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:
-
8x^2+5y^2+15x+6y-7=0
- 0x^2+xy+0y^2+2x+y+2=0
- 4x^2+y+20x-11=0
- 𝑥1^2+𝑥2^2+5𝑥32−6𝑥1𝑥2+2𝑥1𝑥3−2𝑥2𝑥3
- Plot:
- atan(y)=-atan(x)+3
- z=sqrt(x*y)
- x^2+y^2+10z^2=100
-
|y-1|-|x-2|=x+1
-
Identical expressions
- 8x^ two +5y^ two +15x+6y- seven = zero
- 8x squared plus 5y squared plus 15x plus 6y minus 7 equally 0
- 8x to the power of two plus 5y to the power of two plus 15x plus 6y minus seven equally zero
- 8x2+5y2+15x+6y-7=0
- 8x²+5y²+15x+6y-7=0
- 8x to the power of 2+5y to the power of 2+15x+6y-7=0
- 8x^2+5y^2+15x+6y-7=O
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Similar expressions
- 8x^2+5y^2-15x+6y-7=0
- 8x^2+5y^2+15x+6y+7=0
- 8x^2+5y^2+15x-6y-7=0
- 8x^2-5y^2+15x+6y-7=0
8x^2+5y^2+15x+6y-7=0 canonical form
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The solution
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2 2 -7 + 5*y + 6*y + 8*x + 15*x = 0
$$8 x^{2} + 15 x + 5 y^{2} + 6 y - 7 = 0$$
8*x^2 + 15*x + 5*y^2 + 6*y - 7 = 0
Detail solution
Given line equation of 2-order:
$$8 x^{2} + 15 x + 5 y^{2} + 6 y - 7 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = 0$$
$$a_{13} = \frac{15}{2}$$
$$a_{22} = 5$$
$$a_{23} = 3$$
$$a_{33} = -7$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}8 & 0\\0 & 5\end{matrix}\right|$$
$$\Delta = 40$$
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
$$8 x_{0} + \frac{15}{2} = 0$$
$$5 y_{0} + 3 = 0$$
then
$$x_{0} = - \frac{15}{16}$$
$$y_{0} = - \frac{3}{5}$$
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
$$a'_{33} = \frac{15 x_{0}}{2} + 3 y_{0} - 7$$
$$a'_{33} = - \frac{2533}{160}$$
then equation turns into
$$8 x'^{2} + 5 y'^{2} - \frac{2533}{160} = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{12665}}{80}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{5066}}{40}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( 1, \ 0\right)$$
$$\vec e_{2} = \left( 0, \ 1\right)$$
$$8 x^{2} + 15 x + 5 y^{2} + 6 y - 7 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = 0$$
$$a_{13} = \frac{15}{2}$$
$$a_{22} = 5$$
$$a_{23} = 3$$
$$a_{33} = -7$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}8 & 0\\0 & 5\end{matrix}\right|$$
$$\Delta = 40$$
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
$$8 x_{0} + \frac{15}{2} = 0$$
$$5 y_{0} + 3 = 0$$
then
$$x_{0} = - \frac{15}{16}$$
$$y_{0} = - \frac{3}{5}$$
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
$$a'_{33} = \frac{15 x_{0}}{2} + 3 y_{0} - 7$$
$$a'_{33} = - \frac{2533}{160}$$
then equation turns into
$$8 x'^{2} + 5 y'^{2} - \frac{2533}{160} = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{12665}}{80}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{5066}}{40}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
-15 (----, -3/5) 16
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:
$$8 x^{2} + 15 x + 5 y^{2} + 6 y - 7 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = 0$$
$$a_{13} = \frac{15}{2}$$
$$a_{22} = 5$$
$$a_{23} = 3$$
$$a_{33} = -7$$
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
$$I_{1} = 13$$
$$I_{3} = \left|\begin{matrix}8 & 0 & \frac{15}{2}\\0 & 5 & 3\\\frac{15}{2} & 3 & -7\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}8 - \lambda & 0\\0 & 5 - \lambda\end{matrix}\right|$$
$$I_{1} = 13$$
$$I_{2} = 40$$
$$I_{3} = - \frac{2533}{4}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 13 \lambda + 40$$
$$K_{2} = - \frac{625}{4}$$
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
$$\lambda^{2} - 13 \lambda + 40 = 0$$
Solve this equation
$$\lambda_{1} = 8$$
$$\lambda_{2} = 5$$
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
$$8 \tilde x^{2} + 5 \tilde y^{2} - \frac{2533}{160} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{12665}}{80}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{5066}}{40}\right)^{2}} = 1$$
- reduced to canonical form
$$8 x^{2} + 15 x + 5 y^{2} + 6 y - 7 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = 0$$
$$a_{13} = \frac{15}{2}$$
$$a_{22} = 5$$
$$a_{23} = 3$$
$$a_{33} = -7$$
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
$$I_{1} = 13$$
|8 0|
I2 = | |
|0 5|$$I_{3} = \left|\begin{matrix}8 & 0 & \frac{15}{2}\\0 & 5 & 3\\\frac{15}{2} & 3 & -7\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}8 - \lambda & 0\\0 & 5 - \lambda\end{matrix}\right|$$
| 8 15/2| |5 3 |
K2 = | | + | |
|15/2 -7 | |3 -7|$$I_{1} = 13$$
$$I_{2} = 40$$
$$I_{3} = - \frac{2533}{4}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 13 \lambda + 40$$
$$K_{2} = - \frac{625}{4}$$
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
$$\lambda^{2} - 13 \lambda + 40 = 0$$
Solve this equation
$$\lambda_{1} = 8$$
$$\lambda_{2} = 5$$
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
$$8 \tilde x^{2} + 5 \tilde y^{2} - \frac{2533}{160} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{12665}}{80}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{5066}}{40}\right)^{2}} = 1$$
- reduced to canonical form
The graph