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+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^2=y
-
x^3(y-x)=x^3+1
-
x^2/4+y^2/2=16
-
Identical expressions
- -31x^ two - two 4xy+14y^2= zero
- minus 31x squared minus 24xy plus 14y squared equally 0
- minus 31x to the power of two minus two 4xy plus 14y squared equally zero
- -31x2-24xy+14y2=0
- -31x²-24xy+14y²=0
- -31x to the power of 2-24xy+14y to the power of 2=0
- -31x^2-24xy+14y^2=O
-
Similar expressions
- 31x^2-24xy+14y^2=0
- -31x^2+24xy+14y^2=0
- -31x^2-24xy-14y^2=0
-31x^2-24xy+14y^2=0 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
2 2 - 31*x + 14*y - 24*x*y = 0
$$- 31 x^{2} - 24 x y + 14 y^{2} = 0$$
-31*x^2 - 24*x*y + 14*y^2 = 0
Detail solution
Given line equation of 2-order:
$$- 31 x^{2} - 24 x y + 14 y^{2} = 0$$
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
$$a_{11} = -31$$
$$a_{12} = -12$$
$$a_{13} = 0$$
$$a_{22} = 14$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-31 & -12\\-12 & 14\end{matrix}\right|$$
$$\Delta = -578$$
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
$$- 31 x_{0} - 12 y_{0} = 0$$
$$- 12 x_{0} + 14 y_{0} = 0$$
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
$$- 31 x'^{2} - 24 x' y' + 14 y'^{2} = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{15}{8}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{15}{8} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{8}{17}$$
$$\cos{\left(2 \phi \right)} = \frac{15}{17}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{4 \sqrt{17}}{17}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{17}}{17}$$
substitute coefficients
$$x' = \frac{4 \sqrt{17} \tilde x}{17} - \frac{\sqrt{17} \tilde y}{17}$$
$$y' = \frac{\sqrt{17} \tilde x}{17} + \frac{4 \sqrt{17} \tilde y}{17}$$
then the equation turns from
$$- 31 x'^{2} - 24 x' y' + 14 y'^{2} = 0$$
to
$$14 \left(\frac{\sqrt{17} \tilde x}{17} + \frac{4 \sqrt{17} \tilde y}{17}\right)^{2} - 24 \left(\frac{\sqrt{17} \tilde x}{17} + \frac{4 \sqrt{17} \tilde y}{17}\right) \left(\frac{4 \sqrt{17} \tilde x}{17} - \frac{\sqrt{17} \tilde y}{17}\right) - 31 \left(\frac{4 \sqrt{17} \tilde x}{17} - \frac{\sqrt{17} \tilde y}{17}\right)^{2} = 0$$
simplify
$$- 34 \tilde x^{2} + 17 \tilde y^{2} = 0$$
$$34 \tilde x^{2} - 17 \tilde y^{2} = 0$$
Given equation is degenerate hyperbole
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{34}}{34}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{\sqrt{17}}{17}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{4 \sqrt{17}}{17}, \ \frac{\sqrt{17}}{17}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{17}}{17}, \ \frac{4 \sqrt{17}}{17}\right)$$
$$- 31 x^{2} - 24 x y + 14 y^{2} = 0$$
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
$$a_{11} = -31$$
$$a_{12} = -12$$
$$a_{13} = 0$$
$$a_{22} = 14$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-31 & -12\\-12 & 14\end{matrix}\right|$$
$$\Delta = -578$$
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
$$- 31 x_{0} - 12 y_{0} = 0$$
$$- 12 x_{0} + 14 y_{0} = 0$$
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
$$- 31 x'^{2} - 24 x' y' + 14 y'^{2} = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{15}{8}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{15}{8} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{8}{17}$$
$$\cos{\left(2 \phi \right)} = \frac{15}{17}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{4 \sqrt{17}}{17}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{17}}{17}$$
substitute coefficients
$$x' = \frac{4 \sqrt{17} \tilde x}{17} - \frac{\sqrt{17} \tilde y}{17}$$
$$y' = \frac{\sqrt{17} \tilde x}{17} + \frac{4 \sqrt{17} \tilde y}{17}$$
then the equation turns from
$$- 31 x'^{2} - 24 x' y' + 14 y'^{2} = 0$$
to
$$14 \left(\frac{\sqrt{17} \tilde x}{17} + \frac{4 \sqrt{17} \tilde y}{17}\right)^{2} - 24 \left(\frac{\sqrt{17} \tilde x}{17} + \frac{4 \sqrt{17} \tilde y}{17}\right) \left(\frac{4 \sqrt{17} \tilde x}{17} - \frac{\sqrt{17} \tilde y}{17}\right) - 31 \left(\frac{4 \sqrt{17} \tilde x}{17} - \frac{\sqrt{17} \tilde y}{17}\right)^{2} = 0$$
simplify
$$- 34 \tilde x^{2} + 17 \tilde y^{2} = 0$$
$$34 \tilde x^{2} - 17 \tilde y^{2} = 0$$
Given equation is degenerate hyperbole
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{34}}{34}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{\sqrt{17}}{17}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
(0, 0)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{4 \sqrt{17}}{17}, \ \frac{\sqrt{17}}{17}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{17}}{17}, \ \frac{4 \sqrt{17}}{17}\right)$$
Invariants method
Given line equation of 2-order:
$$- 31 x^{2} - 24 x y + 14 y^{2} = 0$$
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
$$a_{11} = -31$$
$$a_{12} = -12$$
$$a_{13} = 0$$
$$a_{22} = 14$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = -17$$
$$I_{3} = \left|\begin{matrix}-31 & -12 & 0\\-12 & 14 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 31 & -12\\-12 & 14 - \lambda\end{matrix}\right|$$
$$I_{1} = -17$$
$$I_{2} = -578$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} + 17 \lambda - 578$$
$$K_{2} = 0$$
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
$$\lambda^{2} + 17 \lambda - 578 = 0$$
$$\lambda_{1} = 17$$
$$\lambda_{2} = -34$$
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
$$17 \tilde x^{2} - 34 \tilde y^{2} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{17}}{17}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{\sqrt{34}}{34}\right)^{2}} = 0$$
- reduced to canonical form
$$- 31 x^{2} - 24 x y + 14 y^{2} = 0$$
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
$$a_{11} = -31$$
$$a_{12} = -12$$
$$a_{13} = 0$$
$$a_{22} = 14$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = -17$$
|-31 -12|
I2 = | |
|-12 14 |$$I_{3} = \left|\begin{matrix}-31 & -12 & 0\\-12 & 14 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 31 & -12\\-12 & 14 - \lambda\end{matrix}\right|$$
|-31 0| |14 0|
K2 = | | + | |
| 0 0| |0 0|$$I_{1} = -17$$
$$I_{2} = -578$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} + 17 \lambda - 578$$
$$K_{2} = 0$$
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
$$\lambda^{2} + 17 \lambda - 578 = 0$$
$$\lambda_{1} = 17$$
$$\lambda_{2} = -34$$
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
$$17 \tilde x^{2} - 34 \tilde y^{2} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{17}}{17}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{\sqrt{34}}{34}\right)^{2}} = 0$$
- reduced to canonical form