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:
-
3x^2+10xy+3y^2-2x-14y=13
- 5x^2-2xy+5y^2+10x-2y+1=0
- 4x^2+9y^2=25
- xy+4=0
- Plot:
- tan(sin(e−1x))−x=0
- (9,27-(x+y+z))*x*y*z=9,27
-
4x^2+4y^2+8x*y+4x+4y+1==0
-
x²+(y-x⅔)²=1
-
Identical expressions
- 5x^ two - two xy+5y^2+ one zero x-2y+1=0
- 5x squared minus 2xy plus 5y squared plus 10x minus 2y plus 1 equally 0
- 5x to the power of two minus two xy plus 5y squared plus one zero x minus 2y plus 1 equally 0
- 5x2-2xy+5y2+10x-2y+1=0
- 5x²-2xy+5y²+10x-2y+1=0
- 5x to the power of 2-2xy+5y to the power of 2+10x-2y+1=0
- 5x^2-2xy+5y^2+10x-2y+1=O
-
Similar expressions
- 5x^2-2xy+5y^2+10x-2y-1=0
- 5x^2+2xy+5y^2+10x-2y+1=0
- 5x^2-2xy+5y^2+10x+2y+1=0
- 5x^2-2xy-5y^2+10x-2y+1=0
- 5x^2-2xy+5y^2-10x-2y+1=0
5x^2-2xy+5y^2+10x-2y+1=0 canonical form
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The solution
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2 2 1 - 2*y + 5*x + 5*y + 10*x - 2*x*y = 0
$$5 x^{2} - 2 x y + 10 x + 5 y^{2} - 2 y + 1 = 0$$
5*x^2 - 2*x*y + 10*x + 5*y^2 - 2*y + 1 = 0
Detail solution
Given line equation of 2-order:
$$5 x^{2} - 2 x y + 10 x + 5 y^{2} - 2 y + 1 = 0$$
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
$$a_{11} = 5$$
$$a_{12} = -1$$
$$a_{13} = 5$$
$$a_{22} = 5$$
$$a_{23} = -1$$
$$a_{33} = 1$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}5 & -1\\-1 & 5\end{matrix}\right|$$
$$\Delta = 24$$
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
$$5 x_{0} - y_{0} + 5 = 0$$
$$- x_{0} + 5 y_{0} - 1 = 0$$
then
$$x_{0} = -1$$
$$y_{0} = 0$$
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
$$a'_{33} = 5 x_{0} - y_{0} + 1$$
$$a'_{33} = -4$$
then equation turns into
$$5 x'^{2} - 2 x' y' + 5 y'^{2} - 4 = 0$$
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
$$\phi = \frac{\pi}{4}$$
$$\sin{\left(2 \phi \right)} = 1$$
$$\cos{\left(2 \phi \right)} = 0$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
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
$$5 x'^{2} - 2 x' y' + 5 y'^{2} - 4 = 0$$
to
$$5 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right)^{2} - 2 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right) \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right) + 5 \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right)^{2} - 4 = 0$$
simplify
$$4 \tilde x^{2} + 6 \tilde y^{2} - 4 = 0$$
Given equation is ellipse
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
$$5 x^{2} - 2 x y + 10 x + 5 y^{2} - 2 y + 1 = 0$$
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
$$a_{11} = 5$$
$$a_{12} = -1$$
$$a_{13} = 5$$
$$a_{22} = 5$$
$$a_{23} = -1$$
$$a_{33} = 1$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}5 & -1\\-1 & 5\end{matrix}\right|$$
$$\Delta = 24$$
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
$$5 x_{0} - y_{0} + 5 = 0$$
$$- x_{0} + 5 y_{0} - 1 = 0$$
then
$$x_{0} = -1$$
$$y_{0} = 0$$
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
$$a'_{33} = 5 x_{0} - y_{0} + 1$$
$$a'_{33} = -4$$
then equation turns into
$$5 x'^{2} - 2 x' y' + 5 y'^{2} - 4 = 0$$
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
$$\phi = \frac{\pi}{4}$$
$$\sin{\left(2 \phi \right)} = 1$$
$$\cos{\left(2 \phi \right)} = 0$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
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
$$5 x'^{2} - 2 x' y' + 5 y'^{2} - 4 = 0$$
to
$$5 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right)^{2} - 2 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right) \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right) + 5 \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right)^{2} - 4 = 0$$
simplify
$$4 \tilde x^{2} + 6 \tilde y^{2} - 4 = 0$$
Given equation is ellipse
2 2
\tilde x \tilde y
--------- + ---------- = 1
2 2
/ 1 \ // ___\\
|-----| ||\/ 6 ||
\2*1/2/ ||-----||
|\ 6 /|
|-------|
\ 1/2 / - reduced to canonical form
The center of canonical coordinate system at point O
(-1, 0)
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:
$$5 x^{2} - 2 x y + 10 x + 5 y^{2} - 2 y + 1 = 0$$
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
$$a_{11} = 5$$
$$a_{12} = -1$$
$$a_{13} = 5$$
$$a_{22} = 5$$
$$a_{23} = -1$$
$$a_{33} = 1$$
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
$$I_{1} = 10$$
$$I_{3} = \left|\begin{matrix}5 & -1 & 5\\-1 & 5 & -1\\5 & -1 & 1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}5 - \lambda & -1\\-1 & 5 - \lambda\end{matrix}\right|$$
$$I_{1} = 10$$
$$I_{2} = 24$$
$$I_{3} = -96$$
$$I{\left(\lambda \right)} = \lambda^{2} - 10 \lambda + 24$$
$$K_{2} = -16$$
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:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 10 \lambda + 24 = 0$$
$$\lambda_{1} = 6$$
$$\lambda_{2} = 4$$
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
$$6 \tilde x^{2} + 4 \tilde y^{2} - 4 = 0$$
- reduced to canonical form
$$5 x^{2} - 2 x y + 10 x + 5 y^{2} - 2 y + 1 = 0$$
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
$$a_{11} = 5$$
$$a_{12} = -1$$
$$a_{13} = 5$$
$$a_{22} = 5$$
$$a_{23} = -1$$
$$a_{33} = 1$$
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
$$I_{1} = 10$$
|5 -1|
I2 = | |
|-1 5 |$$I_{3} = \left|\begin{matrix}5 & -1 & 5\\-1 & 5 & -1\\5 & -1 & 1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}5 - \lambda & -1\\-1 & 5 - \lambda\end{matrix}\right|$$
|5 5| |5 -1|
K2 = | | + | |
|5 1| |-1 1 |$$I_{1} = 10$$
$$I_{2} = 24$$
$$I_{3} = -96$$
$$I{\left(\lambda \right)} = \lambda^{2} - 10 \lambda + 24$$
$$K_{2} = -16$$
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:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 10 \lambda + 24 = 0$$
$$\lambda_{1} = 6$$
$$\lambda_{2} = 4$$
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
$$6 \tilde x^{2} + 4 \tilde y^{2} - 4 = 0$$
2 2
\tilde x \tilde y
---------- + --------- = 1
2 2
// ___\\ / 1 \
||\/ 6 || |-----|
||-----|| \2*1/2/
|\ 6 /|
|-------|
\ 1/2 / - reduced to canonical form