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- 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
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How to use it?
- Canonical form:
- x^2+y^2+z^2=2x+2y-1
- y^2-x-4y+5=0
- -31x^2-24xy+14y^2=0
- 9x^2+4xy+6y^2+40x+20y-50=0
- Plot:
- y=-2*sqrt(|x|)+4
- x+y=1
- x^2=y
-
x^3(y-x)=x^3+1
-
Identical expressions
- 9x^ two +4xy+6y^ two +4 zero x+20y- fifty =0
- 9x squared plus 4xy plus 6y squared plus 40x plus 20y minus 50 equally 0
- 9x to the power of two plus 4xy plus 6y to the power of two plus 4 zero x plus 20y minus fifty equally 0
- 9x2+4xy+6y2+40x+20y-50=0
- 9x²+4xy+6y²+40x+20y-50=0
- 9x to the power of 2+4xy+6y to the power of 2+40x+20y-50=0
- 9x^2+4xy+6y^2+40x+20y-50=O
-
Similar expressions
- 9x^2+4xy+6y^2+40x+20y+50=0
- 9x^2+4xy+6y^2+40x-20y-50=0
- 9x^2-4xy+6y^2+40x+20y-50=0
- 9x^2+4xy+6y^2-40x+20y-50=0
- 9x^2+4xy-6y^2+40x+20y-50=0
9x^2+4xy+6y^2+40x+20y-50=0 canonical form
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The solution
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2 2 -50 + 6*y + 9*x + 20*y + 40*x + 4*x*y = 0
$$9 x^{2} + 4 x y + 40 x + 6 y^{2} + 20 y - 50 = 0$$
9*x^2 + 4*x*y + 40*x + 6*y^2 + 20*y - 50 = 0
Detail solution
Given line equation of 2-order:
$$9 x^{2} + 4 x y + 40 x + 6 y^{2} + 20 y - 50 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 2$$
$$a_{13} = 20$$
$$a_{22} = 6$$
$$a_{23} = 10$$
$$a_{33} = -50$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}9 & 2\\2 & 6\end{matrix}\right|$$
$$\Delta = 50$$
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
$$9 x_{0} + 2 y_{0} + 20 = 0$$
$$2 x_{0} + 6 y_{0} + 10 = 0$$
then
$$x_{0} = -2$$
$$y_{0} = -1$$
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
$$a'_{33} = 20 x_{0} + 10 y_{0} - 50$$
$$a'_{33} = -100$$
then equation turns into
$$9 x'^{2} + 4 x' y' + 6 y'^{2} - 100 = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{3}{4}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{3}{4} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{4}{5}$$
$$\cos{\left(2 \phi \right)} = \frac{3}{5}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{2 \sqrt{5}}{5}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{5}}{5}$$
substitute coefficients
$$x' = \frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}$$
$$y' = \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}$$
then the equation turns from
$$9 x'^{2} + 4 x' y' + 6 y'^{2} - 100 = 0$$
to
$$6 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right)^{2} + 4 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right) + 9 \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right)^{2} - 100 = 0$$
simplify
$$10 \tilde x^{2} + 5 \tilde y^{2} - 100 = 0$$
Given equation is ellipse
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{2 \sqrt{5}}{5}, \ \frac{\sqrt{5}}{5}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{5}}{5}, \ \frac{2 \sqrt{5}}{5}\right)$$
$$9 x^{2} + 4 x y + 40 x + 6 y^{2} + 20 y - 50 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 2$$
$$a_{13} = 20$$
$$a_{22} = 6$$
$$a_{23} = 10$$
$$a_{33} = -50$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}9 & 2\\2 & 6\end{matrix}\right|$$
$$\Delta = 50$$
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
$$9 x_{0} + 2 y_{0} + 20 = 0$$
$$2 x_{0} + 6 y_{0} + 10 = 0$$
then
$$x_{0} = -2$$
$$y_{0} = -1$$
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
$$a'_{33} = 20 x_{0} + 10 y_{0} - 50$$
$$a'_{33} = -100$$
then equation turns into
$$9 x'^{2} + 4 x' y' + 6 y'^{2} - 100 = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{3}{4}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{3}{4} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{4}{5}$$
$$\cos{\left(2 \phi \right)} = \frac{3}{5}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{2 \sqrt{5}}{5}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{5}}{5}$$
substitute coefficients
$$x' = \frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}$$
$$y' = \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}$$
then the equation turns from
$$9 x'^{2} + 4 x' y' + 6 y'^{2} - 100 = 0$$
to
$$6 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right)^{2} + 4 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right) + 9 \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right)^{2} - 100 = 0$$
simplify
$$10 \tilde x^{2} + 5 \tilde y^{2} - 100 = 0$$
Given equation is ellipse
2 2
\tilde x \tilde y
----------- + ---------- = 1
2 2
// ____\\ // ___\\
||\/ 10 || ||\/ 5 ||
||------|| ||-----||
|\ 10 /| |\ 5 /|
|--------| |-------|
\ 1/10 / \ 1/10 / - reduced to canonical form
The center of canonical coordinate system at point O
(-2, -1)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{2 \sqrt{5}}{5}, \ \frac{\sqrt{5}}{5}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{5}}{5}, \ \frac{2 \sqrt{5}}{5}\right)$$
Invariants method
Given line equation of 2-order:
$$9 x^{2} + 4 x y + 40 x + 6 y^{2} + 20 y - 50 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 2$$
$$a_{13} = 20$$
$$a_{22} = 6$$
$$a_{23} = 10$$
$$a_{33} = -50$$
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
$$I_{1} = 15$$
$$I_{3} = \left|\begin{matrix}9 & 2 & 20\\2 & 6 & 10\\20 & 10 & -50\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & 2\\2 & 6 - \lambda\end{matrix}\right|$$
$$I_{1} = 15$$
$$I_{2} = 50$$
$$I_{3} = -5000$$
$$I{\left(\lambda \right)} = \lambda^{2} - 15 \lambda + 50$$
$$K_{2} = -1250$$
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
$$\lambda^{2} - 15 \lambda + 50 = 0$$
$$\lambda_{1} = 10$$
$$\lambda_{2} = 5$$
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
$$10 \tilde x^{2} + 5 \tilde y^{2} - 100 = 0$$
- reduced to canonical form
$$9 x^{2} + 4 x y + 40 x + 6 y^{2} + 20 y - 50 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 2$$
$$a_{13} = 20$$
$$a_{22} = 6$$
$$a_{23} = 10$$
$$a_{33} = -50$$
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
$$I_{1} = 15$$
|9 2|
I2 = | |
|2 6|$$I_{3} = \left|\begin{matrix}9 & 2 & 20\\2 & 6 & 10\\20 & 10 & -50\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & 2\\2 & 6 - \lambda\end{matrix}\right|$$
|9 20 | |6 10 |
K2 = | | + | |
|20 -50| |10 -50|$$I_{1} = 15$$
$$I_{2} = 50$$
$$I_{3} = -5000$$
$$I{\left(\lambda \right)} = \lambda^{2} - 15 \lambda + 50$$
$$K_{2} = -1250$$
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
$$\lambda^{2} - 15 \lambda + 50 = 0$$
$$\lambda_{1} = 10$$
$$\lambda_{2} = 5$$
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
$$10 \tilde x^{2} + 5 \tilde y^{2} - 100 = 0$$
2 2
\tilde x \tilde y
----------- + ---------- = 1
2 2
// ____\\ // ___\\
||\/ 10 || ||\/ 5 ||
||------|| ||-----||
|\ 10 /| |\ 5 /|
|--------| |-------|
\ 1/10 / \ 1/10 / - reduced to canonical form