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:
- 2*x-3/4*x-8=0
- 4x^2+25y^2-16x+50y-67=0
- x^2+y^2-xy+x+y=0
- x^2-y^2-6x-4y+1=0
- Plot:
- (y+4)^2+(x+2)^2=25
-
(x-1)^2+(y-1)^2=4
- ((|y|)+2)^2+((|x|)-4)^2=8
- y=-x2-4*x-4
-
Identical expressions
- 4x^ two + two 5y^2-16x+5 zero y- sixty-seven =0
- 4x squared plus 25y squared minus 16x plus 50y minus 67 equally 0
- 4x to the power of two plus two 5y squared minus 16x plus 5 zero y minus sixty minus seven equally 0
- 4x2+25y2-16x+50y-67=0
- 4x²+25y²-16x+50y-67=0
- 4x to the power of 2+25y to the power of 2-16x+50y-67=0
- 4x^2+25y^2-16x+50y-67=O
-
Similar expressions
- 4x^2+25y^2-16x-50y-67=0
- 4x^2+25y^2+16x+50y-67=0
- 4x^2+25y^2-16x+50y+67=0
- 4x^2-25y^2-16x+50y-67=0
4x^2+25y^2-16x+50y-67=0 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
2 2 -67 - 16*x + 4*x + 25*y + 50*y = 0
$$4 x^{2} - 16 x + 25 y^{2} + 50 y - 67 = 0$$
4*x^2 - 16*x + 25*y^2 + 50*y - 67 = 0
Detail solution
Given line equation of 2-order:
$$4 x^{2} - 16 x + 25 y^{2} + 50 y - 67 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = -8$$
$$a_{22} = 25$$
$$a_{23} = 25$$
$$a_{33} = -67$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & 0\\0 & 25\end{matrix}\right|$$
$$\Delta = 100$$
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
$$4 x_{0} - 8 = 0$$
$$25 y_{0} + 25 = 0$$
then
$$x_{0} = 2$$
$$y_{0} = -1$$
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
$$a'_{33} = - 8 x_{0} + 25 y_{0} - 67$$
$$a'_{33} = -108$$
then equation turns into
$$4 x'^{2} + 25 y'^{2} - 108 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{1}{2 \frac{\sqrt{3}}{18}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{5 \frac{\sqrt{3}}{18}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$4 x^{2} - 16 x + 25 y^{2} + 50 y - 67 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = -8$$
$$a_{22} = 25$$
$$a_{23} = 25$$
$$a_{33} = -67$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & 0\\0 & 25\end{matrix}\right|$$
$$\Delta = 100$$
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
$$4 x_{0} - 8 = 0$$
$$25 y_{0} + 25 = 0$$
then
$$x_{0} = 2$$
$$y_{0} = -1$$
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
$$a'_{33} = - 8 x_{0} + 25 y_{0} - 67$$
$$a'_{33} = -108$$
then equation turns into
$$4 x'^{2} + 25 y'^{2} - 108 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{1}{2 \frac{\sqrt{3}}{18}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{5 \frac{\sqrt{3}}{18}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(2, -1)
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:
$$4 x^{2} - 16 x + 25 y^{2} + 50 y - 67 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = -8$$
$$a_{22} = 25$$
$$a_{23} = 25$$
$$a_{33} = -67$$
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
$$I_{1} = 29$$
$$I_{3} = \left|\begin{matrix}4 & 0 & -8\\0 & 25 & 25\\-8 & 25 & -67\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & 0\\0 & 25 - \lambda\end{matrix}\right|$$
$$I_{1} = 29$$
$$I_{2} = 100$$
$$I_{3} = -10800$$
$$I{\left(\lambda \right)} = \lambda^{2} - 29 \lambda + 100$$
$$K_{2} = -2632$$
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
$$\lambda^{2} - 29 \lambda + 100 = 0$$
$$\lambda_{1} = 25$$
$$\lambda_{2} = 4$$
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
$$25 \tilde x^{2} + 4 \tilde y^{2} - 108 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{5 \frac{\sqrt{3}}{18}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{2 \frac{\sqrt{3}}{18}}\right)^{2}} = 1$$
- reduced to canonical form
$$4 x^{2} - 16 x + 25 y^{2} + 50 y - 67 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = -8$$
$$a_{22} = 25$$
$$a_{23} = 25$$
$$a_{33} = -67$$
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
$$I_{1} = 29$$
|4 0 |
I2 = | |
|0 25|$$I_{3} = \left|\begin{matrix}4 & 0 & -8\\0 & 25 & 25\\-8 & 25 & -67\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & 0\\0 & 25 - \lambda\end{matrix}\right|$$
|4 -8 | |25 25 |
K2 = | | + | |
|-8 -67| |25 -67|$$I_{1} = 29$$
$$I_{2} = 100$$
$$I_{3} = -10800$$
$$I{\left(\lambda \right)} = \lambda^{2} - 29 \lambda + 100$$
$$K_{2} = -2632$$
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
$$\lambda^{2} - 29 \lambda + 100 = 0$$
$$\lambda_{1} = 25$$
$$\lambda_{2} = 4$$
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
$$25 \tilde x^{2} + 4 \tilde y^{2} - 108 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{5 \frac{\sqrt{3}}{18}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{2 \frac{\sqrt{3}}{18}}\right)^{2}} = 1$$
- reduced to canonical form