x^2+2(√8)xy-5y^2
- 2x2+2(sqrt8)xy-5y2
- 2x2+2sqrt8xy-5y2
- 2x²+2(sqrt8)xy-5y²
- 2x to the power of 2+2(sqrt8)xy-5y to the power of 2
- 2x^2+2sqrt8xy-5y^2
-
Similar expressions
- 2x^2+2(sqrt8)xy+5y^2
- 2x^2-2(sqrt8)xy-5y^2
2x^2+2(sqrt8)xy-5y^2 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
2 2 ___ - 5*y + 2*x + 4*x*y*\/ 2 = 0
$$2 x^{2} + 4 \sqrt{2} x y - 5 y^{2} = 0$$
2*x^2 + 4*sqrt(2)*x*y - 5*y^2 = 0
Detail solution
Given line equation of 2-order:
$$2 x^{2} + 4 \sqrt{2} x y - 5 y^{2} = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 2 \sqrt{2}$$
$$a_{13} = 0$$
$$a_{22} = -5$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}2 & 2 \sqrt{2}\\2 \sqrt{2} & -5\end{matrix}\right|$$
$$\Delta = -18$$
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
$$2 x_{0} + 2 \sqrt{2} y_{0} = 0$$
$$2 \sqrt{2} x_{0} - 5 y_{0} = 0$$
then
$$x_{0} = 0$$
$$y_{0} = 0$$
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
$$a'_{33} = 0$$
$$a'_{33} = 0$$
then equation turns into
$$2 x'^{2} + 4 \sqrt{2} x' y' - 5 y'^{2} = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{7 \sqrt{2}}{8}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{7 \sqrt{2}}{8} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{4 \sqrt{2}}{9}$$
$$\cos{\left(2 \phi \right)} = \frac{7}{9}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{2 \sqrt{2}}{3}$$
$$\sin{\left(\phi \right)} = \frac{1}{3}$$
substitute coefficients
$$x' = \frac{2 \sqrt{2} \tilde x}{3} - \frac{\tilde y}{3}$$
$$y' = \frac{\tilde x}{3} + \frac{2 \sqrt{2} \tilde y}{3}$$
then the equation turns from
$$2 x'^{2} + 4 \sqrt{2} x' y' - 5 y'^{2} = 0$$
to
$$- 5 \left(\frac{\tilde x}{3} + \frac{2 \sqrt{2} \tilde y}{3}\right)^{2} + 4 \sqrt{2} \left(\frac{\tilde x}{3} + \frac{2 \sqrt{2} \tilde y}{3}\right) \left(\frac{2 \sqrt{2} \tilde x}{3} - \frac{\tilde y}{3}\right) + 2 \left(\frac{2 \sqrt{2} \tilde x}{3} - \frac{\tilde y}{3}\right)^{2} = 0$$
simplify
$$3 \tilde x^{2} - 6 \tilde y^{2} = 0$$
Given equation is degenerate hyperbole
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{\sqrt{6}}{6}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{2 \sqrt{2}}{3}, \ \frac{1}{3}\right)$$
$$\vec e_2 = \left( - \frac{1}{3}, \ \frac{2 \sqrt{2}}{3}\right)$$
$$2 x^{2} + 4 \sqrt{2} x y - 5 y^{2} = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 2 \sqrt{2}$$
$$a_{13} = 0$$
$$a_{22} = -5$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}2 & 2 \sqrt{2}\\2 \sqrt{2} & -5\end{matrix}\right|$$
$$\Delta = -18$$
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
$$2 x_{0} + 2 \sqrt{2} y_{0} = 0$$
$$2 \sqrt{2} x_{0} - 5 y_{0} = 0$$
then
$$x_{0} = 0$$
$$y_{0} = 0$$
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
$$a'_{33} = 0$$
$$a'_{33} = 0$$
then equation turns into
$$2 x'^{2} + 4 \sqrt{2} x' y' - 5 y'^{2} = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{7 \sqrt{2}}{8}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{7 \sqrt{2}}{8} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{4 \sqrt{2}}{9}$$
$$\cos{\left(2 \phi \right)} = \frac{7}{9}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{2 \sqrt{2}}{3}$$
$$\sin{\left(\phi \right)} = \frac{1}{3}$$
substitute coefficients
$$x' = \frac{2 \sqrt{2} \tilde x}{3} - \frac{\tilde y}{3}$$
$$y' = \frac{\tilde x}{3} + \frac{2 \sqrt{2} \tilde y}{3}$$
then the equation turns from
$$2 x'^{2} + 4 \sqrt{2} x' y' - 5 y'^{2} = 0$$
to
$$- 5 \left(\frac{\tilde x}{3} + \frac{2 \sqrt{2} \tilde y}{3}\right)^{2} + 4 \sqrt{2} \left(\frac{\tilde x}{3} + \frac{2 \sqrt{2} \tilde y}{3}\right) \left(\frac{2 \sqrt{2} \tilde x}{3} - \frac{\tilde y}{3}\right) + 2 \left(\frac{2 \sqrt{2} \tilde x}{3} - \frac{\tilde y}{3}\right)^{2} = 0$$
simplify
$$3 \tilde x^{2} - 6 \tilde y^{2} = 0$$
Given equation is degenerate hyperbole
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{\sqrt{6}}{6}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
(0, 0)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{2 \sqrt{2}}{3}, \ \frac{1}{3}\right)$$
$$\vec e_2 = \left( - \frac{1}{3}, \ \frac{2 \sqrt{2}}{3}\right)$$
Invariants method
Given line equation of 2-order:
$$2 x^{2} + 4 \sqrt{2} x y - 5 y^{2} = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 2 \sqrt{2}$$
$$a_{13} = 0$$
$$a_{22} = -5$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = -3$$
$$I_{3} = \left|\begin{matrix}2 & 2 \sqrt{2} & 0\\2 \sqrt{2} & -5 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & 2 \sqrt{2}\\2 \sqrt{2} & - \lambda - 5\end{matrix}\right|$$
$$I_{1} = -3$$
$$I_{2} = -18$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} + 3 \lambda - 18$$
$$K_{2} = 0$$
Because
$$I_{3} = 0 \wedge I_{2} < 0$$
then by line type:
this equation is of type : degenerate hyperbole
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} + 3 \lambda - 18 = 0$$
$$\lambda_{1} = 3$$
$$\lambda_{2} = -6$$
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
$$3 \tilde x^{2} - 6 \tilde y^{2} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{\sqrt{6}}{6}\right)^{2}} = 0$$
- reduced to canonical form
$$2 x^{2} + 4 \sqrt{2} x y - 5 y^{2} = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 2 \sqrt{2}$$
$$a_{13} = 0$$
$$a_{22} = -5$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = -3$$
| ___|
| 2 2*\/ 2 |
I2 = | |
| ___ |
|2*\/ 2 -5 |$$I_{3} = \left|\begin{matrix}2 & 2 \sqrt{2} & 0\\2 \sqrt{2} & -5 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & 2 \sqrt{2}\\2 \sqrt{2} & - \lambda - 5\end{matrix}\right|$$
|2 0| |-5 0|
K2 = | | + | |
|0 0| |0 0|$$I_{1} = -3$$
$$I_{2} = -18$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} + 3 \lambda - 18$$
$$K_{2} = 0$$
Because
$$I_{3} = 0 \wedge I_{2} < 0$$
then by line type:
this equation is of type : degenerate hyperbole
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} + 3 \lambda - 18 = 0$$
$$\lambda_{1} = 3$$
$$\lambda_{2} = -6$$
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
$$3 \tilde x^{2} - 6 \tilde y^{2} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{\sqrt{6}}{6}\right)^{2}} = 0$$
- reduced to canonical form