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
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How to use it?
- Canonical form:
- x^2+y^2+z^2+6xy+2xz+2yz=0
- x=-sqrt(y)
- z=x^2+y^2+3xy-3x-2y
- 4*x^2+4*y^2-8*z^2-10*x*y+4*y*z+4*z*x-16*x-16*y+10*z-2=0
- Plot:
- (x^(2)+y^(2)-1)^(3)-x^(2)*y^(3)=0
- abs((a*x-b)*x+c)*x-d;
- 2ix+yi
-
(x^2+y^2-1)^3-(x^2)*(y^3)=0
- x^2-4y^2
- Sum of series:
-
16
- Derivative of:
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16
- Integral of d{x}:
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16
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Identical expressions
- x^ two -4y^ two = sixteen
- x squared minus 4y squared equally 16
- x to the power of two minus 4y to the power of two equally sixteen
- x2-4y2=16
- x²-4y²=16
- x to the power of 2-4y to the power of 2=16
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Similar expressions
- x^2+4y^2=16
- 9x^2-4y^2-36z^2-18x+72z+9=0
x^2-4y^2=16 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given line equation of 2-order:
$$x^{2} - 4 y^{2} - 16 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -4$$
$$a_{23} = 0$$
$$a_{33} = -16$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & -4\end{matrix}\right|$$
$$\Delta = -4$$
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
$$x_{0} = 0$$
$$- 4 y_{0} = 0$$
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
$$a'_{33} = -16$$
$$a'_{33} = -16$$
then equation turns into
$$x'^{2} - 4 y'^{2} - 16 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{16} - \frac{\tilde y^{2}}{4} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$x^{2} - 4 y^{2} - 16 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -4$$
$$a_{23} = 0$$
$$a_{33} = -16$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & -4\end{matrix}\right|$$
$$\Delta = -4$$
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
$$x_{0} = 0$$
$$- 4 y_{0} = 0$$
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
$$a'_{33} = -16$$
$$a'_{33} = -16$$
then equation turns into
$$x'^{2} - 4 y'^{2} - 16 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{16} - \frac{\tilde y^{2}}{4} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(0, 0)
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
Invariants method
Given line equation of 2-order:
$$x^{2} - 4 y^{2} - 16 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -4$$
$$a_{23} = 0$$
$$a_{33} = -16$$
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
$$I_{1} = -3$$
$$I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & -4 & 0\\0 & 0 & -16\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & - \lambda - 4\end{matrix}\right|$$
$$I_{1} = -3$$
$$I_{2} = -4$$
$$I_{3} = 64$$
$$I{\left(\lambda \right)} = \lambda^{2} + 3 \lambda - 4$$
$$K_{2} = 48$$
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
$$\lambda^{2} + 3 \lambda - 4 = 0$$
$$\lambda_{1} = 1$$
$$\lambda_{2} = -4$$
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
$$\tilde x^{2} - 4 \tilde y^{2} - 16 = 0$$
$$\frac{\tilde x^{2}}{16} - \frac{\tilde y^{2}}{4} = 1$$
- reduced to canonical form
$$x^{2} - 4 y^{2} - 16 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -4$$
$$a_{23} = 0$$
$$a_{33} = -16$$
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
$$I_{1} = -3$$
|1 0 |
I2 = | |
|0 -4|$$I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & -4 & 0\\0 & 0 & -16\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & - \lambda - 4\end{matrix}\right|$$
|1 0 | |-4 0 |
K2 = | | + | |
|0 -16| |0 -16|$$I_{1} = -3$$
$$I_{2} = -4$$
$$I_{3} = 64$$
$$I{\left(\lambda \right)} = \lambda^{2} + 3 \lambda - 4$$
$$K_{2} = 48$$
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
$$\lambda^{2} + 3 \lambda - 4 = 0$$
$$\lambda_{1} = 1$$
$$\lambda_{2} = -4$$
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
$$\tilde x^{2} - 4 \tilde y^{2} - 16 = 0$$
$$\frac{\tilde x^{2}}{16} - \frac{\tilde y^{2}}{4} = 1$$
- reduced to canonical form