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^2+4*y^2+8*z^2-16
- 3x^2+6x-15y+15=0
- 16x^2+24xy+9y^2+4x+3y=0
- 13x^2+18xy+37y^2
- Plot:
-
x^2/4+y^2/9=1
-
x^2/36-y^2=1
- x^2-2=y
- x^2+y^2=9
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Identical expressions
- 13x^ two +18xy+37y^ two
- 13x squared plus 18xy plus 37y squared
- 13x to the power of two plus 18xy plus 37y to the power of two
- 13x2+18xy+37y2
- 13x²+18xy+37y²
- 13x to the power of 2+18xy+37y to the power of 2
-
Similar expressions
- 13x^2+18xy-37y^2
- 13x^2-18xy+37y^2
13x^2+18xy+37y^2 canonical form
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The solution
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2 2 13*x + 37*y + 18*x*y = 0
$$13 x^{2} + 18 x y + 37 y^{2} = 0$$
13*x^2 + 18*x*y + 37*y^2 = 0
Detail solution
Given line equation of 2-order:
$$13 x^{2} + 18 x y + 37 y^{2} = 0$$
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
$$a_{11} = 13$$
$$a_{12} = 9$$
$$a_{13} = 0$$
$$a_{22} = 37$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}13 & 9\\9 & 37\end{matrix}\right|$$
$$\Delta = 400$$
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
$$13 x_{0} + 9 y_{0} = 0$$
$$9 x_{0} + 37 y_{0} = 0$$
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
$$13 x'^{2} + 18 x' y' + 37 y'^{2} = 0$$
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
$$\cot{\left(2 \phi \right)} = - \frac{4}{3}$$
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
$$13 x'^{2} + 18 x' y' + 37 y'^{2} = 0$$
to
$$37 \left(- \frac{\sqrt{10} \tilde x}{10} + \frac{3 \sqrt{10} \tilde y}{10}\right)^{2} + 18 \left(- \frac{\sqrt{10} \tilde x}{10} + \frac{3 \sqrt{10} \tilde y}{10}\right) \left(\frac{3 \sqrt{10} \tilde x}{10} + \frac{\sqrt{10} \tilde y}{10}\right) + 13 \left(\frac{3 \sqrt{10} \tilde x}{10} + \frac{\sqrt{10} \tilde y}{10}\right)^{2} = 0$$
simplify
$$10 \tilde x^{2} + 40 \tilde y^{2} = 0$$
Given equation is degenerate ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{10}}{10}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{10}}{20}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{3 \sqrt{10}}{10}, \ - \frac{\sqrt{10}}{10}\right)$$
$$\vec e_2 = \left( \frac{\sqrt{10}}{10}, \ \frac{3 \sqrt{10}}{10}\right)$$
$$13 x^{2} + 18 x y + 37 y^{2} = 0$$
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
$$a_{11} = 13$$
$$a_{12} = 9$$
$$a_{13} = 0$$
$$a_{22} = 37$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}13 & 9\\9 & 37\end{matrix}\right|$$
$$\Delta = 400$$
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
$$13 x_{0} + 9 y_{0} = 0$$
$$9 x_{0} + 37 y_{0} = 0$$
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
$$13 x'^{2} + 18 x' y' + 37 y'^{2} = 0$$
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
$$\cot{\left(2 \phi \right)} = - \frac{4}{3}$$
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
$$13 x'^{2} + 18 x' y' + 37 y'^{2} = 0$$
to
$$37 \left(- \frac{\sqrt{10} \tilde x}{10} + \frac{3 \sqrt{10} \tilde y}{10}\right)^{2} + 18 \left(- \frac{\sqrt{10} \tilde x}{10} + \frac{3 \sqrt{10} \tilde y}{10}\right) \left(\frac{3 \sqrt{10} \tilde x}{10} + \frac{\sqrt{10} \tilde y}{10}\right) + 13 \left(\frac{3 \sqrt{10} \tilde x}{10} + \frac{\sqrt{10} \tilde y}{10}\right)^{2} = 0$$
simplify
$$10 \tilde x^{2} + 40 \tilde y^{2} = 0$$
Given equation is degenerate ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{10}}{10}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{10}}{20}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
(0, 0)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{3 \sqrt{10}}{10}, \ - \frac{\sqrt{10}}{10}\right)$$
$$\vec e_2 = \left( \frac{\sqrt{10}}{10}, \ \frac{3 \sqrt{10}}{10}\right)$$
Invariants method
Given line equation of 2-order:
$$13 x^{2} + 18 x y + 37 y^{2} = 0$$
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
$$a_{11} = 13$$
$$a_{12} = 9$$
$$a_{13} = 0$$
$$a_{22} = 37$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = 50$$
$$I_{3} = \left|\begin{matrix}13 & 9 & 0\\9 & 37 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}13 - \lambda & 9\\9 & 37 - \lambda\end{matrix}\right|$$
$$I_{1} = 50$$
$$I_{2} = 400$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 50 \lambda + 400$$
$$K_{2} = 0$$
Because
$$I_{3} = 0 \wedge I_{2} > 0$$
then by line type:
this equation is of type : degenerate ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 50 \lambda + 400 = 0$$
$$\lambda_{1} = 40$$
$$\lambda_{2} = 10$$
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
$$40 \tilde x^{2} + 10 \tilde y^{2} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{10}}{20}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{10}}{10}\right)^{2}} = 0$$
- reduced to canonical form
$$13 x^{2} + 18 x y + 37 y^{2} = 0$$
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
$$a_{11} = 13$$
$$a_{12} = 9$$
$$a_{13} = 0$$
$$a_{22} = 37$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = 50$$
|13 9 |
I2 = | |
|9 37|$$I_{3} = \left|\begin{matrix}13 & 9 & 0\\9 & 37 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}13 - \lambda & 9\\9 & 37 - \lambda\end{matrix}\right|$$
|13 0| |37 0|
K2 = | | + | |
|0 0| |0 0|$$I_{1} = 50$$
$$I_{2} = 400$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 50 \lambda + 400$$
$$K_{2} = 0$$
Because
$$I_{3} = 0 \wedge I_{2} > 0$$
then by line type:
this equation is of type : degenerate ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 50 \lambda + 400 = 0$$
$$\lambda_{1} = 40$$
$$\lambda_{2} = 10$$
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
$$40 \tilde x^{2} + 10 \tilde y^{2} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{10}}{20}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{10}}{10}\right)^{2}} = 0$$
- reduced to canonical form