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:
- =4x^21+(−4b+8)x^22+(−4b−3)x^23+2x1x2−2x1x3+2(4b−1)x2x3
- 4x^2-4xy+y^2+4\sqrt5x+2\sqrt5y+5=0
- (6-6*x)*(x^2-2*x+13)-(9+6*x-3*x^2)*(2*x-2)
- 5x^2-8xy+5y^2+30sqrt(2)x-24sqrt(2)y+81=0
- Plot:
- x^2+(y-((x^2)^(1/(3))))^2=1
-
(x+1)^2+(y+1)^2=4
-
x^2+(y+(x)^(2/3))^2=1
-
tg(x)^arctg(x)=e^((y(x))/x)
-
Identical expressions
- 3x^ two + four xy+4= zero
- 3x squared plus 4xy plus 4 equally 0
- 3x to the power of two plus four xy plus 4 equally zero
- 3x2+4xy+4=0
- 3x²+4xy+4=0
- 3x to the power of 2+4xy+4=0
- 3x^2+4xy+4=O
-
Similar expressions
- 3x^2-4xy+4=0
- 3x^2+4xy-4=0
3x^2+4xy+4=0 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given line equation of 2-order:
$$3 x^{2} + 4 x y + 4 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 2$$
$$a_{13} = 0$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = 4$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}3 & 2\\2 & 0\end{matrix}\right|$$
$$\Delta = -4$$
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
$$3 x_{0} + 2 y_{0} = 0$$
$$2 x_{0} = 0$$
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
$$a'_{33} = 4$$
$$a'_{33} = 4$$
then equation turns into
$$3 x'^{2} + 4 x' y' + 4 = 0$$
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
$$3 x'^{2} + 4 x' y' + 4 = 0$$
to
$$4 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right) + 3 \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right)^{2} + 4 = 0$$
simplify
$$4 \tilde x^{2} - \tilde y^{2} + 4 = 0$$
Given equation is hyperbole
$$\tilde x^{2} - \frac{\tilde y^{2}}{4} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( \frac{2 \sqrt{5}}{5}, \ \frac{\sqrt{5}}{5}\right)$$
$$\vec e_{2} = \left( - \frac{\sqrt{5}}{5}, \ \frac{2 \sqrt{5}}{5}\right)$$
$$3 x^{2} + 4 x y + 4 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 2$$
$$a_{13} = 0$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = 4$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}3 & 2\\2 & 0\end{matrix}\right|$$
$$\Delta = -4$$
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
$$3 x_{0} + 2 y_{0} = 0$$
$$2 x_{0} = 0$$
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
$$a'_{33} = 4$$
$$a'_{33} = 4$$
then equation turns into
$$3 x'^{2} + 4 x' y' + 4 = 0$$
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
$$3 x'^{2} + 4 x' y' + 4 = 0$$
to
$$4 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right) + 3 \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right)^{2} + 4 = 0$$
simplify
$$4 \tilde x^{2} - \tilde y^{2} + 4 = 0$$
Given equation is hyperbole
$$\tilde x^{2} - \frac{\tilde y^{2}}{4} = -1$$
- 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:
$$3 x^{2} + 4 x y + 4 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 2$$
$$a_{13} = 0$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = 4$$
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
$$I_{1} = 3$$
$$I_{3} = \left|\begin{matrix}3 & 2 & 0\\2 & 0 & 0\\0 & 0 & 4\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & 2\\2 & - \lambda\end{matrix}\right|$$
$$I_{1} = 3$$
$$I_{2} = -4$$
$$I_{3} = -16$$
$$I{\left(\lambda \right)} = \lambda^{2} - 3 \lambda - 4$$
$$K_{2} = 12$$
Because
$$I_{2} < 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 3 \lambda - 4 = 0$$
Solve this equation
$$\lambda_{1} = 4$$
$$\lambda_{2} = -1$$
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
$$4 \tilde x^{2} - \tilde y^{2} + 4 = 0$$
$$\tilde x^{2} - \frac{\tilde y^{2}}{4} = -1$$
- reduced to canonical form
$$3 x^{2} + 4 x y + 4 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 2$$
$$a_{13} = 0$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = 4$$
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
$$I_{1} = 3$$
|3 2|
I2 = | |
|2 0|$$I_{3} = \left|\begin{matrix}3 & 2 & 0\\2 & 0 & 0\\0 & 0 & 4\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & 2\\2 & - \lambda\end{matrix}\right|$$
|3 0| |0 0|
K2 = | | + | |
|0 4| |0 4|$$I_{1} = 3$$
$$I_{2} = -4$$
$$I_{3} = -16$$
$$I{\left(\lambda \right)} = \lambda^{2} - 3 \lambda - 4$$
$$K_{2} = 12$$
Because
$$I_{2} < 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 3 \lambda - 4 = 0$$
Solve this equation
$$\lambda_{1} = 4$$
$$\lambda_{2} = -1$$
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
$$4 \tilde x^{2} - \tilde y^{2} + 4 = 0$$
$$\tilde x^{2} - \frac{\tilde y^{2}}{4} = -1$$
- reduced to canonical form
The graph