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:
- 852
- y=2x-4
- 17x^2+12xy+8y^2+0x+0y-20=0
- x^2-y^2-2x-4=0
- Plot:
-
y=(x^2-1)^1/2
- x+y=1
- x^2=y^2
- x^2/9-y^2/4=1
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Identical expressions
- 17x^ two +1 two xy+8y^2+ zero x+0y- twenty =0
- 17x squared plus 12xy plus 8y squared plus 0x plus 0y minus 20 equally 0
- 17x to the power of two plus 1 two xy plus 8y squared plus zero x plus 0y minus twenty equally 0
- 17x2+12xy+8y2+0x+0y-20=0
- 17x²+12xy+8y²+0x+0y-20=0
- 17x to the power of 2+12xy+8y to the power of 2+0x+0y-20=0
- 17x^2+12xy+8y^2+0x+0y-20=O
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Similar expressions
- 17x^2-12xy+8y^2+0x+0y-20=0
- 17x^2+12xy+8y^2-0x+0y-20=0
- 17x^2+12xy+8y^2+0x+0y+20=0
- 17x^2+12xy-8y^2+0x+0y-20=0
- 17x^2+12xy+8y^2+0x-0y-20=0
17x^2+12xy+8y^2+0x+0y-20=0 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
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2 2 -20 + 8*y + 17*x + 12*x*y = 0
$$17 x^{2} + 12 x y + 8 y^{2} - 20 = 0$$
17*x^2 + 12*x*y + 8*y^2 - 20 = 0
Detail solution
Given line equation of 2-order:
$$17 x^{2} + 12 x y + 8 y^{2} - 20 = 0$$
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
$$a_{11} = 17$$
$$a_{12} = 6$$
$$a_{13} = 0$$
$$a_{22} = 8$$
$$a_{23} = 0$$
$$a_{33} = -20$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}17 & 6\\6 & 8\end{matrix}\right|$$
$$\Delta = 100$$
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
$$17 x_{0} + 6 y_{0} = 0$$
$$6 x_{0} + 8 y_{0} = 0$$
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
$$a'_{33} = -20$$
$$a'_{33} = -20$$
then equation turns into
$$17 x'^{2} + 12 x' y' + 8 y'^{2} - 20 = 0$$
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
$$17 x'^{2} + 12 x' y' + 8 y'^{2} - 20 = 0$$
to
$$8 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right)^{2} + 12 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right) + 17 \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right)^{2} - 20 = 0$$
simplify
$$20 \tilde x^{2} + 5 \tilde y^{2} - 20 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{10} \sqrt{5}}{\frac{1}{10} \sqrt{5}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{5} \sqrt{5}}{\frac{1}{10} \sqrt{5}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{2 \sqrt{5}}{5}, \ \frac{\sqrt{5}}{5}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{5}}{5}, \ \frac{2 \sqrt{5}}{5}\right)$$
$$17 x^{2} + 12 x y + 8 y^{2} - 20 = 0$$
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
$$a_{11} = 17$$
$$a_{12} = 6$$
$$a_{13} = 0$$
$$a_{22} = 8$$
$$a_{23} = 0$$
$$a_{33} = -20$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}17 & 6\\6 & 8\end{matrix}\right|$$
$$\Delta = 100$$
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
$$17 x_{0} + 6 y_{0} = 0$$
$$6 x_{0} + 8 y_{0} = 0$$
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
$$a'_{33} = -20$$
$$a'_{33} = -20$$
then equation turns into
$$17 x'^{2} + 12 x' y' + 8 y'^{2} - 20 = 0$$
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
$$17 x'^{2} + 12 x' y' + 8 y'^{2} - 20 = 0$$
to
$$8 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right)^{2} + 12 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right) + 17 \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right)^{2} - 20 = 0$$
simplify
$$20 \tilde x^{2} + 5 \tilde y^{2} - 20 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{10} \sqrt{5}}{\frac{1}{10} \sqrt{5}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{5} \sqrt{5}}{\frac{1}{10} \sqrt{5}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(0, 0)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{2 \sqrt{5}}{5}, \ \frac{\sqrt{5}}{5}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{5}}{5}, \ \frac{2 \sqrt{5}}{5}\right)$$
Invariants method
Given line equation of 2-order:
$$17 x^{2} + 12 x y + 8 y^{2} - 20 = 0$$
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
$$a_{11} = 17$$
$$a_{12} = 6$$
$$a_{13} = 0$$
$$a_{22} = 8$$
$$a_{23} = 0$$
$$a_{33} = -20$$
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
$$I_{1} = 25$$
$$I_{3} = \left|\begin{matrix}17 & 6 & 0\\6 & 8 & 0\\0 & 0 & -20\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}17 - \lambda & 6\\6 & 8 - \lambda\end{matrix}\right|$$
$$I_{1} = 25$$
$$I_{2} = 100$$
$$I_{3} = -2000$$
$$I{\left(\lambda \right)} = \lambda^{2} - 25 \lambda + 100$$
$$K_{2} = -500$$
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
$$\lambda^{2} - 25 \lambda + 100 = 0$$
$$\lambda_{1} = 20$$
$$\lambda_{2} = 5$$
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
$$20 \tilde x^{2} + 5 \tilde y^{2} - 20 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{10} \sqrt{5}}{\frac{1}{10} \sqrt{5}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{5} \sqrt{5}}{\frac{1}{10} \sqrt{5}}\right)^{2}} = 1$$
- reduced to canonical form
$$17 x^{2} + 12 x y + 8 y^{2} - 20 = 0$$
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
$$a_{11} = 17$$
$$a_{12} = 6$$
$$a_{13} = 0$$
$$a_{22} = 8$$
$$a_{23} = 0$$
$$a_{33} = -20$$
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
$$I_{1} = 25$$
|17 6|
I2 = | |
|6 8|$$I_{3} = \left|\begin{matrix}17 & 6 & 0\\6 & 8 & 0\\0 & 0 & -20\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}17 - \lambda & 6\\6 & 8 - \lambda\end{matrix}\right|$$
|17 0 | |8 0 |
K2 = | | + | |
|0 -20| |0 -20|$$I_{1} = 25$$
$$I_{2} = 100$$
$$I_{3} = -2000$$
$$I{\left(\lambda \right)} = \lambda^{2} - 25 \lambda + 100$$
$$K_{2} = -500$$
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
$$\lambda^{2} - 25 \lambda + 100 = 0$$
$$\lambda_{1} = 20$$
$$\lambda_{2} = 5$$
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
$$20 \tilde x^{2} + 5 \tilde y^{2} - 20 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{10} \sqrt{5}}{\frac{1}{10} \sqrt{5}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{5} \sqrt{5}}{\frac{1}{10} \sqrt{5}}\right)^{2}} = 1$$
- reduced to canonical form