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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
-
How to use it?
- Canonical form:
- 4*x^2+4*y^2-4x+8y+59=0
- 4x^2-12xy+13y^2-36x+62y+69=0
- 3(x+2)^2+7(y-1)=36
- 0x^2+5y^2-3z^2+14yz−2y+10z-3=0.
- Plot:
-
x^2+(y-|x|^(1/2))^2=1
- x^2/36+y^2/36=1
- x^4-y^4=x*y
-
((x+5/2)^2-(y-11^(1/2)/2)^2+2.5)*((x+5/2)^2-(y+11^(1/2)/2)^2+2.5)=444-36*x^2
-
Identical expressions
- 4x^ two -1 two xy+13y^2-36x+62y+ sixty-nine = zero
- 4x squared minus 12xy plus 13y squared minus 36x plus 62y plus 69 equally 0
- 4x to the power of two minus 1 two xy plus 13y squared minus 36x plus 62y plus sixty minus nine equally zero
- 4x2-12xy+13y2-36x+62y+69=0
- 4x²-12xy+13y²-36x+62y+69=0
- 4x to the power of 2-12xy+13y to the power of 2-36x+62y+69=0
- 4x^2-12xy+13y^2-36x+62y+69=O
-
Similar expressions
- 4x^2-12xy+13y^2-36x-62y+69=0
- 4x^2-12xy+13y^2+36x+62y+69=0
- 4x^2-12xy+13y^2-36x+62y-69=0
- 4x^2-12xy-13y^2-36x+62y+69=0
- 4x^2+12xy+13y^2-36x+62y+69=0
4x^2-12xy+13y^2-36x+62y+69=0 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
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2 2 69 - 36*x + 4*x + 13*y + 62*y - 12*x*y = 0
$$4 x^{2} - 12 x y - 36 x + 13 y^{2} + 62 y + 69 = 0$$
4*x^2 - 12*x*y - 36*x + 13*y^2 + 62*y + 69 = 0
Detail solution
Given line equation of 2-order:
$$4 x^{2} - 12 x y - 36 x + 13 y^{2} + 62 y + 69 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = -6$$
$$a_{13} = -18$$
$$a_{22} = 13$$
$$a_{23} = 31$$
$$a_{33} = 69$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & -6\\-6 & 13\end{matrix}\right|$$
$$\Delta = 16$$
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
$$4 x_{0} - 6 y_{0} - 18 = 0$$
$$- 6 x_{0} + 13 y_{0} + 31 = 0$$
then
$$x_{0} = 3$$
$$y_{0} = -1$$
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
$$a'_{33} = - 18 x_{0} + 31 y_{0} + 69$$
$$a'_{33} = -16$$
then equation turns into
$$4 x'^{2} - 12 x' y' + 13 y'^{2} - 16 = 0$$
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
$$4 x'^{2} - 12 x' y' + 13 y'^{2} - 16 = 0$$
to
$$13 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right)^{2} - 12 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right) + 4 \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right)^{2} - 16 = 0$$
simplify
$$\tilde x^{2} + 16 \tilde y^{2} - 16 = 0$$
Given equation is ellipse
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{2 \sqrt{5}}{5}, \ \frac{\sqrt{5}}{5}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{5}}{5}, \ \frac{2 \sqrt{5}}{5}\right)$$
$$4 x^{2} - 12 x y - 36 x + 13 y^{2} + 62 y + 69 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = -6$$
$$a_{13} = -18$$
$$a_{22} = 13$$
$$a_{23} = 31$$
$$a_{33} = 69$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & -6\\-6 & 13\end{matrix}\right|$$
$$\Delta = 16$$
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
$$4 x_{0} - 6 y_{0} - 18 = 0$$
$$- 6 x_{0} + 13 y_{0} + 31 = 0$$
then
$$x_{0} = 3$$
$$y_{0} = -1$$
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
$$a'_{33} = - 18 x_{0} + 31 y_{0} + 69$$
$$a'_{33} = -16$$
then equation turns into
$$4 x'^{2} - 12 x' y' + 13 y'^{2} - 16 = 0$$
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
$$4 x'^{2} - 12 x' y' + 13 y'^{2} - 16 = 0$$
to
$$13 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right)^{2} - 12 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right) + 4 \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right)^{2} - 16 = 0$$
simplify
$$\tilde x^{2} + 16 \tilde y^{2} - 16 = 0$$
Given equation is ellipse
2 2
\tilde x \tilde y
--------- + --------- = 1
2 2
/ 1\ / 1 \
\4 / |-----|
\4*1/4/ - reduced to canonical form
The center of canonical coordinate system at point O
(3, -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:
$$4 x^{2} - 12 x y - 36 x + 13 y^{2} + 62 y + 69 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = -6$$
$$a_{13} = -18$$
$$a_{22} = 13$$
$$a_{23} = 31$$
$$a_{33} = 69$$
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
$$I_{1} = 17$$
$$I_{3} = \left|\begin{matrix}4 & -6 & -18\\-6 & 13 & 31\\-18 & 31 & 69\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & -6\\-6 & 13 - \lambda\end{matrix}\right|$$
$$I_{1} = 17$$
$$I_{2} = 16$$
$$I_{3} = -256$$
$$I{\left(\lambda \right)} = \lambda^{2} - 17 \lambda + 16$$
$$K_{2} = -112$$
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
$$\lambda^{2} - 17 \lambda + 16 = 0$$
$$\lambda_{1} = 16$$
$$\lambda_{2} = 1$$
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
$$16 \tilde x^{2} + \tilde y^{2} - 16 = 0$$
- reduced to canonical form
$$4 x^{2} - 12 x y - 36 x + 13 y^{2} + 62 y + 69 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = -6$$
$$a_{13} = -18$$
$$a_{22} = 13$$
$$a_{23} = 31$$
$$a_{33} = 69$$
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
$$I_{1} = 17$$
|4 -6|
I2 = | |
|-6 13|$$I_{3} = \left|\begin{matrix}4 & -6 & -18\\-6 & 13 & 31\\-18 & 31 & 69\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & -6\\-6 & 13 - \lambda\end{matrix}\right|$$
| 4 -18| |13 31|
K2 = | | + | |
|-18 69 | |31 69|$$I_{1} = 17$$
$$I_{2} = 16$$
$$I_{3} = -256$$
$$I{\left(\lambda \right)} = \lambda^{2} - 17 \lambda + 16$$
$$K_{2} = -112$$
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
$$\lambda^{2} - 17 \lambda + 16 = 0$$
$$\lambda_{1} = 16$$
$$\lambda_{2} = 1$$
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
$$16 \tilde x^{2} + \tilde y^{2} - 16 = 0$$
2 2
\tilde x \tilde y
--------- + --------- = 1
2 2
/ 1 \ / 1\
|-----| \4 /
\4*1/4/ - reduced to canonical form